Trigonometry Solutions Manual

INSTRUCTOR'S SOLUTIONS 

MANUAL 

to accompany 

Trigonometry 

Seventh Edition 

Margaret L. Lial 

American River College 

John Homsby 

University ofNew Orleans 

David I. Schneider 

University ofMaryland 

EXCELSIOR EDUCATION CENTER 

7151 SVL BOX 

12217 SPRING VALLEY PARKWAY 

VICTORVILLE, CA 92392 

... 

A 

---- 

Addison 

Wesley 

Boston San Francisc:o New York 

London Toronto Sydney Tokyo Singapore Madrid 

Mexico City Munich Paris Cape Town Hong Kong Montreal

Reproduced by Addison Wesley Longman from camera-ready copy supplied by the authors. 

Copyright © 2001 Addison Wesley Longman. 

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in 

any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written 

permission of the publisher. Printed in the United States of America. 

ISBN 0-321-05761-9 

6 7 8 9 10 VO 03 02 01

CONTENTS 

1 THE TRIGONOMETRIC FUNCTIONS 

1.1 Basic Concepts 1 

1.2 Angles 9 

1.3 Angle Relationships and Similar Triangles 15 

1.4 Definitions of the Trigonometric Functions 21 

1.5 Using the Definitions of the Trigonometric Functions 27 

Chapter 1 Review Exercises 

33 

Chapter 1 Test Exercises 

41 

2 ACUTE ANGLES AND RIGHT TRIANGLES 

2.1 Trigonometric Functions of Acute Angles 43 

2.2 Trigonometric Functions of Non-Acute Angles 49 

2.3 Finding Trigonometric Function Values Using a Calculator 56 

2.4 Solving Right Angles 62 

2.5 Further Applications of Right Triangles 69 


74 


81 

3 RADIAN MEASURE AND THE CIRCULAR FUNCTIONS 

3.1 Radian Measure 84 

3.2 Applications of Radian Measure 88 

3.3 Circular Functions of Real Numbers 95 

3.4 Linear and Angular Velocity 1..03 


107 


112 

4 GRAPHS OF THE CIRCULAR FUNCTIONS 

- " 

4.1 Graphs of the Sine and Cosine Functions 114 

4.2 Translations of the Graphs of the Sine and Cosine Functions 121 

4.3· Graphs of the Other Circular Functions 129 


137 


142

5 TRIGONOMETRIC IDENTITIES 

5.1 Fundamental Identities 144 

5.2 Verifying Trigonometric Identities 153 

5.3 Sum and Difference Identities for Cosine 162 

5.4 Sum and Difference Identities for Sine and Tangent 170 

5.5 Double-Angle Identities 184 

5.6 Half-Angle Identities 195 


205 


218 

6 INVERSE TRIGONOMETRIC FUNCTIONS AND 

TRIGONOMETRIC EQUATIONS 

6.1 Inverse Trigonometric Functions 221 

6.2 Trigonometric Equations I 234 

6.3 Trigonometric Equations n 243 

6.4 Equations Involving Inverse Trigonometric Functions 249 


257 


266 

7 APPLICATIONS OF TRIGONOMETRY AND VECTORS 

7.1 Oblique Triangles and the Law of Sines 269 

7.2 The Ambiguous Case of the Law of Sines 277 

7.3 The Law of Cosines 282 

7.4 Vectors 291 

7.5 Applications of Vectors 300 


307 

Chapter 7 Test 

315 

8 COMPLEX NUMBERS, POLAR EQUATIONS, AND 

PARAMETRIC EQUATIONS 

8.1 Complex Numbers 318 

8.2 Trigonometric (Polar) Form of Complex Numbers 323 

8.3 The Product and QuotientTheorems 328 

8.4 Powers and Roots of Complex Numbers 333 

8.5 Polar Equations 343 

8.6 Parametric Equations 356 


364 


372

9 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 

9.1 Exponential Functions 376 

9.2 Logarithmic Functions 385 

9.3 Evaluating Logarithms and the Change-of-Base Theorem 392 

9.4 Exponential and Logarithmic Equations 399 


405 


411

Trigonometry Solutions Manual 

1.1 Basic Concepts 

1 

CHAPTER 1 THE TRIGONOMETRIC 

FUNCTIONS 

Section 1.1 

Connections (page 5) 

2. To graph (-7, 6), go seven units from the origin to 

the left along the x-axis, and then go six units up 

parallel to the y-axis. The point (-7, 6) is in 

quadrant II since the x-coordinate is negative and 

the y-coordinate is positive. 

y 

8+(-9) -1 1 

1. x= =-=- 

2 2 2 

-4+6 2 

y=--=-=I 

2 2 

Midpoint: (_..!., I) 

2 x 

2. endpoint: (2, -8) 

midpoint: (-1,-3) 

other endpoint: (x2'Y2) 

Exercises 

-I =2+x2 

2 

-2 = 2+x 2 

3. To graph (-7, -4), go seven units from the origin 

-4=x2 

to the left along the x-axis , and then go four units 

-3= -8+ Y2 down parallel to the y-axis. The point (-7 , -4) is 

2 in quadrant III since the .r- and y-coordinates are 

-6=-8+ Y2 both negative. 

2= Y2 

y 

(xi' Y2) =(-4,2) 

1. To graph (3, 2), go three units from the origin to 

the right along the x-axis, and then go two units 

up parallel to the y-axis. The point (3, 2) is in 

quadrant I since the x- and y-coordinates are both 

positive. 

y 

If i ' rrr: i 

- rn'-r'-'---"'-U 

. ;t +-1--Hi 

I-I :- , :: :; , i ! i I i ! 

r-,_+ ' !; , : ' I 

rt"p-h -+t · ··r- + t- + 

, , ,! ~ ; - t-"r l--+- i" ' 

I i I ! I i i : : -l + !--+-H-!-LU... 

I ! T r j i 

J : ;-iJ ;; ~ ! -t-rtq. i - ' . 2).TH-i 

, j I 1 

U+l· l-- q -q -~ - 

! I ! I I ! • ! ; I -f--r r r f-r-jt--t"! 

till'H-t :t Q TTT i T n I r! 

~ _ ' i i ' ~ 1 , I • I I 

j I I I : I 

! ; 

I I ; ; 

, I I 

, 

! . ! 

I , ! I I H=1 

: I Ii i I . 

1 

f-H : I I 

I . i i I 

4 

H-+ t·t-+ t--t-t rr: : r+H:i j 

I 

i t ! , I ! ! 

l 

I 

l i l 

I I i I 

· 

fIlIi i i I 

l i : j 

· : : 

IV 

, ; i I · I u.LYJ 

x 

x 

1


2 Chapter 1 The Trigonometric Functions 

4. To graph (8, -5), go eight units from the origin to 

the right along the x-axis, and then go five units 

down parallel to the y-axis. The point (8, - 5) is in 

quadrant IV since the x-coordinate is positive and 

the y-coordinate is negative . 

y 

6. To graph (-8,0), go eight units from the origin to 

the left along the x-axis, Do not move up or down 

parallel to the y-axis since the y-coordinate is O. 

The point (-8, 0) is not in any quadrant; it lies on 

the x-axis. 

y 

x 

x 

5. To graph (0, 5), do not move along the x-axis at 

all since the x-coordinate is O. Move five units up 

along the y-axis. The point (0, 5) is not in any 

quadrant since it is on the y-axis. 

y 

7. To graph (4.5, 7), go four and one half units from 

the origin to the right along the x-axis, then go 

seven units up parallel to the y-axis. The point 

(4.5,7) is in quadrant I since the x- and 

y-coordinates are both positive. 

y 

x 

2



3 

8. To graph (-7.5,8), go seven and one-half units 

from the origin to the left along the x-axis , and 

then go eight units up paralIel to the y-axis. The 

point (-7.5,8) is in quadrant II since the 

x-coordinate is negative and the y-coordinate is 

positive. 

y 

~ '~- il-~ --'-. - r 

IJ~ :(-7 .U) ! ;: 

t+: ~t- t-rH+' #~f~~[ i~J~ 

_c-+-+_+_ 

~+H+ ~ 

: ;+-! ! : : 

=tJ-+-, ' 

=I=f _1 1 1 ' _ 

1 

I . 1-+ -1- '-+ -i- '" 

1 ' I l l ! ! r~)-f 1-' ! i i-I 

Iii ' iii , 

! ; l i i x 

' ~ 

.. 

I 

filiI ! :: : i 

~- +f 

- ·-r , imffi 

-' ! , i 

! J ~ 

, 

i I I 

; 

!· L .l_ L __ ....!.-J 

.._.rrrrrtr _ 

L .L. 

-W--+IVJ 

9. (-5, n) lies in quadrant II since the x-coordinate is 

negative ad the y-coordinate (n= 3.14) is positive. 

10. (rc, -3) lies in quadrant IV since the x-coordinate 

(n= 3.14) is positive and the y-coordinate is 

negative. 

11. (-J2, - 2.fi) lies in quadrant III since the x- and 

y-coordinates are both negative. 

12. ( 1+../3, ~) lies in quadrant I since the x - and 

y-coordinates are both positive. 

15. Since xy =1, and 1 is a positive number attained 

by the product of two variables x and y, then 

either x and yare both positive (quadrant I), or . 

both negative (quadrant III). The graph of xy =1 

will lie in quadrants I and III. 

16. (a) Since (a, b) lies in quadrant II, a < 0, b > O. 

Therefore -a > 0 and (-a, b) would lie in 

quadrant I. 

(b) Since (a, b) lies in quadrant II, a < 0, 

b > O. Therefore-a > 0 and -b < 0, and 

(-a, -b) would lie in quadrant IV. 

(c) Since (a, b) lies in quadrant Il, a < 0, b » O. 

Therefore -b < 0, and (a, -b) would lie in 

quadrant III. 

In Exercises 17-24, the formula 

d =~(X2 - Xl)2 +(Y2 - Yl )2 is used to find the distance 

between (xl'Yl) and (x2, Y2)' 

17. (xl'YI) =(2, -1) and (X2'Y2) = (-3, -4) 

d= ~(-3 - 2)2 + [-4 _(_1)]2 

=~(_5)2+(_3)2 

=·J25+9 

=..[34 

18. (xl' Yl) =(-5,2) and (x2' Y2) = (3, -7) 

~[3 - (_5)]2 + (-7 - 2)2 

= ~82 +(_9)2 

= ..)64+ 81 

= ..)145 

19. (Xj, Yj) = (-1,0) and (x2' Y2) = (-4, - 5) 

d =~[-4-(-l)f +(-5-0)2 

= ~(_3)2 + (_5)2 

= ..)9+25 

=..[34 

20. (xl, Yl) = (-2, -3) and (x2' Y2) = (-6, 4) 

d= ~[-6 _(_2)]2 + [4 - (_3)]2 

=~(-4)2+72 

= ..)16+49 

=../65 

21. (xl' Yl) = (.fi, -.J5) and (x2, Y2) = (3.fi, 4..)5) 

d= ~(3.fi _.fi)2 +[4..)5 -(-.J5)t 

= ~(2.fi)2 + (5..)Sf 

=..)8+125 

= ..)133 

22. (Xl' Yl) = (5.,fi, -.J3) and 

(x2' Y2) = (-.,fi, 8.J3) 

d= ~(-J7- 5.,fif + [8.J3 - (-.J3)t 

=~(-6.,fi)2 + (9.J3)2 

= ..)252+243 

= ..)495 

=.J9 .../55 

= 3../55 

3



23. Let (xI' YI) = (5, - 6). Let (x2' Y2) = (5, 0) since 31. (a, b, e) = (5, 10, 15) 

we want the point on the x-axis with x-value 5. 

a 2 + b 2 = 52 + 10 2 

d = ~(5 _5)2 +[0 _(-6)]2 

=25+100 

= ~02 +6 2 

=125 

:;t 15 

=J36 

2 

2 

=6 

24. Let (xI' YI)= (5, -6). Let (x2' Y2) = (0, -6) since 

we want the point on the y-axis with y-value-6. 

32. (a, b, e)= (7, 24, 25) 

d =~(O - 5)2 + [-6 - (-6)f 

a 2 + b 2 =7 2 + 24 2 

=~(_5)2 +0 

= 49+576 

=..fi5 

=625 

=5 

25. In the distance formula, we find the square root of 

a 2 + b 2 =c 2 . 

(x2 _ XI)2 +(Y2 - Yl)2. From the display, 

x2 = -2, xI = 3, Y2 = 1. and Yl = -n. Therefore, 

the two points (xI' YI) and (x2' Y2) are (3, -n) 

and (-2, 1). 

26. Since the point (a, b) lies in quadrant I, that is, has 

positive x- and y-coordinates, the point (-a, -b) 

must have negative x- and y-coordinates and will 

lie in quadrant III. Since (a, b) is located 6 units 

from the origin, (-a. -b) will also be located 

6 units from the origin (but in the opposite 

direction). 

a = ~[1-(-2)]2 +(5 _5)2 

29. (a, b, e) =(9, 12, 15) 

a 2 + b 2 =9 2 + 12 2 

= ~32 +0 2 

=81+144 

=../9 

=225 

=3 

=15 2 

b = ~r-(l-_-1)-'-2-+-(9---5)-2 

2 

=e 

=~02 +4 2 

This triple is a Pythagorean triple since 

a 2 +b 2 = e 2. 

=.J16 

=4 

30. (a, b, e) = (6, 8, 10) 

e =~r-[1-_-(_-2-)]::-2+-(-9---5)-=-2 

a 2 +b 2 =6 2 +8 2 

=36+64 

=~32+4 2 

=100 

=10 2 =e 2 

=.J9+16 


=..fi5 

a 2 +b 2 = e 2. 

=e 

This triple is not a Pythagorean triple since 

a 2 +b 2 :;te 2. 

=25 2 = c 2 


33. If, for positive integers a, b, and c, a 2 + b 2 = e 2, 

then it is not necessarily true that a + b = e. For 

example, let a = 3, b = 4, and e = 5. 

Then a 2 = 9. b 2 = 16, and e 2 = 25. 

a 2 + b 2 =9+16=25=e 2 

Note that a + b = 3 + 4 = 7 :;t 5 = e . 

35. Are (-2, 5), (1, 5), and (1, 9) the vertices of a right 

triangle Use the distance formula to find the 

lengths of the sides connecting the points. 

=5 

a 2+b2 =3 2 +4 2 =9+16=25=e 2 

Yes, the given points are vertices of a right 

triangle. 

4



5 

36. Are (-9. -2). (-I, -2). and (-9. 11) the vertices of 

38. (4-.../3.-2.J3}(2-.J3.-./3), (3-.J3.-2.../3) 

a right triangle 

a 2 =[-9-(-1)]2 +[-2-(-2)]2 

a 

2 

= [(4-.../3)-(2-.../3)t + [-2.J3 -(-./3)t 

= (_8)2 + 0 2 = 64 

b 2 =[-9-(-)] 9 2 +(-2-11) 2 

=2 2+(-.../3)2 =7 

= 0 2 + (-13)1 = 169 b 2 = [(4-.J3)-(3-.../3)Y +[-2.../3 -(-2../3)y 

c 2 =[-1-(-9)]2+(-2-11)2 

= 1 2 +0 2 =I 

= 8 2 + (_13)2 = 233 

2 c = [(2-.J3)-(3-.../3)Y +[-./3 -(-2.../3)t 

64 + 169 = 233, so a 2 + b 2 2 

= c . 

Yes. the given points are vertices of a right = (_1)2 + (../3)2 = 4 

triangle. 

Since no two of a 2. b 2 or c 2 add up to the third. 

37. (.../3.2.../3+3). (.../3+4. -.../3+3} the given points are not vertices of a right 

triangle. 

(2.../3. 2.../3+ 4) 

39. Find x such that the distance between (x, 7) and 

a= ~(.../3 +4 _.../3)2 +[-.../3 +3-(2.../3 +3)t (2. 3) is 5. Use the distance formula. and solve for 

x. 

=~42+(_M)2 

5 = ~(2-x)2 +(3-7)2 

=../16+27 

= ~(2 - x)2 + (-4)2 

=~ =~(2-x)2+16 

Square both sides . 

b = ~(2.../3 _.../3)2 +[2.../3 + 4 -(2.../3 +3)Y 25 =(2-x)2 +16 

= ~(.../3)2 + 1 2 9=(2-x)2 

Take the square root of both sides. 

=../3+1 3=2-x or -3=2-x 

x=-I x=5 

=-14 

=2 40. Find ysuch that the distance between (5.y) and 

c = ~r-[ 2.../3-3---( .../3-3+-4---=)t-+-[2.../3-3-+-4--(-./3-3-+-3)---=-t (8. -I) is 5. 

d=~(5-8)2+[y_(-l)f =5 

= ~(.../3 - 4)2 + (M+ If ~9+(y+I)2 =5 

= ~19 - s.J3 + 28 + 6.../3 

Square both sides. 

9+(y+I)2 =25 

=~47-2.../3 

(y+I)2=16 

y+1 =±4 

If y + I =4. then y =3. 

=43+4 If y + I = -4. then y = -5. 

2 Therefore. y = 3 or y = -5. 

a2 +b2 =(~)2 +22 

*,c 

or 47-2.../3 

No. the triangle is not a right triangle. 

5



41. Find y such that the distance between (3, y) and 

(-2,9) is 12. 

12 = ~'(--2---3--::)2-+-(-9-_-y)"7""2 

= ~(_5)2 + (9 _ y)2 

= ~25+(9 _ y)2 

~ 

U 

6 

y 

144 = 25+(9- y)2 

119=(9- y)2 

..jIT9 = 9 - y 

y=9-M 

or -M = 9 - y 

y=9+M 

42. Find x such that the distance between (x, 11) and 

(5, -4) is 17. 

d = ~(x - 5)2 + [11- (-4)]2 = 17 

x= 130rx =-3 

~(x-5)2+152 =17 

(x - 5)2 + 225 = 289 

(x _5)2 = 64 

x-5=±8 

43. Let P =(x, y) be a point 5 units from (0, 0). Then, 

by the distance formula. 

5 =~(x- 0)2 +(y- 0)2 

=~x2+i 

2 

25 = x 2 + y2 or x + i = 25 

This is the required equation. The graph is a circle 

with center (0, 0) and radius 5. 

y 

...y++-H:++++-H-::~ x 

-5 5 

-5 

44. Let P = (x, y) be a point 3 units from (-5, 6). 

Then, by the distance formula, 

d=~[x-(-5)]2 +(y_6)2 =3 

~(x+ 5)2 + (y _6)2 = 3 

Square both sides. 

(x + 5)2 + (y - 6)2 = 9 

This is the equation for all points 3 units from 

(-5,6). 

The graph is a circle with center at (-5, 6) and 

radius 3. 

- 5 

r= 3 

45. By the Pythagorean theorem, 

2 

x =(¥f +(55.4)2 

x = ~(l7 .25)2 + (55.4)2 

= ../297.5625 + 3069.16 

=58.02 ft 

46. Let d = the diagonal by the Pythagorean Theorem 

d 2 = 52 + 12 2 

d = ../25+ 144 

= ../169 

=13ft 

47. AC 2 + BC 2 =AB 2 

(1000)2 + BC 2 = (1000.5)2 

1,000,000+ BC 2 = 1,001,000.25 

2 . 

BC = 1,001,000.25 - 1,000,000 

= 1000.25 

BC = ../1000.25 := 31.6 ft 

48. To find the length AB, apply the Pythagorean 

theorem to triangle ABX. 

(AB)2 =(AX)2 +(BX)2 

=8 2 +8 2 

=128 

AB = ../128:= 11.3137 

The distance across the top of the fabric is 

20 . AB (since there are 20 strips) plus 42 in. for 

the side of the unused triangle at the end. (Since 

this is a 45°- 45° right triangle, it is isosceles and 

its two equal sides are each 42 in.) Thus, the 

distance across the top is 

20(11.3137) + 42 =226.274 + 42 

=268 .274 in. 

Since the price of the material is $10 per linear 

yard, convert from inches to yards to find the 

number of linear yards. 

268.274:= 7.452 d 

36 Y 

The cost of the material is 7.452(10) = $74.52 . 

6



7 

49. Let b = the leg of the right triangle with leg hand 

hypotenuse a. 

b 2 +h 2 = a 2 

b 2 = a 2 _h 2 

=(184.7)2 -(144)2 

=13,378 

b =115.66 

The length I of a side is 2b, or 

2( 115.66) =231.3 m. 

50. Let x = the length of the leg of the right triangle 

with leg a and hypotenuse s. Notice that 

x =!... = 231.3 = 115.65. 

2 2 

a 2 +x 2 = s2 

(184.7)2 + (115.65)2 = s2 

34,114 .1+ 13,374.92 = s2 

47,489.02 = s2 

217.9 =s 

The length of the edge of the pyramid labeled s is 

about 217.9 m. 

51. By the Pythagorean theorem. 

BC 2 = AC 2 + AB 2 

AB=~BC2 -AC 2 

= 

r-( r 

( 42 42 

262 260 

=~(213l)2 -(2130)2 

= .J4,541,161- 4.536.900 

'" 65.28 feet 

52. {xix> 6} 

This is an open interval with interval notation 

(6,00). 

53. {pjp ~ 1O} 

This is a half-open interval with interval notation 

[10,00). 

54. {xl-3 < x < 7} 

This is an open interval with interval notation 

(-3,7). 

55. {y/8SyS13} 

This is a closed interval with interval notation 

[8, 13]. 

56. The graph shows an open interval, starting at 2. 

and includes all real numbers greater than 2. 

The interval notation is (2, 00). 

57. The graph shows an open interval. The interval 

does not include 2, but consists of all real 

numbers less than 2, written in interval notation 

as (-00,2). 

58. The graph shows a half-open interval between 

1 and 4. It includes 4, but does not include 1. 

The interval notation is (1, 4]. 

In Exercises 61-66, f(x) = -2x 2 + 4x + 6 . 

61. f(O) = -2(0)2 + 4(0) + 6 

=0+0+6 

=6 

62. f(-2) = -2(-2)2 + 4(-2) + 6 

=-8-8+6 

=-10 

63. f(-I) = -2(-1)2 + 4(-1) + 6 

=-2(1)+4(-1)+6 

=-2+(-4)+6 

=0 

64. f(-m) = -2(_m)2 + 4(-m) + 6 

=-2m 2 -4m+6 

65. f(1+a)=-2(1+a)2 +4(1+a)+6 

= -2(1 +2a + a 2)+ 4(1+a)+6 

=-2-4a -2a 2 +4+4a+6 

=-2a 2 +8 

66. f(2 - p)= -2(2- p)2 +4(2- p)+6 

= -2(4-4p+ p2)+8-4p+6 

=-8+8p-2p2+8-4p+6 

=_2p2 +4p+6 

67. (a) We want the value ofJ(or l) when X is 2. 

When X = 2 in the table, Y 1 = 1.2, so 

/(2) = 1.2. 

(b) Now we want the value of X for whichf(or 

l) equals -2.4. When f(x) = Y 1 = - 2.4, 

X = 5 in the table. 

(c) The graph of y = j(x) crosses the y-axis at the 

y-intercept (0, 3.6). 

(d) The graph of y = j(x) crosses the x-axis at the 

x-intercept (3, 0). 

7



68. (a) When x = 0 on the graph, y = 11, so 

f(O) = 11. 

(b) When x = 6 on the graph, y = 9, so f(6) = 9 

(c) The value of y =f(a) is 0 when x = -2, so 

a=-2. 

(d) When x = 2 on the graph, f(x) =y =10. 

Also f(7) =10, and f(8) =10. So f(x) =10 

for x =2, 7, and 8. 

(e) 

(8,f(8» =(8, 10) =(xI' YI) and 

(10, f(1O» = (10, 1)= (x2' Y2) 

d = ~(1O - 8)2 + (1-10)2 

=~22 +(_9)2 

=.J4+81 

=~ 

69. (a) P is only a function of a when the range is 

restricted to (0, 00). If the range is not 

restricted, then P = ±,J;;3 and is not a 

function. 

(b) p2 = (1.524)3 

(c) 

p 2 =:: 3.5396 

p =:: 1.88 

Since P is time in years, we only consider the 

principal square root in the answer. The 

orbital period of Mars is approximately 

1.88 years . 

3 

19.18 2 = a 

3 

367.8724 = a 

7.165 =:: a 

70. Y = 4x-3 

Domain: (-00,00). 

Range: (-00,00). 

Y = 4x - 3 is a function because each x-value 

gives exactly one y-value. 

71. 2x + 5y = 10 

Domain: (-00,00) 

Range : (-00, 00) 

This is a function because each value of x gives 

exactly one value ofy . 

72. y=x 2 +4 

Domain: (-00,00) 

Range: The smallest value of x 2 is 0, so the 

smallest value of y =x 2 +4 =0 + 4 =4 . 

So the range is [4,00). 

Y = x 2 + 4 is a function because each x-value 


73. y=2x 2 - 5 

Domain: (-00,00) 

2 

Range: Since x ~ 0, 2x 2 - 5 ~ -5, and the range 

is [-5,00). 


exactly one value of y, 

74. y =-2(x - 3)2 + 4 

Domain: (-00,00) 

Range: Since 

(x - 3)2 ~ 0 

-2(x-3)2 s 0 

-2(x - 3)2 +4:S; 4. 

The range is (-00, 4]. 

y =-2(x - 3)2 + 4 is a function because each 

x-value gives exactly one y-value. 

75. y =3(x+ 1)2_ 5 

Domain: (-00,00) 

Range: Since (x + 1)2 ~ 0, 2(x + 1)2 - 5 ~ -5. 

The range is [-5,00). 


exactly one value of y . 

76. x = y2 

Domain: i ~ 0, so the domain is [0,00). 

Range: (-00,00) 

X = y2 is not a function because all x-values 

(except x = 0) give two y-values. For example, if 

x = 4, y = 2 or y = -2. 

77. y =.J4+x 

Domain: Because we can only take the square 

root of a nonnegative number, 

4+x~0 

x ~ -4 

the domain is [-4,00). 

Range: For all values of x ~ -4, .J4 + x ~ 0, so 

y ~ O. The range is [0, 00). 

y =.J4 + x is a function because each x-value 


8


1.2 Angles 

9 

78. y=~x2+1 

Domain: (-00,00) 

Range: For all x-values, x 2 2: 0, x 2 + 1 2: 1, and 

~x2 + 1 ~ 1. The range is [1,00). · 

y =~ is a function because each x-value 

gives only one y-value. 

79. y=~1-x2 

2 

Domain: 1- x ~ 0 

_x 2 ~-1 

x 2 $1 

-1 s x $1 

The domain is [-1, 1). 

Range: [0, 1] 

This is a function because each x-value gives only 

one y-value. 

80. Domain: [-5,4] 

Range: [-2,6] 

This is a function because any vertical line 

intersects the graph in at most one point. 

81. Domain: [-5,00) 

Range: [0,00) 

This is a function because any vertical line 

intersects the graph in at most one point. 

82. Domain: (-oo,-3]u[3,00) 

Range: (-00, 00) 

This is not a function because some vertical lines, 

for example x = 4, intersect the graph in two 

points. 

83. Domain: [-4,4] 

Range: [-3,3] 

This is not a function because some vertical lines 

intersect the graph in two points, for example, 

x=O. 

1 

84. y= x 

The denominator cannot be zero, so x ~ 0 

Domain: (-00,0) U (0,00) 

-1 

86. y=-;=== 

~x2 +25 

The denominator must not be 0 and the 

expression under the radical sign must be 

nonnegative. 

2 

Since x ~ 0 for all real numbers x, 

x 2 +25 ~25 

~x2 +25 ~5>0. 

Thus, the denominator is not 0 for any real 

number x and the expression x 2 + 25 is 

nonnegative for any x. 

The domain is (-00,00). 

87. From the figure, the length of a side of the large 

square is a + b, so its area is (a + b)2 or 

a 2 +2ab+b 2 . 

88. The area of each right triangle is (~fb, so the 

sum of the areas of the four right triangles is 

4(~Jab or 2ab. The smaller square has side of 

length c, so its area is C • cor c 2 . 

89. The sum of the areas of the four right triangles 

and the smaller square is 2ab + c 2 • 

90. Setting the expressions from exercises 87 and 89 

equal to each other we obtain 

a 2 + 2ab+b 2 =2ab +c 2. 

91. Simplifying the expression in exercise 90 results 

in the Pythagorean theorem' a 2 + b 2 =c 2 . 

Section 1.2 

Exercises 

2. 1° represents _1_ of a rotation. 

360 

Therefore, 45° represents 

45 1 fl ' 

360 =8" 0 a comp ete rotation. 

-2 

85. y=- 

x+1 

The denominator cannot be zero. 

x+1 ~O 

x~ -1 

Domain: (-00, -1) u (-1, 00) 

3. Let x =the measure of the angle. 

If the angle is its own complement. 

x+x=9O 

2x=90 

x=45 

A 45° angle is its own complement. 

9



. 

o · 

7m+3=73 

4. Let x = the measure of the angle. 

Ifthe angle is its own supplement, 

x+x = 180 

2x = 180 

x=90. 

A 90° angle is its own supplement 

25 minutes x degrees 

5. = ;x = 150 0 

60 minutes 360 degrees 

6. The hour hand is ~ of the way between the 1 and 

4 

2, and therefore is at 8.75 minutes past the 12. 

The minute hand is 15 minutes before the 12. The 

smaller angle formed by the hands of the clock is 

(15+ 8.75) minutes x 

--;x = 142.5° 

60 minutes 360° 

7. 7x+ 11x = 180 

18x = 180 

x=1O 

The measures of the two angles are 

(7x)0 = 7(10°) = 70° and (l lx)" = 1l(100) = 110°. 

8. The two angles form a right angle. 

4y+2y=90 

6y=90 

y=15 

The two angles have measures of 4(15°) = 60° 

and 2(15°) = 30°. 

9. (5k + 5) + (3k + 5) = 90 

8k+1O=9O 

8k=80 

k =10 


(5k + 5)° = 5(10°) + 5° = 55° and 

(3k + 5)° = 3(10°) + 5° = 35°. 

10. The sum of the measures of two supplementary 

angles is 180°. 

1Om+7+ 7m+3 = 180 

17m+1O=180 

17m =170 

m=1O 

10m +7 = 107 

The angle measures are 107° and 73°. 

11. 6x -4+8x -12 = 180 

14x-16=180 

14x = 196 

x =14 


(6x _4)° = 6(14°) _4° = 80° and 

(8x - 12)° = 8(14°) - 12° = 100°. 

12. The sum of the measures of two complementary 


9z+6+3z=9O 

12z+6=9O 

12z = 84 

z=7 

9z + 6 = 9(7) + 6 = 69 

3z = 3(7) = 21 

The angle measures are 69° and 21°. 

13. Ifan angle measures x degrees, and two angles are 

complementary if their sum is 90°, then the 

complement of an angle of XO is 

(90-x)0. 

14. If an angle measures x degrees, and two angles are 

supplementary if their sum is 180°, then the 

supplement of an angle of XO is 

(180-x)0. 

15. The first negative angle cotenninal with x 

between 0° and 60° is (x - 360)° . 

16. The first positive angle coterminal with x between 

0° and --600 is (x + 360)°. 

17. 62° 18' + 21° 41' =83° 59' 

18. 75° 15' + 83° 32' = 158° 47' 

19. 71° 18' - 47°29' = 700 78' - 47°29' 

=23°49' 

20. 47°23' - 73° 48' = -(73° 48' - 47° 23') 

= -26°25' 

21. 90° -51°28' = 89°60' - 51°28' = 38° 32' 

22. 180° -124° 51' = 179°60' -124° 51' 

= 55°9' 

23. 90° - 72° 58'11" = 89° 59'60"-72°58'11" 

= 17°1' 49" 

24. 90°-36°18'47" = 89°59'60" -36°18' 47" 

= 53° 41' 13" 

25. 

20054' = 200+ 54° 

60 

=20°+.900° 

=20.900° 

26. 38°42' = 380+ 42° 

60 

=38.700° 

10


1.2 Angles 11 

0 0 

27. 91 035'54"=910+ 35 + 54 

60 3600 

= 91.598 0 

28. 

34051'35"=340+510 + 35° 

60 3600 

= 34° + .85° + .0097° 

= 34.860° 

29. 274018'59" = 2740+~+ 59° 

60 3600 

=274.316° 

30. 165051'9" =1650+ 51° +~ 

60 3600 

= 165° + .85° + .0025° 

= 165.853° 

31. 31.4296° = 31° + (.4296)60' 

=31° + 25.776 

=31° + 25' + (.776)(60") 

= 31°25'47" 

32. 59.0854° =59° + .0854° 

= 59° + (.0854)(60') 

=59° + 5.124' 

= 59° 5' + (.124)(60") 

=59°5'7" 

33. 89.9004° = 89° + (.9004)(60') 

= 89° + 54.024' 

=89° + 54' +(.024)(60") 

=89°54'1" 

34. 102.3771°= 102° + (.3771)(60') 

= 102° + 22.626' 

= 120°22' + (.626)(60") 

= 102° 22' 38" 

35. 178.5994° = 178° + (.5994)(60') 

=178°+35.964' 

=178° + 35' + (.964)(60") 

= 178°35'58" 

36. 122.6853° = 122° + (.6853)(60') 

= 122° + 41.118' 

= 122° 41' + (.118)(60") 

= 122°41'7" 

39. -40° is coterminal with -40° + 360° = 320°. 

43. 539° is coterminal with 539 0 - 360° = 179°. 

44. 699° is coterminal with 699 0 - 360° = 339°. 

45. 850° is coterminal with 

850° - 2(360°) =850° - 720 0 =130°. 


1000° - 2 . 360 0 = 1000° - 720° 

=280°. 

47. 30° 

A coterminal angle can be obtained by adding an 

integer multiple of 360°. 

30° + n : 360° 

48. 45° 



45° + n . 360° 

49. 135° 



135° + n ' 360° 

50. 270° 



270° + n . 360 0 

51. -90° 


integer multiple of 360 0 . 

-90° + n . 360° 

52. -135° 

A coterminal angle can be obtained by adding 


-135 0 + n . 360° 

54. (a) 360° + r"is coterminal with r" because you 

are adding an integer multiple of 360° to r", 

r" + 1 . 360°. 

(b) r" - 360° is coterminal with r" because you 

are adding an integer multiple of 360° to r", 

r" + (-1) . 360° . 

(c) 360° - r" is not coterminal with r" because 

you are not adding an integer multiple of 

360 0 to r, 

360° - r" :t: r" + n . 360° 

41. -125° is coterminal with -125° + 360 0 = 235°. 

42. -203° is coterminal with 360° + (-203°) = 157°. 

11



(d) ,0 + 180° is not cotenninal with ,0 because 

you are not adding an integer multiple of 

360° to ,0. You are adding .!. (360°). 

2 

Choices (c) and (d) are not cotenninal with ,0. 

55. lJ = 3~C:o)-in{3:0)) 

X=~o 

YI = 36{(- 3:~)-in{- 3:~)) 

=320° 

59. 

174 0 o 

y 

534°; -186°; 

quadran I II 

174° is cotenninal with 174° + 360° = 534° and 

174°- 360° = -186°. These angles are in 

quadrant II. 

56. lJ =31(3:0)- intC:o)) 

X=-98° 

lJ =31(- :6~)-in{- :~)) 

=26r 

60. A positive angle coterminal with 234° is 

234° + 360° = 594°. 

A negative angle coterminal with 234° is 

234° - 360° = -126°. 

There angles are in quadrant Ill. 

y 

For Exercises 57-64 angles other than those given are 

possible. 

57. A positive angle cotenninal with 75° is 

75° + 360° = 435°. 

A negative angle cotenninal with 75° is 

75° - 360° = -285°. 

These angles are in quadrant I. 

y 

594°; -126°; 

quadrant III 

61. A positive angle coterminal with 300° is 

300° + 360° = 660°. 

A negative angle coterminal with 300° is 

300° - 360° = -60°. 

These angles are in quadrant IV. 

y 

435°; -285°; 

quadrant I 

58. 

o 

89 0 

660°; -60°; 

quadrant IV 

449°; -271°; 

quadrant 1 

89° is cotenninal with 89° + 360° = 449° and 

89° - 360° = -271°. 

These angles are in quadrant I. 

12


1.2 Angles 13 

62. 

y 

65. 

y 

152°;-208°; 

quadrantII 

A positive angle cotenninal with 512° is 

512° - 360° = 152°. 

A negative angle cotenninal with 512° is 

512° - 2(360') = -208°. 

These angles are in quadrant II. 

63. -61° is cotenninal with 299° and -421°; these 

angles are in quadrant IV. 

y 

66. 

Points: (-3, -3), (0, 0) 

r = ~(-3 _0)2 +(-3 _0)2 

= .../9+9 

=..ff8 

=F2 

=M 

(- 5. 2) 

y 

64. 

299°; -421°; 

quadrantIV 

y 

Points: (-5.2), (0. 0) 

r = ~(-5 _0)2 +(2-0)2 

=.../25+4 

=.fi9 

67. 

y 

-H-t+hI-::+++++-. JC 

201°; -519°; 

quadrantIII 

A positive angle cotenninal with -159° is 

-159° + 360° = 201°. . 

A negative angle cotenninal with -159° is 

-159°-360° = -519°. 

These angles are in quadrant III. 

Points: (-3. -5). (0. 0) 

r = ~(-3- 0)2 + (-5 _0)2 

= .../9+25 

=../34 

13



68. 

y 

--+--+--::11'+....-. x 

o 

2 

71. 90 revolutions per minute 

=-90 revo I unons ' per second 

60 

= 1.5 revolutions per second 

72. 45 revolutions per minute 

45 I' d 

=60 revo unon per secon 

69. 

Points: (.,fj, 1), (0, 0) 

r = ~(.,fj - 0)2 + (1- 0)2 

=..)3+1 

=.,[4 

=2 

y 

--+-t-t-+:::l-+-II-+-++ x 

-2 

Points: (-2, 2.,fj), (0, 0) 

r=~(-2-0)2 +(2..J3 _0)2 

=..)4+ 12 

=.J16 

=4 

=l revolution per second 

4 

73. 600 rotations per minute 

600. d 

= 60 rotations per secon 

=10 rotations per second 

=5 rotations per .!.. second 

2 

= 5(360°) 

1 

= 1800° per - second 

2 

74. If the propeller rotates 1000 times per minute, 

then i 1000 16 2. d 

en It rotates --= - times per secon . 

60 3 

Each rotation is 360 0 , so the total number of 

degrees a point rotates in one second is 

(36O{16~) = 6000°. 

75. 75° per minute = 75°(60) per hour 

= 4500° per hour 

4500 . h 

= 360 rotations per our 

=12.5 rotations per hour 

70. Y 

_H-H::-Id-t++++++ x 

Points: (4.,fj,-4), (0, 0) 

r = ~(4.,fj - 0)2 + (-4 - 0)2 

=..)48+16 

=-./64 

=8 

76. 74.25° = 74° +(.25)(60') = 74° 15' 

74°20' -74°15' = 5' 

74020' = 740 + 20° = 74.330 

60 

74 .33° -74.25° = .08° 

The difference in measurements is 5' to the _ ____ 

nearest minute or .08 0 to the nearest hundredth of 

a degree. 

77. (J = 0.763" 

90° - (J = 90° - 0.763" 

= 89° 59'60" - 0.763" 

= 89° 59'59.237" 

78. (J = 0.292" 

90° - (J = 90° - 0.292" 

= 89° 59' 60" - 0.292" 

= 89° 59'59 .708" 

14



79. The earth rotates 360° in 24 hours. 360° is equal 

to 360(60) minutes. or 21.600' . 

24 hours x 

---= 

21.600' 

l' 

x= o.ooi . 

It should take the motor 0.001 hours or 3.6 =: 4 

seconds to rotate the telescope one minute. 

SO. Since we have five central angles that comprise a 

full circle. we find : 

5 · 26 = 360° 

106=360 

6=36° 

The angle of each point of the five-pointed star 

measures 36°. 

Section 1.3 

Exercises 

1. In geometry. opposite angles are called vertical 

angles. 

3. 5x - 129 = 2x - 21 since vertical angles are equal. 

3x-129=-21 

3x= 108 

x=36 

5x -129 = 5(36) -129 

= 180-129 

=51 

2x-21 :::;;2(36)-21 

=72-21 

=51 

Both angles are 51 0. 

4. l l,r - 37 = Tx + 27 since vertical angles are 

equal. 

4x-37=27 

4x=64 

x=16 

llx-37:::;; 11(16)-37 

:::;; 176-37 

:::;; 139 

7x+27:::;; 7(16)+ 27 

=112+27 

:::;; 139 

Both angles are 139°. 

5. x + (x + 20) + (210 - 3x):::;; 180 since the sum of 

the angles of a triangle is 180°. 

-x+230:::;;180 

-x:::;;-50 

x:::;;50 

x + 20 = 50 + 20 :::;; 70 

210 - 3x:::;; 210 - 3(50) 

:::;; 210-150 

=60 

The three angles are 50°. 70°. and 60°. 

6. (lOx-20) + (x + 15) + (x + 5) = 180 

since the sum of the angles of a triangle are 180°. 

12x = 180 

x:::;; 15 

lOx - 20:::;; 10(15)- 20 

:::;; 150-20 

:::;; 130 

x + 15 :::;; 15+ 15 = 30 

x+5:::;; 15+5:::;;20 

The three angles are 130°. 30°. and 20°. 

7. (x - 30) + (2x -120) + (~x + 15) = 180 since the 

sum of the angles of a triangle is 180°. 

L, -135:::;; 180 

2 

7 

-x:::;; 315 

2 

2 . 

x:::;; (315)-=90 

7 

x - 30 =90 - 30 =60 

1 1 

-x+ 15:::;; -(90)+ 15 

2 2 

=45 + 15 

:::;;60 

2x -120:::;; 2(90) -120 

:::;;180-120 

=60 

All three angles are 60°. 

8. x + x + x = 180 since the sum of the angles of a 

triangle are 180°. 

3x =180 

x=60 

All three angles are 60°. 

15


16 Chapter 1 The Trjgonometrle Functions 

9. (3x + 5) + (5x + 15) = 180 since these two angles 

are supplementary. 

8x+20= 180 

8x= 160 

x=20 

3x + 5 = 3(20) + 5 

=60+5 

=65 

5x + 15 = 5(20) + 15 

= 100+ 15 

=115 

The two angles are 65° and 115°. 

10. (5x - 1) + (2x) = 90 since the angles are 

complementary. 

7x-l=90 

7x=91 

x=13 

5x -1 = 5(13) -1 = 65 -1 = 64 

2x = 2(13) = 26 

The two angles are 64° and 26°. 

11. 2x - 5 = x + 22 since alternate interior angles are 

equal. 

x-5 =22 

x=27 

2x - 5 = 2(27) - 5 = 54 - 5 = 49 

x + 22 = 27 + 22 = 49 


12. 2x + 61 = 6x - 51 since alternate exterior angles 

are equal. 

61=4x-51 

112=4x 

28=x 

2x+61 =2(28) +61 =56+61 =117 

6x-51 =6(28)- 51 = 168 -51 = 117 


13. (x + 1) + (4x - 56) = 180 since the sum of interior 

angles on the same side of the transversal is 180°. 

5x -55 =180 

5x=235 

x=47 

x + 1 = 47 + 1 = 48 

4x-56=4(47)-56= 188-56= 132 

The angles are 48° and 132°. 

14. lOx + 11 = 15x- 54 since alternate exterior 

angles are equal. 

11 =5x-54 

65=5x 

13=x 

lOx + 11 = 10(13)+ 11 

=130+11 

= 141 

15x - 54 = 15(13)- 54 

= 195-54 

=141 


IS. Let x =the measure of the third angle. 

37° + 52° + x = 180° 

89°+x =180° 

x=91° 

16. Let x = the measure of the third angle. 

x + 29° + 104° =180° 

x+133°=180° 

x=47° 


147°12' + 30°19' + x =180° 

177°31'+x =180° 

x =179°60' -177°31' 

x =2°29' 


x + 136°50' + 41° 38' =180° 

x + 178°28' =180° 

x =179°60' -178°28' 

x=1°32' 


74.2° + 80.4° + x =180° 

154.6° + x =180° 

x=25.4° 


29.6° + 49.7° + x =180° 

79.3° + x =180° 

x =100.7° 

16



23. angle 1 = 55° (vertical angle with 55°) 

angle 5 = 180° - 120° = 60° (interior angles on 

same side of transversal) 

angle 4 = 180° - 60° - 55° = 65° 

(supplemental angle with angle 5 and 55°) 

angle 2 = 65° (vertical angle with angle 4) 


angle 6 = 120° (vertical angles) 

angle 7 = 180° - 120° = 60° (supplemental angle 

with angle 6) 


angle 10= 180° - 60° - 65° = 55° (sum of angles 

of triangle is 180°) . 


24. Given x + y = 40. 

(x + 2y) + l1x = 180 since the angles are 

supplementary . 

x+ y = 40 (1) 

12x + 2y = 180 (2) 

Take -2 times equation (1) + equation (2). 

-2x-2y=-80 

12x +2y = 180 

lOx =100 

x=1O 

Substitute 10 for x in equation (1). 

lO+y=40 

y=30 

x + 2y = 10+ 2(30) = 10 + 60 = 70 

llx = 11(10) = 110 


25. The triangle has a right angle, but each side has a 

different measure. The triangle is a right triangle 

and a scalene triangle. 

26. The triangle has one obtuse angle and three 

unequal sides, so it is obtuse and scalene. 

27. The triangle has three acute angles and three equal 

sides, so it is acute and equilateral. 

28. The triangle has two equal sides and all angles are 

acute, so it is acute and isosceles. 

29. The triangle has a right angle and three unequal 

sides, so it is right and scalene . 

30. The triangle has one obtuse angle and two equal 

sides, so it is obtuse and isosceles. 

31. The triangle has a right angle and two equal sides, 

so it is right and isosceles. 

32. The triangle has a right angle with three unequal 

sides, so it is a scalene, right triangle. 

33. The triangle has one obtuse angle and three 

unequal sides, so it is obtuse and scalene. 

34. This triangle has three equal sides and all angles 

are acute, so it is an acute, equilateral triangle. 

35. The triangle has three acute angles and two equal 

sides, so it is acute and isosceles. 

36. This triangle has a right angle with three unequal 

sides, so it is a scalene, right triangle. 

40. 

Connect the right end of the semi-circle to the 

point where the arc crosses the semi-circle. Since 

the setting of the compass never changed, the 

triangle is equilateral. Therefore, each of its 


41. Corresponding angles are A and P, Band Q, C 

and R; corresponding sides are AB and PQ, CB 

and RQ, AC and PRo 

42. Corresponding angles are A and P, B and Q, C 

and R; Corresponding sides are AC and PR, BC 

and QR, AB and PQ. 

43. Corresponding angles are A and C, ABE and CBD, 

E and D; Corresponding sides are AB and CB, EB 

and DB,AE and CD. 

44. Corresponding angles are Hand F, K and E, HGK 

and FGE; corresponding sides are HK and FE, 

GK and GE, HG and FG. 

45. Q = A so Q = 42°. 

A + B + C= 180° so B = 180° - 90° - 42° = 48°. 

B = R so R = 48°. 

P = C so P = 90°. 

46. A + B + C = 180° 

so A = 180° - 46° - 78° = 56°. 

M=BsoM=46°. 

N = A so N = 56°. 

P= CsoP = 78°. 

47. K=BsoB= 106°. 

A + B + C= 180° so A = 180°.- 30° - 106° = 44°. 

A=MsoM=44°. 

48. Y = V so Y = 28°. 

X + Y + Z = 180° so Z = 180° - 74° - 28° = 78°. 

T=XsoT=74°. 

W=Zso W=78°. 

49. X+ Y+Z= 180° soX= 180°-90°-38°=52° 

X = M so M = 52°. 

17



50. P + Q + R = 180° so R =180° - 64° - 20° = 96°. 

T = P so T = 20°. 

U =R so U =96° . 

V=QsoV=64°. 

In Exercises 51-56, corresponding sides of similar 

triangles are proportional. 

57. Let T =the height of the tree. 

The triangle formed by the tree and its shadow is 

similar to the triangle formed by the stick and its 

shadow . 

L::::dT .6J' 

45 3 

25 a 

51. -=- T 45 

10 8 

-= 

2 3 

200= lOa 

T 

20=a -=15 

25 b 

2 

-=- T=30 

10 6 The tree is 30 m high. 

150 = lOb 

15=b 58. 

a 75 b 75 

52. -=- -= 

10 25 20 25 

b 

~=3 -=3 

10 20 

a=30 b=60 

180 

X 9 

6 a 

53. -= 

-= 

12 12 

180 15 

72 = 12a 

15x =1620 

6=a 

x=108 

b 6 

The height of the tower is 108 ft. 

-= 

15 12 59. Let x = the middle side of the actual triangle, 

b 1 

-= 

and y =the longest side of the actual triangle. 

15 2 The triangles in the photograph and on the actual 

2b=15 piece of land are similar. The shortest side on the 

15 1 land corresponds to the shortest side in the 

54. -= 

b=-or7 

2 2 

photograph. 

400 x 

-=a 

3 4 5 

6 9 100 =!. 

9a=18 5 

a=2 x=500 

400 y 

-=x 

9 

55. -= 

4 7 

4 6 

loo=L 

x 3 7 

-= 

4 2 y=700 

2x=12 The other two sides are 500 m and 700 m long. 

x=6 

m 21 

56. -= 

12 14 

14m =252 

m=18 

18


60. Let x = the height of the lighthouse. x 100+ 120 

63. =--- 

50 100 

x~ x 220 

-= 

28 50 100 

x 11 

-= 

50 . 5 

1.75~ 5x =550 

3.5 

x=!lO 

X 28 

--=- 

64. L= 40 

1.75 3.5 

3.5x =49 

60 40+160 

y 40 

x=14 -= 

60 200 

The lighthouse is 14 m tall. 200y =2400 

61. Let h = the height of the building. 

y=12 

The triangle formed by the house and its shadow 

is similar to the triangle formed by the building 

c 

65. 

10+90 

-=-- 

and its shadow. 100 90 

c 100 

15~ -= 

100 90 

40 c 10 

-= 

100 9 


1000 

c=--:::lll.1 

9 

Map : 

T 17 cm Y 75 

8c~cm 

m=p 

3 

m=- 

256 

300 

66. The larger right triangle (with hypotenuse m) and 

15 h 

-= 

the smaller right triangle (with hypotenuse 80) are 

40 300 similar. 

4500 = 40h m 75+5 

112.5= h -=- 

80 75 

The height of the building is 112.5 ft. m 80 

-= 

80 75 

62. Let a =the distance between Phoenix and Tucson 

75m=6400 

and b = the distance between Tucson and Yuma. 

6400 

Actual : 

T b Y 67. Let a = the length of the longest side ofthe first 

quadrilateral, 

x =the length of the shortest side of the second 

~ 

P 

quadrilateral, 

a 230 b 230 and y =the other unknown length . 

-=- -=- 

8 12 17 12 

24 

I8Q' 

12a = 1840 12b = 3910 

a « 153.3 b::: 325.8 

The distance from Phoenix to Tucson is about 

153.3 Ian. The distance from Tucson to Yuma is 

about 325.8 km. 

«>: 

~ 

19



Corresponding sides are in proportion. 

. 32 2 

Smce -=-, 

48 3 

o 2 

-= 

60 3 

30 =120 

0=40 

24 2 

-=y 

3 

2y=72 

y=36 

18 2 

-=x 

3 

2x=54 

x=27 

The unknown side in the first quadrilateral is 

40 em; the unknown sides in the second 

quadrilateral are 36 em and 27 em. 

68. Let x =length of entire body carved into 

mountain. 

.1= 61 

60 x 

-=t 

.If 

60 x 

~x=6OC:) 

~x=380 

4 

x = 1520 = 506~ 

3 3 

Abraham Lincoln's body would be 506~ft tall. 

3 

69. Corresponding sides are in proportion. 

x-5 5 

--= 

15 15 

15(x-5) = 75 

15x-75=75 

15x= 150 

x=to 

x-2y 5 

----"'--= 

(x-2y)+(x+y) 15 

x-2y 5 

--= 

2x - y 15 

15(x- 2y) =5(2x - y) 

15x-30y= tox-5y 

5x= 25y 

Substitute to for x. 

50=25y 

y=2 

Therefore, x = to and y =2. 

70. The two triangles are similar. 

Corresponding angles are equal. 

toy+ 8 =58 

toy=50 

y=5 

Corresponding sides are in proportion. 

x-y 6 1 

--=-=x+ 

y 18 3 

3(x- y) = I(x+ y) 

3x-3y= x+ y 

2x=4y 

x=2y 

Substitute 5 for y. 

x=2(5) = to 

Therefore, x = to and y = 5. 

71. (a) Since a thumb covers about 2 arc degrees or 

about 120 arc minutes, approximately..!. of a 

4 

thumb would cover the moon. 

(b) 20 + 10 =30 arc degrees 

72. (a) Let D s be the Mars-Sun Distance, d s the 

diameter of the sun, D m the Mars-Phobes 

distance, and d m the diameter of the Phobes. 

Then, by similar triangles 

Ds =.!!.L 

D m d m 

D = Dsdm 

m 

d s 

142,000,000(17.4) 

= 

865,000 

=2856 mi 

(b) No. Phobes does not come close enough to 

the surface of Mars. 

73. (a) Let D, be the Jupiter-Sun distance, d, the 

diameter of the sun, D m the Jupiter 

Ganymede distance, and d m the diameter of 

Ganymede. Then, by similar triangles, 

D, d 

-= s 

D m d m 

D = Dsdm 

m d, 

484,000,000(3270) 

= 

865,000 

= 1,830,000 mi 

(b) Yes; Ganymede is usually less than half this 

distance from Jupiter. 

20



74. (a) Since CG and DH are parallel and the 

triangles b.CAG and b.HAD share the 

common angle A, they are similar. 

Additionally, BD II EG with the common 

angle A, so triangles b.BAD and b.EAG are 

similar. 

CG AG . 

(b) Triangles b.ABD and MEG are similar, so 

AG EG 

-=-. However, BD = EF,so 

AD BD 

AG EG 

AD = EF' 

. EG EG EG 

(c) SmceBD= EF= 1 -=-=-. 

, EF BD 1 

Section 1.4 

Exercises 

1. 

y 

8 in Quadrant IV 

y 

SeCtion 1.4 

AG EG 

(d) CG feet = - = - = EG paces 

AD EF 


1. From the figure, we see that b.QOP and MOB 

are similar. 

PQ=y=l.=l.=sin6 (sincer= 1) 

1 r 

- = - . HD = 1, so this becomes 

HD AD 

CG AG 

-= 

1 AD 

OQ= x = ~= ~ =cos6 (since r= 1) 

1 r 

AB OA 

-=y 

x 

. AB 1 y 

SmceOA= 1, -=-, so AB=-=tan6. 

. y x x 

2. 

3. 

y 

y 

2. y 

/ 

8 in Quadranl III 

21



4. y 7. (0,2) 

x=O,y =2 

r=~02 +2 2 

+-+-+-+-+-+-H--+-t-Hi-t;ooIl'I':'t"'!t+x 

= ..}0+4 

=../4 

=2 

sinO=~= 1 

2 

o 

cosO=-=0 

2 

5. (-3,4) 

x=-3,y =4 

2 

r = ~(_3)2 + 4 2 tan0 = ō undefined 

=..}9+16 o 

cotO = = 0 

=../25 

2 

=5 

2 

sec0 = undefined 

sino=z=i 

r 5 

x 3 

cosO=-=- 

r 5 

y 4 

tanO=-=- 

x 3 

x 3 

cotO=-=- 

Y 4 

r 5 

secO=-=- 

x 3 

r 5 

cscO=-=y 

4 

o 

2 

cscO=-= 1 

2 

8. (-4,0) 

r=4 

sinO=Q=O 

4 

-4 

cosO==-1 

4 

o 

tan=-=O 

-4 

-4 

cot 0 = Undefined 

6. (-4, -3) 4 

secO=-=-1 

x=-4,y=-3 -4 

4 

r = ~(-4)2 + (_3)2 

cscO= Undefined 

o 

= ..}16+9 

=../25 

=5 

. y -3 3 

smO=-=-=- 

r 5 5 

x -4 4 

cosO=-=-=- 

r 5 5 

y -3 3 

tanO=-=-=x 

-4 4 

x -4 4 

cotO=-=-=y 

-3 3 

455 

secO=-=-=- 

x -4 4 

r 5 5 

cscO=-=-=- 

y -3 3 

9. (1. .J3) 

o 

x=l,y=.J3 

r = ~12 +(./3)2 

=.J1+3 

=../4 

=2 

sinO = .J3 

2 

1 

cosO= 2 

22



12. (-5.1021,7.6132) 

tan 8 = ..[3 =..[3 

1 

r "" .J83.99 "" 9.1647 

1 1..[3..[3 

cot8=-=-·-=sin8 

= 7.6132 "" .83071 

..[3 ..[3..[3 3 

9.1647 

2 cos8 = -5.1021 "" -.55671 

sec8=-=2 9.1647 

1 

2 2..[3 2..[3 tan 8 = 7.6132 '" -1.4922 

csc8 = - = _ .- =-- -5.1021 

..[3 ..[3..[3 3 

cot 8 = -5.1021 '" -.67016 

7.6132 

10. -2..[3, - 2) 

sec 8 = 9.1647 '" -1.7963 

r = ~r-(--2..[3-3--::)2-+-(_-2)-2 

-5.1021 

9.1647 

esc 8 = --"" 1.2038 

=.J12+4 7.6132 

=.J16 

=4 15. Since the cot 8 is undefined, and cot 8 = ~, 

-2 1 y 

sin8=-=- 

4 2 where (x, y) is a point on the terminal side of 8 , 

y = 0 and x can be any nonzero number. 

-2..[3 ..[3 

cos8=--=- 

4 2 tan 8 = I., which is 0 divided by a nonzero 

-2 1 ..[3 

x 

tan8 = -2..[3 =13 = 3" 

number. Therefore, tan 8 = O. 

-2..[3 r:; 16. Since the terminal side of angleft is in quadrant 

cot8=--=v3 

III and (x, y) is a point on that terminal side, both 

. -2 

4 2 2..[3 x and y must be negative. Since r = ~x2 +i . r 

sec 8 = -2..[3 =-13=--3 is positive . 

4 

csc8=-=-2 

sinf3 = I. 

-2 r 

Since y is negative and r is positive, sinft is 

11. (8.7691, -3.2473) negative. 

x = 8.7691, y = -3.2473 

x 

cosf3 = r 

r = ~(8.7691)2 + (-3.2473)2 

Since x is negative and is positive cosft is 

= .J87.442072 negative. 

=9.3510466 

tanf3 = I. 

sin 8 = I. "" -.34727 

r 

cos8 = ~ "" .93777 

r 

tan8 = I. "" -.37031 

x 

cot 8 = ~ '" -2.7004 

y 

sec 8 =!.. "" 1.0664 

x 

csc8 =!.. "" -2.8796 

y 

x 

Both y and x are negative, so tanft is positive 

cotf3 =- x 

Y 

Both x and y are negative, so cotft is positive. 

r 

secf3 = 

x 

Since r is positive and x is negative, secft is 

negative. 

r 

cscf3 = y 

Since r is positive and y is negative, cscft is 

negative. 

23



In Exercises 17-24, r = ~ x 2 + Y 2 , W hi IC h i IS posinve. . . 

. ., y. . , 

17. In quadrant II, y IS positive so - IS posrtive. 

r 

18. In quadrant II, x is negative so .:::. is negative. 

r 

26. Since x ~ 0, the graph of the line 3x + 5y = ais 

shown to the right of the y-axis , A point on this 

graph is (5, -3) since 3(5) + 5(-3) = O. 

r=~52+(_3)2 

y 

=.../25+9=.J34 

3x + 5y = O. x



28. Since x ::; O. the graph of the line -6x - y = 0 is 30. Since x ~ 0, the graph of the line 6x - 5y = 0 is 

shown to the left of the y-axis, A point on this 

shown to the right of the y-axis , A point on this 

graph is (-1, 6) since -6(-1) - 6 = O. line is (5, 6) since 6(5) - 5(6) = O. 

r = ~(_1)2 +6 2 =.Jl + 36 =.J37 

r=~52+62 =.J25+36=.J6l 

y 

y 

6 

5 

6x-5y =O.x~ 0 

. y 6 6.J37 . y 6 6.J6l 

sm6=-=--=- 

r ~ 37 r .J6l 61 

x -1 .../37 x 5 5.J6l 

cos6=-=--=- 

r .J6l 61 

sm6=-=--=-- 

cos6 = -;:= 'J37 =-37 

y 6 y 6 

tan6=-=-=-6 

tan6=-=x 

-1 x 5 

x -1 1 x 5 

cot6=-=-=-- 

cot6=-=y 

6 6 y 6 

r.J6l 

sec6 = !. = .../37 =-.../37 

sec6=-=- 

s -1 x 5 

r .../37 r .J6l 

csc6=-=-- 

cscf}=-=- 

y 6 y 6 

29. Since x ::; 0, the graph of the line -5x - 3y = 0 is 33. cos 90° + 3 sin 270° = 0 + 3(-1) =-3 

shown to the left of the y-axis. A point on this line 

is (-3, 5) since -5(-3) - 3(5) = O. 

r=~(-3)2+52 =.J9+25 =../34 35. 3 sec 180° - 5 tan 360° = 3(-1) - 5(0) =-3 

(-3.5)' 

5 

36. 4 esc 270° + 3cos 180° 

=4(-1)+3(-1) 

=-7 

-3 37. tan3600+4sinI800+5cos 2180° 

. = 0 + 4(0) + 5(-1)2 

sin6 = 5.J34 38. 2 sec 0° + 4 cot 2 90° + cos 360° 

34 

=2(1) + 4(0)2 + 1 

3../34 

=5 

cos6=--- =3 

34 

5 

tan6=- 

3 

3 

cot6 =-- 40. sin 2 360° + cos 2 360° =0 2 + 1 2 

5 =1 

..J34 

sec6=-- 

3 

..J34 

csc6= 

5 

41. sec21800-3sin23600+2cosI80° 

=(_1)2 _3(0)2 +2(-1) 

=1-0-2 

=-1 

25



42. 5sin 2 90° + 2 cos 2 270° -7 tan 2 360° 

== 5(1)2+ 2(0)2 - 7(0)2 

==5 

43. sin[n · 180°] 

This angle is a quadrantal angle whose terminal 

side lies on either the positive part of the x-axis or 

the negati ve part of the x-axis, Any point on these 

terminal sides would have the form (k, 0), where k 

is any real number k :;t. o. 

sin[n · 180°] ==1. == 0 == 0 

r ~k2 +0 2 

44. cos [(2n + 1) .90°] 


side lies on either the positive part of the y-axis or 

the negative part of the y-axis. Any point on these 

terminal sides would have the form (0, k), where k 

is any real number, k:;t. O. 

x 

cos[(2n + 1).90°] = 

r 

o 

45. tan[(2n + 1) . 90°] 


side lies on either the positive part of the y-axis or 

the negati ve part of the y-axis . 

Any point on these terminal sides would have the 

form (0, k), where k is any real number, k:;t. O. 

tan[(2n + 1).90°] =1. =!5.. 

x 0 

tan[(2n + I) . 90°] is undefined. 

46. tan [n . 180°] 

The angle is a quadrantal angle whose terminal 

side lies on either the positive part of the x-axis or 

the negative part of the x-axis, Any point on these 

terminal sides would have the form (k, 0), where k 

is any real number, k:;t. o. 

y 0 

tann·180 = - = - =0 

k k 

47. Using a calculator, sin 15° = .258819045 and 

cos 75° == .258819045. 

We can conjecture that the sines and cosines of 

complementary angles are equal. Try another pair 

of complementary angles. 

sin 30° = cos 60° 

.5=.5 

Therefore, our conjecture is true. 

48. Using a calculator, tan 25° = .466307658 and 

cot 65° = .466307658. 

We can conjecture that the tangent and cotangent 

of complementary angles are equal. Try another 

pair of complementary angles. 

tan 45° = .cot 45° 

1=1 

Therefore, our conjecture is true. 

49. Using a calculator, sin 10.° = .173648178 and 

sin(-100) =- .173648178. 

We can conjecture that the sines of an angle and 

its negative are opposites of each other. Using a 

circle, an angle ()having the point (x, y) on its 

terminal side has acorresponding angle -() with a 

point (x, -y) on its terminal side. From the 

definition of sine, sin(-(}) = _1. and sin () =1.. 

r 

r 

The sines are negatives of each other. 

50. Using a calculator, cos 20° =.93969262 and 

cos (-20°) =.93969262. 

We can conjecture that the cosines of an angle 

and its negative are equal. Using a circle, and 

angle ()having the point (x, y) on its terminal side 

has a corresponding angle -() with point (x, -y) 

on its terminal side. From the definition of cosine, 

x 

x 

cos(-(}) = - and cos () = -. 

r 

r 

The cosines are equal. 

53. When sin ()== cos (), then 1. = ~ . As a result, the 

r r 

two values of ()must lie along the line y =x, 

which occurs when ()= 45° and ()= 225°. 

54. When tan () = 1, we have 1.. = 1or y = x. The 

x 

two values of () must lie along the line y = x, so 

() = 45° and () = 225°. 

55. Since cos ()== .8 occurs when x = .8, r =1 and 

() == 36.87° , cos ()= -.8 occurs when x = -.8 and 

r= 1. As a result, ()= (180 + 36.87) = 216.87° 

and () == (180 - 36.87) =143.13°. 

56. Since sin ()= .8 occurs when y = .8, r = 1 and 

() == 53 .13, sin () = -.8 occurs when y = -.8 and 

r= 1. As a result () == (180 + 53,13) = 233.13° 

and 8 = (360 - 53.13) =306.87° 

57. cos 20° is about .940, and sin 20° is about .342. 

58. Use the TRACE feature to move around the circle 

in quadrant I. 

cos 40° == .766 so T == 40° . 

26





sin 35° == .574 so T =35° 



cos 45° = sin 45° so T= 45°. 

61. As T increases from 0° to 90°, the cosine 

decreases and the sine increases. 

62. As T increases from 90° to 180°, the cosine 

decreases and the sine decreases. 

63. (a) From the figure in the text and the definition 

of tan (J, we can see that tan (J = I . 

x 

(b) Solve for x. 

tan(J=I 

x 

x tan (J = y 

x=_Ytan(J 

64. The area of a triangle is .!.b .h. The area of one of 

2 

the six equilateral triangles is 

.!.(x{.J3 x)=(.!.x2X.J3) .Sin(J= ¥x = .J3. 

2 2 2 2 x 2 

Section 1.5 

Exercises 

Substituting sin (J for .J3 in the area of the 

2 

triangle gives .!.x 2 sin (J. The area of the hexagon 

2 

is ,~x2 sin(J) = 3x 2 sin(J. 

1. 1 is its own reciprocal. 

sin (J= esc (J= 1 

sin 90° = esc 90° = I 

Therefore, (J = 90°. 

2. -1 is its own reciprocal. 

cos (J = sec (J =-1 

cos 180° = sec 180° =-1 

3. esc (J= 3 

sin (J = _1_ = .!. 

csc(J 3 

1 

5. tanf3=- 

5 

1 

cot f3 =- f3 

tan 

1 

= -t 

=-5 

. 1 1 ~ 

6. sma=--=--=- 

csca ~ 15 

7. secf3=_I_= ~ =~ 

cosf3 -../7 

8. cot (J = _ .J5 

3 

1 

tan(J=- 

cot(J 

1 

=-4 

3 .J5 

=-~. .J5 

3.J5 

=-- 

5 

9. esc (J= 1.42716321 

. (J 1 

SID =- 

csc(J 

1 

=--- 

1.42716327 

=.70069071 

10. cosa = _1_ = 1 = .1019% 57 

seca 9.80425133 

14. Since tan (J =_1_, multiply both sides by cot (J . 

cot (J 

tan (J cot (J = 1 

Now divide both sides by tan (J . 

1 

cot(J=- 

tan(J 

15. cot Y= 2 

1 1 

tany=--=cot 

y 2 

1 1 

16. tan¢' = --=- 

cot¢' 3 

1 1 

4. cosa=--=--=-.4 

seca -2.5 

27



.J3 

17. cotm= 

3 

1 

tanm=- 

cot to 

1 

=../3 

T 

3 

=.J3 

=.J3 

18. tan 9 = _1_ = 1 = 12.,[6 = 12.,[6 =2.,[6 

cot 9 if. .,[6 ..J6 6 

19. cot a = -.01 

1 

tana=- 

cota 

1 

=- 

- .01 

=-100 

1 1 

20. tan/3=--=- =2.5 

cot /3 .4 

1 

21. tan(3B- 4°) =--- 

cot(5B-8°) 

tan(3B - 4°) = tan(5B - 8°) 

3B-4° =5B-8° 

4°=2B 

B=2° 

1 

22. cos(6A + 5°) =---- 

sec (4A + 15°) 

= cos(4A + 15°) 

6A + 5° = 4A + 15° 

2A =10° 

A=5° 

23. sec(2a + 6°)cos(5a + 3°) =1 

cos(5a + 3°) =--- 

sec(2a+6°) 

cos(5a + 3°) = cos(2a + 6°) 

5a+3°=2a+6° 

3a=3° 

a=lo 

24. sin(49+ 2°) esc (39 + 5°) = 1 

sin(49+2 0 ) = 1 

esc (39 + 5°) 

sin (49+ 2°) =sin(39 + 5°) 

49+ 2° = 39+ 5° 

9=3° 

I 

27. sin a> 0 implies a as in quadrant I or II. cos a < 

oimplies a is in quadrant II or III. a is in 

quadrant II. 

28. cosjJ> 0 impliesjJ is in quadrant I or IV. 

tanjJ> 0 impliesjJ is in quadrant I or III. 

jJ is in quadrant I. 

29. tan r> 0 implies ris in quadrant I or III. cot r> 0 

implies ris in quadrant I or III. ris in quadrant I 

or Ill. 

30. tan w < 0 implies w is in quadrant II or IV. cot w 

< 0 implies w is in quadrant II or IV. w is in 

quadrant II or IV. 

31. cos /3< 0 implies /3is in quadrant II or III. 

32. tan 9 > 0 implies 9 is in quadrant I or III. 

In Exercises 3~, the signs + for positive and - for 

negative are given in the following order: sine, cosine, 

and tangent. 

33. 74° is in quadrant I. 

+;+;+ 

34. 129° is in quadrant II. 

+;-; 

35. 183° is in quadrant ill. 

-;-;+ 

36. 298°, is in quadrant IV. 

-;+; 


+;+;+ 


+;+;+ 

39. -82° is in quadrant IV. 

-;+; 

40. -121° is in quadrant ill. 

-;-;+ 

0 sin 30° 

41. tan 30 =-- 

cos 30° 

cos 30° < I, so sin 30° + cos 30° > sin 30° . . 

Therefore, tan 30° is greater. 

42. sin 21° > sin 20° 

sin ()increases as ()increases from 0° to 90°, so 

sin 21° is greater. 

28



43. sec 33° = __1_; since 

cos 33° 

1 

cos33° < 1, ---> cos 33°. 

cos33° 

sin 33° < cos 33° < sec 33° . 

Therefore, sec 33° is greater . 

44. cos 5° < 1 so (cos 5°)2 = cos 2 5° < cos5° 

cos 5° is greater. 

45. cos 26° > cos 27° 

cos ()decreases as ()goes from 0° to 90°, so 

cos 26° is greater. 

46. cot 2° > cos 2° 

cot 2 

0 

=-- 

cos 2° S' 

10cesm 

. 20 

 

1 1 1 

. 

sin 2° sin 2° 

cos2° 20 

-->cos 

Multiplying by cos 2°, sin2° ,so 

cot 2° is greater. 

47. -I:!> sin ()$ 1 for all ()and 2 is not in this range. 

sin ()= 2 is impossible. 

48. cos a = -1.001 is impossible since -1 :!> cos a $ 1. 

49. tan /3can take on all values. 

tan /3= .92 is possible. 

50. cot w = -12.1 is possible since cot w can be any 

real number. 

51. sec a = 1 is possible since sec a $ -lor sec a ~ 1. 

52. tan ()can take on all values. tan ()= 1 is possible. 

53. sin /3+ 1 = .6 means sin /3= -.4, which is possible 

because -1 $ sin /3:!> 1. sin /3+ 1 =.6 is possible. 

54. sec w + 1 = 1.3 gives sec w = .3 which is 

impossible since sec w:!> -1 and sec w ~ 1. 

57. Find tan a if sec a = 3, with a in quadrant IV. 

Start with the identity tan 2 a + 1 = sec 

2 

a, and 

replace sec a with 3. 

tan 2 a +1 = 3 2 

2 

tan a = 8 

tan a = ±..J8 = ±2../2 

Since a is in quadrant IV, tan a < 0, so 

tan a = -2../2. 

58. Find sin a if cos a = _.!. , with a in quadrant II. 

4 

Start with the identity sin 2 a + cos 2 a = 1, and 

replace cos a with _.!.. 

4 

2a+(-±r 

sin =1 

. 2 1 1 

sm a+-= 

16 

. 2 15 

S10 

a= 16 

sina = ± (l5 = ±.Jl5 

fi"6 4 

Since a is in quadrant II, sin a > 0, so 

. .Jl5 

s1Oa=--. 

4 

59. Find esc /3if cot /3 = _.!. and /3is in quadrant IV. 

2 

2 2 

1+ cot /3= csc /3 

csc 2 /3 = 1+ (_.!.)2 = 1+.!. = ~ 

244 

csc/3=± V4 

fI =±4=±-J5 ,,4 2 

/3is in quadrant IV so esc /3 = _ -J5 . 

2 

55. sin a = .!. is possible because -1 $ sin a $ 1. 

2 

Furthermore, when sin a = .!., 

2 

1 1 

csca=--=-=2. 

sina 

t 

sin a = .!. and esc a = 2 is possible. 

2 

56. tan f3 = 2 is possible. but when tan f3 = 2, 

1 1 

cot /3 =--= -. 

tan/3 2 

tan f3 = 2 and cot f3 = -2 is impossible. 

29



60. Find sec oif tan 0 =..fi and eis in quadrant m. 

3 

tan 2 0 + 1 = sec 2 0 

2 (..fi)2 

sec 0= 3 +1 

7 9 

=-+ 

9 9 

16 

= 

9 

(l6 

sec 0 = ±V9 

=+i 

-3 

()is in quadrant III so sec 0 = _i. 

3 

61. Find cos f3, if esc P= -4 with pin quadrant Ill. 

Start with the identity sin f3 =.-.!..-p. 

csc 

sin p = _1_=_.!. 

-4 4 

Now use the identity sin 2 f3 + cos 2 P=1 and 

replace sin f3 with _.!. . 

4 

(-±Y+COS 2 P = 1 

Smce 

..!..-+cos 2 f3 =1 

16 

cos 2 f3 =~ 

16 

cos f3 =± [l5 =± .Jl5 

'116 4 

· f3"IS 10 quadrant III, cos f3 =--- .Jl5 . 

62. Find sin (), if sec ()::: 2, with 0 in quadrant IV. 

Since sec ()=2 and cos 0 =_1_, cos 0 :::.!. . 

sec 0 2 

Since sin 2 ()+ cos 2 ()::: I, and cos 0 ::: .!. . 

2 

sin 2 o+(±Y ::: 1 

sin 2 O+.!.::: 1 

4 

. 2 Il 3 

SIO 17 = 

4 

sinO = ± IT =± .,fj 

'14 2 

4 

Since ()is in quadrant IV, and sin ()is negative in 

quadrant IV, sin 0 = - .,fj . 

2 

63. Find cot a if esc a = -3.5891420, with a in 

quadrant III. 

1+ cot 2 a ::: csc 2 a 

1+ cot 2 a::: (-3.5891420)2 

l+cot 2 a::: 12.8819403 

cot 2 a::: 11.8819403 

cota::: ±3.44701905 

Since a is in quadrant Ill, cot a> 0, so 

cot a= 3.44701905. 

64. Find tan f3, if sin f3::: .49268329, with f3 in 

quadrant II. 

sin 2 f3 + cos 2 P=1 

cos 2 p::: I-sin 2 p 

cos p::: ±~1- sin 2 p 

cos f3 is in quadrant II, and the cosine is negative 

in quadrant II, 

cos f3 = _~rl-_-s-in-:2:-f3- 

=-Jl-(.49268329)2 

= -./1-.24273682 

=-.J.75726318 

= -.87020870 

sin 0 

tan f3 =- 

cosO 

.49268329 

= - .87020870 

= - .56616682 

65. sinX= .8 

sin 2 X +cos 2 X = 1 

cos 2 X = I-sin 2 X 

cos 2 X :::1-(.8)2 

cos 2 X = (cos X)2 = .36 

66. tan X :::2 

tan 2 X + 1::: sec 2 X 

tan 2 X + 1 = (__1 

cos 2-) X 

(2)2 + 1 = (_1_)2 

cosX 

1 

5- (- 

cos X 

)2 

30



67. sin 2 9+cos 2 8 = 1 3 

70. cosa=- 

(.8)2+(-.6)2:b 1 

5 

.64+.36:b 1 

x 

cosa=- soletx=-3,,=5. 

1=1 True r 

x2 + y2 =,2 

(-3)2+1 =5 2 

Yes, there is an angle 9 for which cos 9 = -.6 and 

sin 9= .8. 

68. sin 2 8+ cos 2 8 = 1 

(iY +(~r ~1 

9 4 

16 9 

9+1 =25 

l =25-9 

=16 

y =±.J16 

-+-~1 y=±4 

a is in quadrant ill so y = -4. 

145 :# 1 . 4 

144 

sma=- 

5 

No, there is no angle for which cos 9 = ~ and 3 

3 cosa=- 

5 

• Ll 3 

sm e =-. 

-4 4 

4 tana=-= 

-3 3 

3 

For Exercises 69-76, remember that' is always positive. cota= 4 

15 5 

69. tana=- 

seea=- 

8 

3 

5 

tan a = Z. and a is in quadrant II, so let y = 15 

x 

andx=-8. 

x2+1 =,2 71. tan fJ = .J3 

csca=- 

4 

(_8)2 + 15 2 =,2 tan fJ = Z. and fJ is in quadrant ill, so let y = -./3 

64+225 =,2 

x 

and x=-l. 

289 =,2 x2 + y2 =,2 

17= , 

. y 15 

(_1)2 + (-./3)2 =,2 

sma=-= 

r 17 1+ 3=,2 

x -8 8 

cosa=-=-=- 4 =,2 

, 17 17 

2=, 

y 15 15 

tana=-=-=-- . y -./3 .J3 

x -8 8 smfJ=-=--=- 

r 2 2 

x -8 8 

cota=-=-=-- x -1 1 

y 15 15 

cosf3=-=-=- 

,22 

, 17 17 

seea=-=-=- 

x -8 8 tan f3 = Z. = -./3 =.J3 

x -1 

r 17 

csca=-=- x -1 I .J3 

y 15 cot f3 = Y= -./3 = .J3 = 3 

, 2 

seefJ = - = - =-2 

x -1 

r 2 2.J3 

esc f3 = Y= -./3 = - -3 

31



2 r 

72. esc f) = 2 = - = - , so let y = 1, r = 2. 

1 Y 

x 2 + y2 =r2 

x 2 +1 2 =2 2 

x 2 +1 =4 

x 2 =4-1 =3 

x = ±.J3 

(J is in quadrant II, so x = -J3 . 

. f) 1 

sm = 

2 

.J3 

cosf) =- 

2 

1 .J3 

tanf)=~=-3 

cotf) = - .J3 =-J3 

1 

2 2.J3 

secf)=--::)3 =--3 

2 

cscf)=-=2 

1 

73 sin R = ../5, with tan R> 0 

• fJ 7 fJ 

Since sin {j is positive, {j is in quadrant I or II. 

Since tan {j> 0, {j is in quadrant I or Ill. 

{j is in quadrant I. 

Since sin{j = 1., let y =../5 and r= 7. 

r 

x 2 + y2 = r2 

x 2 +(../5)2 = 7 2 

x 2 +5 =49 

x 2 =44 

x = ±.J44 = ±2.JU 

Since fJ is in quadrant I, x = 2..,fli. 

. y../5 

sm/3=-=r 

7 

x 2.JU 

cos{j=-=- 

r 7 

y ../5 .J55 

tan{j=-=--=x 

2.JU 22 

cot/3 = ~ = 2.JU = 2.../55 

y ../5 5 

r 7 7.JU 

sec/3=;=2Jii=~ 

r 7 7../5 

csc{j=y= ../5 =-5 

.J3 x 

74. cota=-= 

8 y 

sin a> 0 means y > 0, so let x =.J3 y = 8. 

x 2 + y2 =r2 

(-J3Y +8 2 = r 2 

2 

3+64 =r 

r 2 =67 

r =.J67 

8 8.J67 

sin a = .J67 =(jJ 

.J3 .J201 

cosa=-=- 

.J67 67 

8 8.J3 

tan a = .J3 = -3 

.J3 

cota= 

8 

.J67 .J201 

seca= .J3 =-3 

.J67 

csca=- 

8 

75. cot 0 = -1.49586 with 0 in quadrant IV. 

Since cot 0 =~ and 0 is in quadrant IV, let 

y . 

x = 1.49586 and y =-1. 

x 2 +i =r 2 

(1.49586)2 +(_1)2 = r 2 

2 

2.23760 + 1= r 

2 

3.23760 = r 

1.79933= r 

sinO = 1.= -1 = -.555762 

r 1.79933 

o= ~ = 1.49586 _ .831343 

cos r 1.79933 

tan 0 = 1. = -1 = - .668512 

x 1.49586 

_ 

cotO--- x _ 1.49586 

-. _ -1 49586 

y -1 

0 = !-. = 1.79933 = 1.20287 

sec x 1.49586 

cscO =!-. = 1.79933 =-1.79933 

y -1 

32


Chapter 1 Review Exercises 33 

76. sina = .164215 = .164215 

1 

. Y 

sma=r 

so let y = .164215, = 1. 

x 2 + y2 =r2 

x 2 +(.164215)2 = 1 2 

x 2 +.026966 = 1 

2 

x = 1-.026966 

=.973034 

x =±..J.973034 

=±.986425 

a is in quadrant Il, so, x = - .986425. 

sina = .164215 

cos a =-.986425 

tan a =-.166475 

cota = 

.986425 = -6.00691 

.164215 

seca = 1 =-1.01376 

-.986425 

csca = 1 = 6.08958 

.164215 

79. The statement is false. 

sin 30°+ cos30° 1, 1 

.5+ .8660 1, I 

l.3660¢ 1 

Since 1.3660¢ 1, the given statement is false. 

SO. The statement is false. 

For any angle (J, -1 :s; sin (J:S; 1. 

Since sin (J=2, the given statement is false. 

81. Let h =the height of the tree. 

50 ft 

oos7oo=- 

h 

hcos70° =50 

50 50 

h=--=--= 146 feet 

cos 70° 0.3420 

82. (8) r 2 = 52 + 12 2 

r =..J25+ 144 

=..J169 

=13prism diopters 

(b) tan 0 =2 

12 

84. 90° < 0 < 180°;180°< 20 < 360°; 

so sin 20 is negative. 

85. 90° < 0 < 180°'45°



8. Let A = (-2, -2), B = (8, 4) and C = (2, 14). 

Distance from A to B is 

~[8 - (_2)]2 + [4 - (_2)]2 = .Jl00+ 36 

Distance from A to Cis 

= .J136 

=2.../34. 

~(-2-2)2+(-2-14)2=.J16+256 

= .J272 

=4.Jfi. 

Distance from B to Cis 

~(8-2)2 +(4-14)2 =.J36+100 

= .J136 

=2.../34. 

Does (2.../34)2 + (2.../34f = (4.Jfi)2 

Yes, since 136 + 136 = 272. 

Therefore. these points are vertices of a right 

triangle. 

10. f(x)=-x 2+3x+2 

f(-2) = -{_2)2 +3(-2)+2 

=-4-6+2 

=-8 

11. f(x)=-x 2+3x+2 

f(x+ 1)= -{x + 1)2 +3(x+l)+2 

= -{x 2 +2x+ 1)+3x +3 +2 

= _x 2 -2x-l+3x+3+2 

=_x 2 +x+4 

12. (a) When X =1. Yl = 6. so f(l) =6. 

(b) When f(x) =II =-14. X =5 

(c) The graph of y = f(x) crosses the y-axis at 

the y-intercept (0. 6). 

(d) The graph of y = f(x) crosses the x-axis at 

the x-intercept (3. 0). 

13. y = 9x + 2 

Domain: x can take on any value so the domain is 

(-00,00). 

Range: y can take on any value, so the range is 

(-00,00). 

Each x-value gives one y-value so y =9x + 2 is a 

function. 

14. y =/xl 

Domain: x can take on any value. so the domain is 

(-00.00). 

Range: y can be any nonnegative number, so the 

range is 

(0,00). 

Each x-value gives one y-value so this relation is 

a function. 

15. y =.,J; 

Domain: x ~ O. so the domain is [0,00) . 

Range: y can take on any nonnegative value, so 

the range is [0, 00). 

This is a function. 

16. x + I =y2 

Domain: y2 is always ~ O. so x + 1 ~ 0 and x ~ -1 

The domain is [-1, 00). 

Range: y can take on any value. so the range is 

(-00,00). 

Some x-values give more than one y-value. For 

example, if x = 3, y = 2 or y = -2. 

So x + 1 = y2 is not a function. 

17. Domain: Since the values ofx are -5 S x S 5, the 

domain is [-5.5]. Range: Since the values ofyare 

-3 S y S 3. the range is [-3.3]. 

Since there are values of x that correspond to 

more than 1 y-value, this is not a function. . 

18. Domain: Since any 'value of x would have a point 

on the graph. the domain is (-00, 00). 

Range: Since any point on this graph has a 

y-value greater than or equal to 1. the range is 

[1. 00). 

Each x-value gives one y-value, so this is a 

function. 

19. -51° is coterminal with -51° + 360° =309°. 

20. -174° is coterminal with 360° + (-174°) = 186°. 

21. 792° is coterminal with 792° - 2(360°) =72°. 

22. Let n represent any integer. Any angle coterminal 

with 270° would be 270° + n . 360°. 

23. 320 rotations per minute 

=320(360°) per minute 

=115.200° per minute 

115.200° 

= per second 

60 

= 1920° per second 

= 1920 0 ( ~) per ~second 

2 

= 1280° per - second 

3 

34



24. 650 times per minute 

=- 650 =10. 83 times ' per secon d 

60 

In 2.4 seconds, a point rotates 

(2.4)(10.83) = 26 times. 

The point rotates 26· 360° = 9360°. 

25. 

47025'11" = 470 + 25° +...!..!.:.. 

60 3600 

= 47° + .417° + .003° 

=47.420° 

° 3 0 

26. 119°8'3"=119°+!..+- 

60 3600 

= 119.134° 

27. -61.5034° = -61° - (.5034)(60') 

= -61° - 30.204' 

= -61°30' -(.204)(60'') 

= -61°30'12" 

28. 275 .1005°=275°+(.1005)(60') 

= 275° + 6.03' 

= 275°6' + (.03)(60") 

= 275°6'2" 

29. The angles (9x + 4)° and (l2x - 14)° are vertical 

angles, which have equal measure. 

9x+4 =12x-4 

4 =3x-14 

18=3x 

6=x 


30. The sum of the measures of the angles of a 

triangle is 180°. 

8x+6x+4x = 180 

18x = 180 

x=lO 

8x = 8(10) = 80 

6x =6(10) =60 

4x =4(10) =40 

The angles are 40°, 60°, and 80°. 

31. Since a line has 180°, the angle supplementary to 

/3 is 180° - /3. The sum of the angles of a triangle 

is 180°, so 

O+a+(l80° -,8) = 180° 

0= 180-a-1800+ /3 

O=-a+/3 

O=/3-a 

32. Assuming PQ and BA are parallel, 6.PCQ is 

similar to MCB since LPCQ = LACB. Then 

PQ PC 

-= 

BA AC 

1.25 mm 150 mm 

---=-- 

BA 30 Ian 

BA(l50 mm) = (30 1an)(1.25 mm) 

BA = (30 Ian)(1.25) 

150 

=0.25 Ian 

Hint for Exercises 33-38: In similar triangles, 

corresponding sides are in proportion and corresponding 


33. V=X=41° 

z= T= 32° 

The sum of the angles in a triangle is 180° so 

Y = U = 180° - (41° + 32°) = 107°. 

34. Q =N = 12° 

R = P = 82° 

M = S = 86° 

75 3 

35. -= 

50 2 

m 3 

-= 

30 2 

2m=90 

m=45 

n 3 

-= 

40 2 

2n = 120 

n=60 

36. Since the large triangle is equilateral, the small 

triangle is also equilateral. 

p=q=7 

6 7 

37. -=- 

r 11+7 

6 7 

-=r 

18 

7r = 108 

108 

r= 

7 

k 12+9 

38. -=- 

6 9 

k 21 

-= 

6 9 

. .9k = 126 

. k = 14 

35



39. Iftwo triangles are similar. their corresponding 42. x = 1, y =-J3 

sides are proportional and their corresponding 


r=~x2+l 

40. Let x =the shadow of the 30-ft tree. = ~12 + (_$)2 

30ft 

=.Jl+3 

=.J4 

=2 

. y -J3 $ 

smy=-=--=- 

r 2 · 2 

x 1 

cosy=-= 

8 ft r 2 

20 30 

-=- tan y =1..= -.J3 =_$ 

8 x x 1 

20x =240 x 1 $ 

x=12 coty=-=--=- 

y -.J3 3 

The shadow of the 30-ft tree would be 12 ft. 

r 2 

41. x = -3. y =-3 secy=-=-=2 

x 1 

r=~x2 +l 

r 2 2$ 

=~(_3)2+(_3)2 cscy =y=-$ =--3 

= ./9+9 

=..ff8 

43. sin 180° = 0 

cos 180° =-1 

=3-J2 tan 180° = 0 

. y 3 1 .J2 cot 180° is undefined. 

smO=-=---=--=- 

r 3-J2 .J2 2 sec 180° =-1 

3 1 .J2 

esc 180° is undefined. 

cosO = xr =--- =--=- 

3.fi .J2 2 44. 360° is coterminal with 0°. 

tan 0 =1..= -3 =1 

sin 360=0 

x -3 cos 360° = 1 

x -3 tan 360° = 0 

cotO=-=-=1 

y -3 

cot 360° is undefined. 

sec 360° = 1 

esc 360° is undefined. 

sec0 = !..= 3.fi =-.J2 

x -3 

esc 0 =!..=3-J2 =-J2 

y -3 

36


Chapter I Review Exercises 37 

45. (-8, 15), x = -8, y = 15 

r = ~(_8)2 + 15 2 

=.../64+225 

=.../289 

=17 

sinO=I=~ 

r 17 

x 8 

cosO=-=- 

r 17 

tanO=I=~=_~ 

x -8 8 

x 8 

cotO=-=- 

y 15 

r 17 17 

seeO=-=-=- 

x -8 8 

r 17 

cscO=-=y 

15 

46. (3, -4), x =3, y =-4 

r = ~32 + (-4)2 

=.../9+16 

=~ 

=5 

sinO=I= -4 =_i 

r 5 5 

x 3 

cosO=-=r 

5 

tanO=I= -4 =_i 

x 3 3 

x 3 3 

cotO=-=-=- 

y -4 4 

r 5 

seeO=-=x 

3 

r 5 5 

cscO=-=-=- 

y -4 4 

47. (1,-5), x = 1, y =-5 

r = ~(1)2 +(_5)2 

=~ 

. y -5 5../26 

SlnO=-=-=-- 

r ~ 26 

x 1 ../26 

cosO=-=--=- 

r ~ 26 

tanO=I= -5 =-5 

x 1 

xII 

cotO=-=-= - 

y -5 5 

seeO=!.= ~ =../26 

x 1 

r ~ ../26 

cscO=-=-=-- 

y -5 5 

48. (9,-2), x =9, y =-2 

r =~(9)2 + (_2)2 

=.../81+4 

=../85 

. y -2 2~ 

SlnO=;= ../85 =-85 

x 9 9~ 

cosO=-=-=- 

r ../85 85 

tanO=I= -2 =_~ 

x 9 9 

x 9 9 

cot 0 = = =- 

y -2 2 

r ../85 

seeO=-=x 

9 

r../85 ~ 

cscO=-=-=-- 

y -2 2 

37



49. (Wi,-6),x=6$,y=-6 

r = ~(6$)2 +(-6)2 

=..J108+36 

51. Since the terminal side of the angle is defined by 

5x - 3y = 0, x :2: 0, a point on this terminal side 

would be (3, 5) since 5(3) - 3(5) :=0. 

r= ~32 +5 2 =..J9 + 25 =-134 

. y 5 5-J34 

sm9=-=--=- 

=..J144 

r -134 34 

=12 

x 3 3-134 

sin 9 = I = - ~ = _.!. cos9 = -;:= -134 =34 

r 12 2 

x 6$ $ y 5 

cos9=-=--= 

r 12 2 x 3 

tan9=-= 

x 3 

tan9=I= __6_= __I_=_ $ 

cot9=-=y 

5 

x Wi ..[3 3 

r -134 

sec9=-=- 

cot9=~= Wi =-./3 x 3 

y -6 

r .J34 

r 12 2 2..[3 csc9=-=- 

sec9 = -:;= Wi = ..[3=-3- y 5 

r 12 52. If the terminal side of a quadrantal angle lies 

csc9=-=-=-2 

y -6 along the y-axis, a point on the terminal side 

would be of the form (0, k), where k is a real 

number, k*O. 

50. (-2Ji, 2Ji), x =-2Ji,y =2Ji 

r = ~(_2Ji)2 + (2Ji)2 

sin9:=I:= ~ 

r r 

cos 9 = -:= x -:= ° 0 

=..J8+8 r r 

:=...[16 

tan 9 = I :=~ undefined 

=4 x 0' 

. y 2Ji -Ii x 0 

sm 9 =- :=--=- cot9:=-= - = 0 

r 4 ,2 y k 

x -2Ji .J2 r r 

cos9=-=--=-- sec 9 = - = - undefined 

r 4 2 x 0' 

tan9=I= 2~ =-1 

x -2...,2 

x -2Ji 

cot9=y= 2Ji =-1 

sec9=!.-=_4_= -2 =-..J2 

53. 

esc 9 :=!.-:=!. 

y k 

The tangent and secant are undefined. 

(-I, 5) 

x -2Ji Ji y=-5x.x~O 

r 4 t: 

esc 9 = y:= 2Ji :=...,2 

y 

9 

54. Using the point (-1, 5), 

r:= ~(_1)2 + (5)2 :=..J1 + 25 =.../26. 

sin 9 :=5.../26 

26 

.../26 

cos9=-- 

26 

38



5 

55. tan 0= -=-5 

-1 

56. 

4 sec 180 0-2sin2 270 0=4(-1)-2(-1)2 

= -4-2(1) 

=-4-2 

=-6 

57. -cot 2 90 0 + 4sin 270 0 - 3 tan 180° 

= -(0)2 + 4(-1) - 3(0) 

=0-4-0 

=-4 

59. From the screen, cos X = .1. 

sin 2 X +cos 2 X = 1 

sin 2 X = 1-(.1)2 

sin 2 X = .99 

2 2 sin 2 X 

tan X=(tanX) =--2 

cos X 

(tan 

X)2 

=-=99 

.99 

.01 

60. sin 0 ;;;; ~ is possible because -1 s sin 6 s 1 for all 

4 

6. When sin 0 ;;;; ~ then esc = i because 

4 3 

1 

csc(J=--. 

sinO 

sin (J ;;;; ~ and esc 0 = i is possible. 

4 3 

61. For any angle 0, sec (J ~ -lor sec 0 ~ l. 

Therefore, sec 0 = - ~ is impossible . 

3 

1 

62. Ifcos 6 = .25, sec 0 = - = 4, so cos 6 = .25 and 

.25 

sec 6 = -4 is impossible. 

63. tan (J can take on all values. 

tan (J = 1.4 is possible. 

500 sin 50° 

64. tan =-- 

cos50° 

Since cos 50° < I, sin 500+ cos 50° > sin 500. 

Therefore tan50 0 > sin 500. 

65. In quadrant I as (J increases, sin (J increases . 

Therefore sin 44° > sin43°. 

66. Sin 

·0.J3· = -, Sin 0 = Y 

5 r 

x 2 + y2 = r2 

x 2+(.J3)2=52 

x 2 +3=25 

x 2 =25-3 

so y = .J3 and r = 5. 

x 2 =22 

x=±..fii 

cos 6 < 0 means x < 0 so x = -J22. 

sin 0 = 1.;;;; .J3 

r 5 

x ..fii 

cosO=-=-- 

r 5 

y .J3 .J66 

tanO=-=--=-- 

x -fi2 22 

x ..fii .J66 

cotO=-=---=-- 

y .J3 3 

r 5 s..fii 

secO=-=--=-- 

x -J22 22 

r 5 5.J3 

cscO=-=-=- 

y .J3 3 

67. cosy = _., with rin quadrant ill. 

8 

x 

cosr=-, sox=-5andr=8. 

r 

x 2 +i =r 2 

(_5)2 + i = (8)2 

25+ y2 =64 

i=39 

y=±.,[39 

Since r is in quadrant III, y = -.,[39. 

. y -.,[39 .J39 

slnr=-=--=-- 

r 8 8 

x 5 

cosr=-=- 

r 8 

y -.,[39 .J39 

tanr=-=--=- 

x -5 5 

x -5 5.J39 

cotr=-=--;;;;- 

y -.,[39 39 

r 8 8 

secr=-=-=- 

x -5 5 

r 8 8.,[39 

cscr=-=--=-- 

y -.,[39 39 

39



68. tan a = 2 = ~, tana = L with a in quadrant III, 

1 x 

70. sin0 = - ~, sin 0 = I with () in quadrant Ill, so 

5 , 

so let x = -1 and y = -2. let y = -2 and, = 5. 

x2 + y2 = ,2 , x2 + y2 =,2 

(-Ii + (_2)2 = ,2 x 2 +(_2)2 = 52 

1+ 4 = ,2, r = .J5 x 2 +4 =25 

. y 2 2.J5 

sma = -;:= - .J5 = --5 

x 1 .J5 

cosa=-=--=- 

r .J5 5 

y -2 

tana=-=-=2 

x -1 

x -1 1 

cota=-=-=y 

-2 2 

seca =!.. = .J5 =-15 

x -1 

r .J5 .J5 

csca=-=-=- 

y -2 2 

69. sec /3 = -.J5, with /3 in quadrant II. 

see/3 = !.., so ,=.J5 andx= -l. 

x 

x2 + y2 =,2 

(_1)2 + i =(.J5)2 

x 2 =21 

x =±.fij 

Since x is negative in quadrant III, x = -J2j. 

. y -2 2 

sm 0 =- = =- 

,5 5 

x -J2j .fij 

cosO = = -=-- 

r 5 5 

y - 2 2.fij 

tan0 = = -=- 

x -J2j 21 

x ~ .fij 

cot 0 = = -=- 

y -2 2 

r 5 5.fij 

sec8=-=--=-- 

x -J2j 21 

r 5 5 

csc 8 = = = - 

y -2 2 

71. seca =~, with a in quadrant IV. sec a =!.., so 

4 x 

,=5 andx= 4. 

l+i =5 x2 + y2 =,2 

y2 =4 4 2 +y2=5 2 

y=±2 

16+i =25 

Since f3 is in quadrant II, y = 2. 

. y 2 2.J5 

sm /3 = -;:= .J5 = -5 

x -1 .J5 

cos/3=-=-=- 

r.J5 5 

tan/3 = L = l:.... =-2 

x -1 

x -1 1 

cot/3=-=-=- 

y 2 2 

see /3 = !.. = .J5 =-15 

x -1 

,.J5 

csc/3=-= 

,., y 2 

i=9 

y=±3 

Since a is in quadrant IV, y =-3. 

. y -3 3 

sma=-=-=- 

,55 

x 4 

cosa=-= 

, 5 

y -3 3 

tana=-=-=- 

x 4 4 

x 4 4 

cot a = = = - 

y -3 3 

, 5 

seca=-=x 

4 

,55 

csca=-=-=- 

y -3 3 

40


Chapter 1 Test Exercises 41 

72. The sine is negative in quadrants III and IV. The 

cosine is positive in quadrants I and IV. Since 

sin e< 0 and cos e> 0, elies in quadrant IV. 

sinO . 

Since tan 0 =-- and SIn () < 0 and cos ()> 0, 

cosO 

the sign of tan ()is negative (a negative divided 

by a positive is a negative). 

(b) x 2 2: 0, so f(x) 2:-2 . The range is [-2,00). 

(c) f(-2) = .5(-2)2 -2 = 0 

360° 9 

74. ---= - degrees I yr 

26,000 yr 650 

9 

= - (3600) sec I yr 

650 

648 

=- sec/yr 

13 

'" 50 sec I yr 

The celestial north pole moves about 50" each 

year. 

75, Let x bethe depth of the crater Autolycus. Then , 

x 1.3 

--= 

11,000 1.5 

1.5x = 1.3(11,000) 

14,300 

x=-- 

1.5 

x ",9500 

The depth of the crater Autolycus is about 

9500 ft. 

76. Let x be the height of Bradley. Then, 

x 1.8 

--= 

21,000 2.8 

2.8x = 1.8(21,000) 

37,800 

2.8 

x = 13,500 

Bradley is 13,500 ft tall. 


1. d=~(5-2)2+(-2-(-5»2 

=~32 +3 2 

= ../9+9 

=.Jf8 

=3../2 

2. (6400km + 150krn)2 = d 2 + (64ookrn)2 

6550 2 -6400 2 =d 2 

1400'" d 

The distance is about 1400 krn. 

3. (a) x can be any value, so the domain is 

(-00,00) . 

rev 1min 7 5 rev 

6. 450-·--= . 

min 60sec sec 

7.5 rev = 7.5(360°) = 2700° 

The point on the edge of the tire moves 2700° in 

one second. 

7. The sum of the angles of a triangle is 180°. 

60°+ y+90 0 = 180° 

y = 180°-60°_90° 

y=30° 

The angle supplementary to 60° is 120° . 

x + 120°+30° = 180° 

x = 180° -120° - 30° 

x=30° 

8. Since we have two similar triangles, 

20 x 

-= 

30 roo-., 

20(100- x) = 30x 

2000 - 20x =30x 

50x=2000 

x=40 

x=--The lifeguard will enter the water 40 yards east of 

his original position. 

9. (2,-5), x = 2,y =-5 

r=~x2+i 

r=~22 +(_5)2 

r=..fi9 

y -5..fi9 5..fi9 

sinO=;= ..fi9 '129=-29" 

x 2 .J29 2..J29 

cos 0 =; = .J29'.J29 = 29" 

y -5 5 

tan 0 =-=-=- 

x 2 2 

10. If cos 0 < 0, then 0 is in quadrant II or III. If 

cot 0 > 0 , then 0 is in quadrant I or III. Therefore 

o terminates in quadrant III. 

41



4 x 

11. If cos0 = - = - , let x = 4 and r = 5. 

5 r 

2 

x 2 +i =r 

4 2 +i =5 2 

i =25-16 

y = ±../9 

y=±3 

Since 0 is in quadrant IV, y is negative, so 

y=-3. 

. y -3 3 

smO=-=-=- 

r 5 5 

y -3 3 

tanO=-=-=- 

x 4 4 

x 4 4 

cotO=-=-=- 

y -3 3 

r 5 

secO=-=x 

4 

r 5 5 

cscO=-=-=- 

y -3 3 

12. No; when sin 0 and cos 0 are both negative for 

sin O . . . 

th e same va I ue 0 f 0 , tan 0 =- IS posrtrve. 

cosO 

42



CHAPTER 2 ACUTE ANGLFS AND . side opposite 45 

8. smA= :=; 

RIGHT TRIANGLFS hypotenuse 53 

Section 2.1 

side adjacent 28 

cosA= = 

hypotenuse 53 

Exercises side opposite 45 

tan A = = 


For Exercises 1-6, refer to the Special Angles chart. 

cot A :=; side adjacent = 28 

1.,C"sin 30° = .!. 2 side opposite 45 

sec A = hypotenuse = 53 


hypotenuse 53 

csc A= = 

side opposite 45 

. side opposite n 

1 1 9. smA= = 

4. G" sec 60° = = - = 2 

hypotenuse p 

, cos 60° 1 

2 side adjacent m 

cosA= = 

hypotenuse p 

1 1 2 ..[3 2..[3 side opposite n 

5. E; esc 60° = sin 600 = ..{j = ..[3 = "J3 = -3- tan A = = 

side adjacent m 

2 

side adjacent m 

cot A = = 

side opposite n 

sec A = hypotenuse =.E.. 

side adjacent ' m 

hypotenuse p 

esc A = = 

7. sin A = side opposite = E.. side opposite n 

hypotenuse 29 

cos A = side adjacent = 20 

10. 

sin A = side opposite = ! 

hypotenuse 29 hypotenuse z 

side opposite 21 side adjacent y 

tan A = = cos A = = 

side adjacent 20 hypotenuse z 

side adjacent 20 side opposite k 

cot A = = tan A = = 


hypotenuse 29 side adjacent y 

secA= =- cot A = = 

side adjacent . 20 side opposite k 

hypotenuse 29 

csc A= 

hypotenuse z 

= secA= = 


hypotenuse z 

csc A = = 

side opposite k 

11. a =5, b =12 

c 2 =a 2 +b 2 

c 2 =5 2 +12 2 

c 2 =25+144 

2 

c = 169 

c =13 

43


44 Chapter 2 Acute Angles and Right Triangles 

. side opposite 12 . side opposite .Jl3 

Sin B= =- Sin B= =- 

hypotenuse 13 hypolenuse 7 

side adjacent 5 side adjacent 6 

cosB= =- cosB= = 

hypotenuse 13 hypotenuse 7 


tan B = =- side opposite .Jl3 


tanB= =- 



cotB= =- cot B = side adjacent = ~ = 6.Jl3 


side opposite .Jl3 13 

hypotenuse 13 

seeB= =- sec B 


= hypolenuse = 7.. 


hypotenuse 13 

cscB= = 


hypotenuse 7 7.Jl3 

cscB= =--=- 

side opposite .Jl3 13 

12. a =3, b =5 

2 14. b e T, c=12 

c = a 2 +b 2 

2 

2 c =a 2 +b 2 

+7 

c 2 =9+25 

c =32 +5 2 12 2 =a 2 2 

c 2 =34 

144=a 2 + 49 

c =..[34 a 2 =95 

. B side opposite 5 5..[34 

a=.J95 

sinB=£'=2.. 

Sin = hypotenuse = ..[34 = 34"" c 12 

B side adjacent 3 3.J34 a .J95 

cosB=-=- 

cos = hypotenuse =""J34 = 34"" c 12 

tan B = side opposite = ~ b 7 7../95 


tan B=-;; =.J95 =.gs 


cotB= =- a .J95 

side opposite 5 cotB=-=- 

b 7 

sec B = hypotenuse =.J34 

secB=~=~= 12../95 

side adjacent 3 a.J95 95 

esc B = hypotenuse =.J34 

c 12 

cscB=-=side 

opposite 5 b 7 

13. a=6, c=7 15. sin A = cos (90° - A) 

2 cos A = sin (90° - A) 

c =a 2 +b 2 tan A = cot (90° - A) 

7 2 =6 2 +b 2 cot A = tan (90° - A) 

49 =36+b 2 sec A = esc (90° - A) 

13=b 2 esc A = sec (90° - A) 

.Jl3 =b 

16. cot 73° = tan (90° -73°) 

= tan 17° 

17. see 39° = esc (90° - 39°) 

= csc51° 

18. cos (a + 20°) 

= sin [90°- (a + 20°)] 

= sin (70° - a) 

44



19. cot (/3_10°) 28. cot (59 + 2°) = tan (29 + 4°) 

= tan [90° - (/3 _10°)] Since cotangent and tangent are cofunctions, 

= tan (100° - /3) 

(59 + 2°) + (29 + 4°) = 90° 

79+6° = 90° 

20. tan 25.4° = cot (90° - 25.4°) 79 = 84° 

= cot 64.6° 9 = 12°. 

21. sin 38.7° = cos (90° - 38.7°) 29. tan 28° :5 tan 40° 

= cos51.3° tan 9 increases as 9 increases from 0° to 90°. 

Since 

22. A sin(900-A) cosA 40° > 28° 

--~-------+------- 

30° sin 60° = .8660 cos 30° = .8660 

tan 40° > tan 28°. 

Therefore, the given statement is true. 

45° sin 45° = .7071 cos 45° = .707 1 

60° sin 30° =.5 cos 60° =.5 30. sin 50° > sin 40° 

In the interval from 0° to 90°, as the angle 

-60° sin 150° = .5 cos(-67°) = .5 

increases, so does the sine of the angle, so 

120° sin(-300) = -.5 cos 120° = - .5 sin 50° > sin 40° is true. 

In each case, sin (90° -A) = cosA. 31. sin 46° < cos 46° 

sin 9 increases as 9 increases from 0° to 90° . 

23. tan a = cot (a + 10°) Since 

Since tangent and cotangent are cofunctions, this 46° > 44° 

equation is true if the sum of the angles is 90°. 

sin 46° > sin 44°. 

a+(a+100) =90° But sin 44° = cos 46°, 

2a+ 10° =90° so sin 46° > cos 46° . 

2a =80° 

False 

a=40° 

32. cos 28° < sin 28° 

24. cos 9 = sin 29 

cos 45° = sin (90° - 45°) 

Since sine and cosine are cofunctions, this 

= sin 45° 

equation is true if the sum of the angles is 90°, so 

In the interval from 0° to045°, the sine increases 

9+29 = 90° 

from 0 to sin 45° = .J2 , while the cosine 

39=90° 2 

9 = 30°. decreases from 1 to 

25. sin (2r + 10°) = cos (3r - 20°) cos45°=-. 

2 

Since sine and cosine are cofunctions, 

Therefore, cos 28° > sin 28°, and 

(2r + 10°) + (3r - 20°) = 90° cos 28° < sin 28° is false . 

5r-lO° =90° 

5r = 100° 33. tan 41° < cot 41° 

r=20°. 

tan 9 increases as 9 increases from 0° to 90°. 

Since 

26. sec (/3+ 10°) = esc (2/3+ 20°) 

49° > 41° 

tan 49° > tan 41 0. 

Since secant and cosecant are cofunctions, 

But 

(/3 + 10°) + (2/3 + 20°) = 90° 

.... tan 49° = cot 41°, 

3/3+ 30° = 90° so 

3/3=60° cot 41° > tan 41 0. 

/3= 20°. True 

27. tan (3B + 4°) =cot (5B - 10°) 

Since tangent and cotangent are cofunctions, 

(3B + 4°) + (5B _10°) = 90° 

8B-6°=90° 

8B = 96° 

B = 12°. 

.J2 

45



34. cot 30° < tan 40° hypotenuse 

tan (90° - 30°) =tan 60° =cot 30° csc 45 0 = side opposite 

In the interval from 0° to 90°, the tangent 

increases so .J2 

= 

tan 60° > tan 40°. Therefore, 

1 

cot 30° > tan 40°, so 

=.J2 

cot 30° < tan 40° is false. 

40. 

35. 

I~" 

~l 

~ ~ 

..J3 1 

0 side opposite sec 450 = hypotenuse 

tan 30 = side adjacent 

side adjacent 

1 .J2 

=.,[3 

= 

1 

.,[3 

=- 

=.J2 

3 

41. See figure in Exercise 35. 

36. . 600 side opposite 

sm 

= hypotenuse 

.,[3 

= 

2 

42. See the figure in Exercise 36. 

side adjacent 

cos 60 0 = hypotenuse 

side adjacent 

cot 30 0 = 1 

side opposite 

= 

2 

= 

.,[3 

1 

=.,[3 

37. See the figure in Exercise 35. 

. 300 side opposite 

sm = hypotenuse 

1 

= 

2 

38. See the figure in Exercise 36. 44. 

side adjacent 

cos 300 = hypotenuse 

.,[3 

= 

2 

43. 

39. 

LJ5. 

.J2 1 

45· 

1 

46



sin45°=! cos45° =~ 

4 4 

Y = 4sin45° x = 4 cos 45° 

=4 . ...[2 

...[2 

x=4' 

(.70710678, .70710678). This is the point 

2 2 

=2...[2 

=2...[2 

45. The legs of the right triangle provide the 

coordinates of P. 

P is (2...[2, 2...[2). 

46. Steps 1 & 2: 

y 

--1f----f........o...--t-- .. 

2 

51. Graph Y 1 = x and Y 2 = ~1- x in a window such 

as -2 s x s 2, -2 s y s 2. Use the intersection 

function to see that the point of intersection is 

(~,~} 

'," ' ,. 

'. 

'1',,' t , 

-2 :: :"'1~'" 

· : r ~~~ .... 

.': 

H=.70710&71 '=.70710&71 ' 

-2 

These coordinates are the sine and cosine of 45°. 

:; ~ ' , -. . . - .. 

IlIt4:rS4:ctitrl :' ''''' ~.'< ' 

52. y 

Step 3: sin 60 0=! 2 

cos 600=~ 

2 

y= 2 sin 60° x = 2cos 60° 

.J3 

1 

=2· =2· 

2 

2 

=1 

=.J3 

(I, {3) 

-----+lllI:-..r.....&-_....x 

Thus the coordinates of P are (1,.J3). 

47. sin 0° = 0 

sin 45° = .70711 

sin 90° = 1 

tanOo=O 

tan 45° = 1 

tan 90° is undefined. 

Y t is sin x and Y 2 is tan x. 

48. cos 0° = 1 csc O is undefined 

cos600=.5 csc600=1.1547 

cos 90° = 0 esc 90° = 1 

Y 1 is cos x and Y2 is esc x. 

The line passes through (0,0) and (1, .J3). The 

slope is .J3. The equation of the line is y = .J3x. 

53. y 

-----~.....;;.;.........I-\"....x 

49. sin 60° = .8660 for A between 0° and 90° . 

Therefore, a = 60°. 

The line passes through (0, 0) and (.J3, 1). The 

slope is .J3. The equation of the line is 

3 

.J3 

y=-x. 

3 

47



54. The slope of a line is the change in y over the 

c hanze ange i mx, or tan 6 . S' mce y=-x ../3 , 

3 

../3 . ../3 

m = - = tan 6. Since tan 30° = -, 

3 3 

the line y = ../3 x makes a 30° angle with the 

3 

positive x-axis. (See Exercise 53 .) 

55. The slope of a line is the change in y over the 

change in x, or tan 6 . Since y = ../3x, 

m = ../3 = tan 6 . Since tan 60° =../3, the 

line y = ../3x makes a 60° angle with the positive 

x-axis, (See Exercise 52.) 

1 

56. tan 6=- 

cot 6 

tan 6 = cot(90° - 6) 

Tangent and cotangent are both reciprocals and 

cofunctions. 

57. A 

(a) Each angle of every equilateral triangle has a 

measure of 60°. 

(b) The perpendicular bisects the opposite side, 

so the length of each side opposite each 30° 

angle is k. 

(c) Let x equal the length of the perpendicular 

and apply the Pythagorean theorem. 

x 2 + k 2 =(2k)2 

2+k2=4k 

x 

2 

x 2 =3k 2 

x =k.../3 

The length of the perpendicular is k../3. 

(d) In a 30° - 60° right triangle, the hypotenuse 

is always 2 times as long as the shorter leg, 

58. k 

and the longer leg has a length that is ../3 

times as long as that of the shorter leg. Also, 

the shorter leg is opposite the 30° angle, and 

the longer leg is opposite the 60° angle. 

{Zl 

k 

(a) The diagonal forms two isosceles right 

triangles. Each angle formed by a side of the 

square and the diagonal measures 45°. 

2 

(b) k 2 +k 2 = c 

2 

2k 2 =c 

~2k2 =c 

k..fi = c 

(c) 

The length of the diagonal is k..fi. 

In a 45° - 45° right triangle, the hypotenuse 

has a length that is..fi times as long as either 

leg. 

1 . 

59. a = -(24), so a = 12. 

2 

b = a../3, so b = 12../3. 

d = b, so d = 12../3. 

c = d..fi, so c = (12../3X:J2) = 12.J6. 

1 9 

60. y=-(9), soy=-. 

2 . 2 

x = y../3, so x = 2-.J3. 

2 

y =z../3, so 

y t 9../3 3../3 

z= ../3= ../3=-6-=2' 

w =2z , so W =2( 3~J = 3.../3. 

48



7 7../3 

61. 7 = m../3, so m = ../3 = -3-' 

a = 2m, so a = 2( 7~) = 14~ . 

14../3 

n=a, so n=--. 

3 

q= n.,fi, so q= ( 14~).J2 = 14~ . 

62. p = 15 

r = p.J2, so r = 15.J2. 

r 15.J2 f/ 

r=q../3, so q=13= ../3 =5"\16. 

t = 2q, so t = 2(5../6) = lcN6. 

63. Let h be the height of the equilateral triangle. By 

geometry, h bisects the base, s;and forms two 

30°-60° right triangles. 

s s 

:2 :2 

The formula for the area of a triangle is 

1 

Area = -bh. 

2 

In this triangle, b = s. The height h of the triangle 

is the side opposite the 60° angle in either 

30°-60° right triangle. The side opposite the 

30° angle is !... The height is..J3 . !... So the area 

2 2 

of the entire triangle is 

2 

1.s(../3 .!..) = s ../3 . 

2 2 4 

64. Area = 1. base' height 

2 

1 1 2 s2 

=_ ·s·s=-s or 

2 2 2 

65. Yes, the third angle can be fund by subtracting the 

given acute angle from 90°, and the remaining 

two sides can be found using a trigonometric 

function involving the known angle and side. 

68. (a) 55 mlhr = 80 ~ ftlsec and 

3 

30 miJhr = 44 ftlsec 

D = 1.05(\12 - V 2 

2 ) 

64.4(K 1 + K 2 + sin 9) 

Let 9 = 3.5°, K 1 =.4, and K 2 = .02. 

1.05(80.67 2 _ 44 2 ) 

= 64.4(.4 + .02 + sin 3.5°) 

== 155 ft 

(b) D = 1.05(\12 - V2 2 ) 

64.4(K 1 + K 2 + sin 9) 

Now let 9 = _2°. 

1.05(80.67 2 _ 44 2 ) 

= 64.4[.4 + .02 +sin(-2°)] 

'" 194 ft 

69. Using the values for K 1 and K 2 from Exercise 68, 

determine V 2 when D = 200 and 

\1 = 90 mph = 132 ftlsec. 

Section 2.2 

Exercises 

D= 1.05(\12 - Vl) 

64.4(K 1 +K 2 +sin9) 

2 2 

200 = 1.05(132 - V2 ) 

64.4(.4 + .02 + sine-3.5°» 

2 

200 =18,295.2 -1.05V 2 

23.12 

2 23.12(200)-18,295.2 

V 2 = -1.05 

V 2 "" 114 ftlsec 

V 2 ==78 mph 

1. C; 180 - 98 =82° 

(98° is in quadrant II) 

2. F; 212 - 180 = 32° 

(212° is in quadrant III) 

3. A; -135 + 360 =225° 

225 - 180 =45° 

(225° is in quadrant III) 

4. B; -60 + 360 =300° 

360 - 300 = 60° 

(300° is in quadrant IV) 

5. D; 750 - 2(360) =750 - 720 =30° 

(30° is in quadrant I) 

49



6. B; 480- 360 = 120° 

180 - 120 = 60° 

(120° is in quadrant II) 

11. 30°: 

{j 

tan 30° = 3 

cot 30° = {j 

12. 45°: 

13. 60°: 

sin 450 = ..fi 

2 

..fi 

cos 45°= 

2 

sec 45° =..fi 

esc 45° =..fi 

sin 60° = {j 

2 

{j 

cot 60°=_ 

3 

esc 600 = 2{j 

3 

14. 120°: 

1 

cos 120° = -cos 60° =- 2 

{j 

cot 120° = -cot 60°=- 

3 

sec 120° = -sec 60° =-2 

15. 135°: 

tan 135° = - tan 45° =-1 

cot 135° = -cot 45° =-1 

16. 150°: 

sin 150° = sin 30° = ~ 

2 

cotl500 = -cot 30° =-J3 

2{j 

sec 150° = -sec30° = -- 

3 

17. 210°: 

{j 

cos 210° = -cos300=- 

2 

2{j 

sec 210° = -sec30° = -- 

3 

18. 240°: 

tan 240° = tan 60° = {j 

cot 240° = cot 60° = {j 

3 

19. 225° 

Use the method illustrated in Example 2. 

y 

x=-I,y=-I, r=..fi 

. 50 -1 ..fi 

sm22 =-=- 

..fi 2 

-1 ..fi 

cos 225° = - =- 

..fi 2 

-1 

tan 225° = - = 1 

-1 

-1 

cot 225° =- = 1 

-1 

, . ..fi 

sec 225° = - = -J2 

-1 

esc 225° = ..fi =-..fi 

-I . 

20. 300° 


y 

The reference angle is 60°. Since 300° is in 

quadrant IV, the sine, tangent, cotangent, and 

cosecant are negative. 

sin 300° = - sin 60° = - {j 

2 

1 

cos 300° = cos 60° = 

2 

tan 300° = - tan 60° = -J3 

-J3 

cot 300° = -cot60°=- 

3 

sec 300° = sec 60° = 2 

2{j 

csc 300° = - csc 60° =-- 3 

50



21. 315° 


y 

23. 420° 


y 

315° 

60° 

The reference angle is 45° Since 315° is in 



sin315 0;;:-sin450 ;;: 

-Ii 

2 

~ 

cos315°;;: cos 45° ;;: 

2 

tan 315°;;: -tan 45° ;;:-1 

cot 315° ;;:- cot 45° ;;: -1 

sec315°;;: sec 45° ;;:.J2 

csc315°;;: -csc45° =-J2 

22. 405° 


y 


quadrant I, all quadrants are positive. 

sin 420° = sin 60° = .fj 

2 

1 

cos 420° = cos 60° = 

2 

tan 420° = tan 60° =.fj 

.fj 

cot 420° = cot 60° = 

3 

sec 420° = sec 60° = 2 

2.fj 

esc 420° = esc 60° ;;:- 

3 

24. 480° 


y 

480° 

405° 


quadrant I, all functions are positive. 

sin 405° ;;:sin 45° =~ 

2 

~ 

cos 405° = cos 45° = 

2 

tan 405° ;;:tan 45° =1 

cot 405° =cot 45° =1 

sec 405° = sec 45° =.J2 

csc405° = csc45° = ~ 


quadrant II, the cosine, tangent, cotangent, and 

secant are negative. 

sin 480° = sin 60° = .fj 

2 

1 

cos 480° = -cos 60° =- 

2 

tan 480° = - tan 60° = -.J3 

.fj 

cot 480° =-cot60° =- 

3 

sec 480° =-sec60° =-2 

2.fj 

esc 480° = esc 60° =- 

3 

51


S2 

Chapter 2 Acute Angles and Right Triangles 

2S. 495° 


y 


quadrant II, cosine, tangent. cotangent, and secant 

are negative. 

sin 495° = sin 450 = .J2 

2 

.J2 

cos 495° =-cos45°=- 

2 

tan 495° = - tan 45° = -1 

cot 495° =-cot 45° =-1 

sec 495° = - sec 45° = -,fi 

esc 495° =esc 45° = .J2 

26. 570° 


y 

570· 


quadrant III. the sine. cosine, secant, and cosecant 

are negative. 

sin 570° = - sin 30° =_.!. 2 

..fj 

cos 570° = -cos30° = -- . 

2 

tan 570° = tan 30° =..fj 

3 

cot 570° =cot 30° = ..fj 

sec 570° = - sec 300 = _ 2..fj 

3 

esc 570° = -csc30° =-2 

27. 750° is cotenninal with 

750° - 2(360°) = 750° - 720° 

=30°. 

sin 750° = sin 30° =.!. 

2 

..fj 

cos 750° = cos 30° = 

2 

..fj 

tan 750° = tan 30° = 

3 

cot 750° = cot 30° =..fj 

2..fj 

sec 750° =-sec 30°=- 

3 

esc 750° =esc 30° = 2 


1305° - 3(360°) = 1305° -1080° 

=225°. 


quadrant ill, the sine, cosine, secant, and cosecant 

are negative. 

sin 1305° = - sin 450 = _ .J2 

2 

.J2 

cos 1305° = -cos45° =- 2 

tan 1305° = tan 45° = 1 

cot 1305° = cot 45° =1 

sec 1305° = - sec 45° =-,fi 

esc 1305°= -csc45C>=-,fi 


1500° - 4(360°) =1500° -1440° 

=60°. 

sin 1500° = sin 60° = ..fj 

2 

1 

cos 1500° =cos 60° = 

2 

tan 1500° = tan 60° =..fj 

..fj 

cot 1500° =cot 60° = 3 

sec 1500° = sec 60° =2 

2..fj 

esc 1500° = esc 60° =- 3 

52




2670° -7(360°) =150°. 



secant are negative . 

sin 2670° =sin 30° =.!. 

2 

.J3 

cos 2670° =- cos 30° =- 

2 

.J3 

tan 2670° =- tan 30° =- 

3 

cot 2670° =- cot 30° =-fj 

2.J3 

sec 2670° =-sec 30° =-- 

3 

esc 2670° =esc 30° =2 

31. -390° is coterminal with 

":'390° - (-2)(360°) = 330°. 

The reference angle is 30°. Since -390° is in 



sin(-3900) =-sin 30° =_.!. 

2 

.J3 

cos(- 390°) = cos 30° = 

2 

.J3 

tan(-390°) =- tan 30° = - 

3 

cot(-390°) = - cot 30° =-.J3 

2.J3 

sec(-390°) = sec 30° =- 

3 

csc(-3900) =-csc30° =-2 

32. _510° is coterminal with 

-510° - (-2)(360°) = -510° + 720° 

=210°. 


quadrant III, the sine, cosine, and secant and 


sin(-510°) =- sin 30° =_.!. 

2 

.J3 

cos(-5100) =-cos 30° =- 

2 

.J3 

tan(-510°) = tan 30° = 

3 

cot(-510°) = cot 30° = .J3 

2.J3 

sec(-510°) =- sec 30° =-- 

3 

csc(-510°) = - csc 30° = -2 


-1020° - (-3)(360°) = -1020° + 1080° 

=60°. 

sin(-1020°) =sin 600 =.J3 

2 

1 

cos(-1020°) =cos 60° = 

2 

tan(-1020°) =tan 60° =.J3 

.J3 

cot(-1020°) = cot 60° = 

3 

sec(-1020°) =sec 60° = 2 

2.J3 

csc(-1020°) =esc 60° = - 

3 


-1290°-(-4)(360°) = -1290°+ 1440° 

=150°. 


quadrant II, the cosine, tangent. cotangent, and 


sin(-1290°) =sin 30° =.!. 

2 

.J3 

cos(- 1290°) = - cos 30° =- 

2 

tan(-1290°) = - tan 300 =_ .J3 

3 

cot(-1290°) = - cot 30° =-fj 

sec(-1290°) = - sec 300 = ~ 2.J3 

3 

csc(-1290°) = esc 30° = 2 

35. 225° is in quadrant III, so the reference angle is 

225°- 180° = 45°. 

cos 45° = ~ 

sin 45° = L 

r 

r 

x = r cos45° y = r sin 45° 

= 10 . ..fi =5..fi =10 . ..[i =5..J2 

2 2 

Since 225° is in quadrant 1lI. both the x and y 

coordinate will be negative . The coordinate of 

Pare: (-5..fi, -5..[i). 

36. 150° is in quadrant II, so the reference angle is 

180 -150 =30° 

cos 30° =~ sin 30° =L 

r 

r 

x = r cos 30° y = r sin 30° 

=6· ~ =3.J3 = 6 .±=3 

Since 150° is in quadrant n. x will be negative 

and y will be positive. The coordinates of Pare: 

(-3.J3, 3). 

53



37. For every angle O. sin 0 =L and cos 0 

, 

=~. 

, 

Thus. for every angle 0, 

(sin 0)2 + (cos 0)2 =(-;r +(:;f 

i+x 

= 2 

2 

r 

,2 

=2=1. 

r 

Since(.6)2 + (_.8)2 =.36 + .64 =1. 

there is an angle 0 for which cos 0 =.6 and 

sin 0= -.8. Since cos 0> 0 and sin 0< O. it is an 

angle in quadrant IV. 

38. For every angle O. sin 0 =L and cos 0 =~ . 

r 

r 

Thus for every angle O. 

(sin0)2 +(cos 0)2 =(-;r+ (-:;J 

i+x 2 

=~-;;-- 

,2 

,2 

=2 

r 

=1. 

Since 

( 

~) 2 +(3.)2 =~+i= 145 *1 

4 3 16 9 144 • 

there is no angle 0 for which cos 0 =3. 

3 

andsinO=~ . 

4 

39. If 0 is in the interval (90°. 180°). 

90° < 0< 180°. 

Dividing by 2 gives 45° < ~ < 90°. 

2 

Thus, ~ is a quadrant I angle and sin ~ is 

2 2 

positive. 

41. If 0 is in the interval (90°, 180°). 

90°< 0 < 180°. 

Adding 180° gives 270° < 0+ 180° < 360°. 

Thus. 0 + 180° is a quadrant IV angle and 

cot (0 + 180°) is negative. 

42. If 0 is in the interval (90°. 180°), 

90° < 0 < 180°. 

Adding 180° gives 270° < 0 + 180° < 360°. 

Thus. 0 + 180° is a quadrant IV angle and 

sec (0 + 180°) is positive. 


90° < 0 < 180°. 

Multiplying by -1 gives -90° > -0 > -180°. 

Thus. -0 is a quadrant III angle and cos (-0) is 

negative. 


90° < 0 < 180°. 

Multiplying by -1 gives _90° > -0 > -180°. 

Thus. -0 is a quadrant III angle and sin (-8) is 

negative. 

47. The reference angle is 65°. Since 115° is in . 

quadrant II. the cosine is negative. Cos 0 

decreases on the interval (90°. 180°) from 

oto -1. Therefore. cos 115° is closest to -.4. 

48. The reference angle is 65°. Since 115° is in 

quadrant II the sine is positive. Sin 0 decreases 

on the interval (90°, .180°) from 1 to O. Therefore, 

sin 115° is closest to .9. 

49. When 0 =45°. sin 0 =cos O. Sine and cosine are 

opposites in quadrants II and IV. Thus 

180° - 0 =180° - 45° =135° in quadrant II. 

360° - 0 =360° - 45° =31Y in quadrant IV. 

50. When 0 =45°, sin 0 = cos O. Sine and cosine are 

both positive in quadrant I and both negative in 

quadrant III. 

o + 180° = 45° + 180° =225° 

Thus 45° is the quadrant I angle. and 225° is the 

quadrant III angle. 

40. If 0 is in the interval (90°. 180°), 

90° < 0 < 180°. 

Dividing by 2 gives 45° < ~ < 90°.. 

2 

Thus.~ is a quadrant I angle and cos~ is 

2 2 

positive. 

54



51. sin 30° + sin 60° J, sin (30° + 60°) 

Evaluate each side to determine whether this 

statement is true or false. 

sin 30° + sin 600 = .!.+ .J3 

2 2 

1+.J3 

=- 

2 

sin(30° + 60°) = sin 90° 

=1 

. 1+.J3 1 h . . f 1 

Smce-- * • t e given statement IS a se. 

2 

52. sin (30° + 60°) J, sin 30° . cos 60° 

+ sin 60° . cos 30° 


equation is true or false. 

sin (30° + 60°) = sin 90° = 1 

sin 30° . cos 60° + sin 60° . cos 30° 

=_._+_. 

1 1 .J3.J3 

2 2 2 2 

1 3 

=-+-=1 

4 4 

Since. 1 = 1. the statement is true. 

53. cos 60°J, 2cos 2300-1 



1 

cos600= 

2 

2COS 2300-1=2(v;f -1 

=2(~)-1 

=(~)-1 

1 

= 

2 

S' 11 . 

mce - = -. the statement IS true. 

2 2 

54. cos 60°J, 2 cos 30° 



1 

cos60o= 

2 

2cos30° = 2( V;) =.J3 

Since .!. *.J3. the statement is false. 

2 

55. sin 120°J, sin 150° - sin 30° 


statement is true or false . 

sin 1200= .J3 

2 

sin 150° - sin 30° =.!. _.!. = 0 

2 2 

Since .J3 * O. the statement is false. 

2 

56. sin 210° J,. sin 180° + sin 30° 



sin210"=-.!. 

2 

sin 180°+ sin 30° = 0 +.!. = .!. 

2 2 

Since · - - 11 * -. th e statement IS . fa Ise. 

2 2 

57. sin 120°J, sin 180° .cos 60° - sin 60° .cos 180° 


statement is true or false . 

sin 1200= .J3 

2 

sin 180°· cos 60° - sin 60° . cos 180° 

=O(~)-( V;}-I) 

=0-(- V;)= V; 

Since · -.J3 =-. .J3 th e statement IS-true. . 

2 2 

58. cos 300° J,. cos 240° . cos 60° 

- sin 240° .sin 60° 



1 

cos3000= 

2 

cos 240° . cos 60° - sin 240° .sin 60° 

=(-~)(~)-(- V;)( V;) 

=-±-(-~) 

1 

= 

2 

· 11 . 

Since - = - • the statement IS true. 

2 2 

55



2 

L :::; ~[~4 _-..:....( ---,3)-,,-]3_36__ 

59. (a) 

200(1 .9 + 336 tan .9°) 

L '" 550ft 

(b) 

Section 2.3 

Exercises 

L:::; [4-(-3)]336 2 

200(1.9 + 336 tan 1.5°) 

"'369ft 

1. The caution at the beginning of this section 

suggests verifying that a calculator is in degree 

mode by finding sin 90°. Ifthe calculator is in 

degree mode, the display should be I. 

2. A calculator gives approximate values of almost 

all trigonometric functions. 

3. To find trigonometric values of cotangent, secant 

and cosecant with a calculator, it is necessary to 

find the reciprocal of the reciprocal function 

value. 

4. The reciprocal is used before the inverse function 

key when finding the angle, but after the function 

key when finding the trigonometric function 

value. 

For the following exercises, be sure your calculator is in 

degee mode. If your calculator accepts angles in 

degrees, minutes, and seconds, it is not necessary to 

change angles to decimal degrees. Keystroke sequences 

may vary on the type and/or model of calculator being 

used . 

5. tan 29° 30' 

29°30' :::; (29 + :)0 

:::; 29.5° 

Enter: 29.5 8 

or 8 29.5 ~ 

Display: .5657728 

6. sin 38° 42' 

38 042':::;(38+ :~)" 

:::; 38.7° 

Enter: 38.7 ~ 

or ~ 38 .7 ~ 

Display: .6252427 

7. cot 41 ° 24' 

42°24':::; (41 + ~~y 

:::; 41.4° 

Enter: 41.4 8 Gill 

or CD 8 41.4 

Display: 1.1342773 

8. sec 13° 15' 

13°15':::; (13+ ~~r 

CD 0 

:::; 13.25° 

Enter: 13.25 @!) Gill 

or CD §) 13.25 CD 

E§) 

Display: 1.0273488 

9. esc 44° 30' 

44°30':::; ( 44+ :r 

:::; 44.5° 

Enter: 44.5 ~ Gill 

or CD ~ 44.5 CD 

Display: 1.4267182 

10. esc 145° 45' 

145 045':::;(145+ :r 

:::; 145.75° 

Enter: 145.75 ~ Gill 

or CD ~ 145.75 CD 

E§) 

Display: 1.7768146 

11. cot 183° 48' 

:::; 183.8° 

183°48':::; (183+ :r 

Enter: 183.8 8 (§] 

or CD ~ 183.8 CD 

E§) 

Display: 15.0557227 

12. cos 421 ° 30' 

421°30':::; (421+ :)0 

:::; 421.5° 

Enter: 421 .5 @!) 

or @D 421.5 ~ 

Display: .4771588 

0 

E§) 

0 ~ 

0 

0 

56



13. sec 312° 12' 

312°12' = (312 + ~~ )O 

19. = sin 514°24' 

csc514°24' 

514°24' =(514+ ~~ y 

= 312.2° 

Enter: 312.2 §J § = 514.4° 

or CD §J 312.2 CD 0 

Enter: 514.4 ~ 

~ 

or ~ 514.4 ~ 

Display: 1.4887142 

Display: .4320857 

14. tan (- 80° 6') 

20. ---= tan 23.4° 

-80°6' = _(80 + 6~ ) cot 23.4° 

0 

Enter: 23.4 8 

= -80.1° or 8 23.4 ~ 

Enter: 80.1 EJ ~ Display: .4327386 

or ~ (ill 80.1 ~ 

Display: -5.7297416 

21. --=tan 

sin 33° 330 

cos 33° 

15. sin (-317° 36') Enter: 33 8 

or 8 33 E!§) 

-317 ~~y 036'=-(317+ Display: .6494076 

0 

= -317.6 cos 77° 770 

Enter: 317.6 EJ ~ 22. = cot 

or 

sin 77° 

~ (ill 317.6 ~ 

Enter: 77 8 (@) 

Display: .6743024 

or CIJ ~ 77 CD 0 ~ 

Display: .2308682 

16. cot (-512° 20') 

23. cos (90° - 3.69°) = sin 3.69° 

-512°20' = -(512+ ~y 

Enter: 3.69 ~ 

:::: -512.3333333° 

or ~ 3.69 ~ 

Enter: 512.3333333 EJ 8 

Display: .0643581 

(@) 

or CD 8 (B) 512.3333333 CD 0 

24. cot (90° - 4.72°) = tan 4.72° 

~ 

Display: 1.9074147 

Enter: 4.72 8 

or 8 4.72 ~ 

Display: .0825664 

17. cos (-15') 

_15'=_(~)O 25. sec 2 47.8°-1 = tan 2 47.8° 

Enter: 47.8 8 IZl 

=-.25° or CD ~ 47.8 CD I£J ~ 

Enter: .25 EJ (§ Display: 1.2162701 

or ~ (ill .25 ~ 

Display: .9999905 

18. ---= cos 14.8° 

29. sin ()= .84802194 

secl4.8° 

INV 

Enter: 14.8 ~ 

or ~ 14.8 ~ 

or 

Display: .9668234 Enter: .84802 194 ARCSIN 

or 

1·..·'1 

or (E) ~ .84802194 ~ 

Display : 57.997172 

(}:::: 57.997172° 

57



30. tan (} = 1.4739716 

' INV 

Enter: 1.4739716 

or 

ARCTAN 

or 

[,...-'] 

or ~ EJ 1.4739716 ~ 

Display : 55.845496 

(} == 55.845496° 

31. tan (}= 6.4358841 

Enter: 6.4358841{INV AR~~AN 

or 

[T...-'I 

or ~ EJ 6.4358841 ~ 

Display: =81.168073 

(}== 81.168073° 

32. sin (} =.27843196 

INV 

Enter: .27843196 

or 

ARCSIN 

or 

[... ,] 

or ~ ~ .27843196 ~ 

Display: 16.166641 

(} == 16.166641° 

33. sec (}=1.1606249 

Enter: 1.1606249 l:!2D 

!NV 

or 

ARCCOS 

or 

[""".,] 

or ~ @!) 1.1606249 ~ ~ 

Display: 30.502748 

(} == 30.502748° 

34. cot (}= 1.2575516 

Enter 1.2575516 l:!2D 

INV 

or 

ARCTAN 

or 

1'...-'1 

or B IE] 1.2575516 0 ~ 

Display: 38.491580 

(} == 38.491580° 

35. esc (}= 1.3861147 

Enter: 1.3861147 ~ 

INV 

or ~ ~ 1.3861147 ~ 

Display: 46.173582 

() == 46.173582° 

36. sec (}= 2.7496222 

2.7496222 l:!2D {INV AR~~OS 

or 

[""".,] 

or ~ @!) 2.7496222 ~ 

Display : 68.673241 

(}== 68.673241° 

37. tan A = 1.482560969 

or 

ARCSIN 

or 

["' ~I 

Enter : 1.482560969{INV AR~~AN 

or 

[''''-'1 

or ~ EJ J .482560969 ~ 

Display: 56 

A == 56° 

38. A =sin 22° 

Enter: 22 ~ 

or~22~ 

Display : .3746065934 

A == .3746065934° 

In Exercises 39-42, be sure that your calculator is in 

degree mode. Keystroke sequences may vary based on 

the type and/or model of calculator being used. 

39. sin 35° cos 55° + cos 35° sin 55° 

Enter: IT) 35 ~ CD 0 IT) 55 

§) CO ITJ IT) 35 §) CD 0 

CD 55 ~ ITJ @] 

or ~ 35 0 @D 55 0 rn @D 

35 CO 

EJ 55 ITJ ~ 

Display: 1 

sin 35° cos 55° + cos 35° sin 55° = 1 

40. cos 100° cos 80° - sin 100° sin 80° 

Enter: 100 @D 0 80 @!) G 

100 ~ 0 80 EJ @] 

or @D 100 CD @D 80 CO G 

EJ 100 CO ~ 80 CO ~ 

Display: -1 

cos 100° cos 80° - sin 100° sin 80° =-1 

58


2.3 Finding Trigonometric Function Values Using a CalcuJator 59 

41. cos 75° 29' cos 1431' 44. 0) = 39°, 9z = 28° 

- sin 75° 29' sin 14° 31' ~= sin9\ 

Enter: CD 75 rn 29 CD 60 OJ Cz 

~ 

sin9z 

o OJ 14 rn 31 CD 60 OJ ~ 

c) sinOz 



48. cl =3 x 10 8 m per sec, 55. ~ sin 40° =sin ~ (40°) 

c2 =2.254 x 10 8 m per sec 

9 1 =90° - 29.6° =60.4° 

2 2 

Using a calculator gives 

~sin4O° =.32139380 and 

2.254 x 10 8 (sin 60.4°) 2 

sin9 2 = 

3x 10 8 sin~(400) =.34202014. 

9 2 = 40.8° 2 

Light from the object is refracted at an angle of 

False 

40.8° from the vertical. Light from the horizon is 

refracted at an angle of 48.7° from the vertical. 56. sin 39° 48' + cos 39° 48' =1 

Therefore, the fish thinks the object lies at an 


angle of 48.7° - 40.8° =7.9° above the horizon. sin 39° 48' +cos 39° 48' =1.40839322. 

49. cos 40° J, 2 cos 20° 

False 


57. F= Wsin9 

cos 40° =.766044 and 

F = 2400 sin (-2.4°) 

2 cos 20° =1.8793852. = -100.5 Ib 

False 

F is negative because the car is traveling 

SO. sin 10°+sin 10°J, sin 20° 

downhill. 

Using a calculator gives 58. F= Wsin9 

sin 10°+sin 10° =.34729636 and 

F =2100sin 1.8 0 

sin 20° =.34202014. 

=65.% Ib 

False 

59. F=Wsin9 

51. cos 70 0 J, 2 cos 2 35 0 -1 150 = 3000 sin 9 

Using a calculator gives 150 . 9 

cos 70° =.34202014 and 

--=sm 

3000 

2cos 2 35 0 -1 = .34202014. 

.05 =sin9 

True 

9 = 2.87° 

52. sin 50° J, 2 sin 25° . cos 25° 60. F= Wsin9 


120 =W sin 1.5° 

sin50° = .76604444 and 

120 

=W 

2sin25° ·cos25° = .76604444. 

sin 1.5° 

True 

W=45841b 

53. 2 cos 38° 22' =cos 76° 44' 61. F=Wsin9 


-145 = Wsin(-3°) 

2 cos 38°22' = 1.5681094 and -145 

cos 76° 44' =.22948353. 

=W 

sin(-3°) 

False 

W=2771Ib 

54. cos 40° = 1-2sin 2 80 0 62. F= Wsin9 


-130 =2600 sin 9 

cos 40° = .76604444 and 

-130 . 

1-2sin 2 80° = -.93969262. 

--=sm9 

2600 

False 

-.05 =sin9 

9 =-2.87° 

60



63. F= WsinO 65. For parts (a) and (b), a = 3°, g = 32.2, and 

F=2200sin2° /=.14. 

F = 76.77889275 Ib 

F= WsinO (a) Since 45 rni/hr = 66 ft/sec, 

F = 2000sin 2.2° 

F = 76.77561818 Ib 

The 2200-lb car on a 2° uphill grade has the 

greater grade resistance. 

y2 

R=--- 

g(f + tan a) 

66 2 

32.2(.14 + tan 3°) 

reO 

64. 0 sin 0 tan 0 - '" 703 ft 

180 

(b) Since 70 mi/hr = 102213 ft/sec, 

0° .0000 .0000 .0000 

y 2 

.5° .0087 .0087 .0087 

R=--- 

g(f+tana) 

1° .0175 .0 175 .0175 102.67 2 

"'----- 

1.5° .0262 .0262 .0262 32.2(.14 + tan 3°) 

'" 1701 ft 

2° .0349 .0349 .0349 

2.5° .0436 .0437 .0436 

(c) Intuitively, increasing a would make it easier 

to negotiate the curve at a higher speed much 

3° .0523 .0524 .0524 like is done at a race track. Mathematically, a 

3.5° .0610 .0612 .0611 

larger value ofa (acute) will lead to a larger 

value for tan a. Iftan a increases, then the 

4° .0698 .0699 .0698 ratio determining R will decrease. Thus , the 

radius can be smaller and the curve sharper if 

(a) From the table we see that if 0 is small, 

a is increased. 

• Ll Ll reO 

sm 17 "" tan 17 '''''-. 

180 

(b) 

(c) 

(d) 

F = Wsin 0 :=: Wtan 0 '" WreO 

180 

4 

tanO=-=.04 

100 

y2 

R=--- 

g(/ +tana) 

for small O. 66 2 

""----- 

32.2(.14 + tan 4°) 

'" 644 ft 

y2 

R=--- 

g(/+tana) 

F"" WtanO 

= 2000(.04) 102.67 2 

=801b 

32.2(.14+ tan 4°) 

WreO 

Use F"" - from part (b) 

180 

Let 0= 3.75 and W = 1800. 66. From Exercise 65, 

F "" 18001t(3.75) 

180 

:=: 117.81Ib 

"" 1559 ft 

As predicted, both values are less. 

y2 

R=--- 

g(f + tan a) 

Let R = 1150, g = 32.2,/= .14, anda= 2.1°. 

Then 

v = "/""-'R-g("""'/:-+-ta-n-a""'-) 

= ../1150(32.2)(.14 + tan 2.1°) 

"" 80.9 ftlsec 

80.9 ftlsee . 3600 seclhr . 1 mi/5280 ft:=: 55 mph 

It should have a 55 mph speed limit. 

61



v 2 sin (}cos()+ vcos(}~(vsin (})2 + 64h 

67. (a) D =-------....!------.;-- 

32 

Let v =44 ftlsee and h = 7 ft. 

For () =40°, 

Section 2.4 

44 2 sin40°cos400+44cos400~(44sin400)2 +64(7) 

D =: ----------........,;.----'---"""""--' 

32 

""67.00 ft 

For () =42°, 

D =: 44 2 sin 42° cos 42° + 44 cos 42°~( 44 sin 42°)2 + 64(7) 

32 

==67.14ft 

For () =: 45°, 

D =: 44 2 sin 45° cos 45° + 44 cos 45°~( 44 sin 45 0)2 + 64(7) 

32 

== 66.84 ft 

D increases and then decreases. 

(b) Let h =7 ft and () =42°. 

For v =43 ft/sec, 

D =43 2 sin 42° cos 42° + 43cos42°~(43sin42o)2 +64(7) 

32 

== 64.40 ft 

For v =44 ft/sec, 

D= 44 2 sin42°cos42°+44cos42°~(44sin42o)2 +64(7) 

32 

==67.14ft 

For v =45 ftlsee, 

D =: 45 2 sin 42°cos42°+ 45cos42°~(45sin42o)2+64(7) 

32 

== 69 .93 ft 

D increases. 

(c) v; the shotputter should concentrate on achieving as large a value of v as possible. 


Our Steps 1 and 2 correspond to Polya's Steps 1 and 2. Our Step 3 corresponds to Poyla's Steps 3 and 4. 

Exercises 

1. 16,454.5 to 16,455.5 

2. 28,999.5 to 29,000.5 

3. 546.5 to 547.5 

4. 8958.5 to 8959.5 

8. A measurement of 26.0 lb indicates a weight between 25.95 and 26.05 lb. 

62


9. If h is the actual height of a building and the 

height is measured as 58.6 ft, then 

Ih-58.6Is .05. 

10. If w is the actual weight of a car and the weight is 

measured as 15.00 x 10 21b, then Iw -15001 S .5. 

11. A = 36°20', c = 964 m 

A+B=90° 

B=900-A 

B = 90° - 36°20' 

= 89°60' - 36°20' 

B= 53°40' 

sin A =~ 

c 

a =c sin A 

a = 964 sin 36°20' 

Use a calculator and round answer to three 

significant digits. 

a =571 m 

b 

cosA= c 

b =ccosA 

b = 964 cos 36°20' 



b=777m 

12. A = 31°40', a = 35.9 krn 

B=900-A 

B = 90° - 31°40' = 58°20' 

tan A =~ 

b 

tan 31°40' = 35.9 

b 

35.9 

tan31°4Q' 



= 58.2 krn 

. B b 

SID 

b=--- 

sin A =~ 

c 

SID 

. 31040' =- 35.9 

c 

35.9 

sin 31°4Q' 



= 68.4 krn 

13. N = 51.2°, m = 124 m 

M+N=9O° 

M=900-N 

M = 90° - 51.2° 

M = 38.8° 

tanN=~ 

m 

n=mtanN 

n = 124 . tan 51.2° 

n = 154 m 


cos N =!!!. 

p 

m 

p=- 

cosN 

124 

p=-- 

cos51.2° 

p = 198 m 

14. X = 47.8 °, z = 89.6 cm 

Y = 90° - 47.8° = 42.2° 

~=sin47 .8° 

89.6 

x = 89.6sin47.8° 

=66.4 cm 

_Y-=cos47.8° 

89.6 

y = 89.6 cos 47.8° 

=60.2 em 

15. B = 42.0892°, b = 56.851 ern 

A+B=9O° 

A =90 0-B 

A = 90° - 42.0892° 

=47.9108° 

= c 

b 

c=- 

. sinB 

56.851 

c=---- 

sin 42.0892° 

c= 84.816 em 

b 

tan B=a 

b 

a=- 

c=--- tanB 

56.851 

a=---- 

tan 42.0892° 

a=62.942 em 

63



16. B = 68.5142°, c = 3579.42 m 24. B = 46.00°, c = 29.7 m 

A = 90° - 68.5142° = 21.4858° 

B 

a 

---= cos 68.5142° 

3579.42 

a = 3579.42 cos 68.5142° 

= 1311.04 m 

90· 

b = 3579.42 sin 68.5142° 

Abe 

=3330.68 m 

A = 90° - 46.00° = 44.00° 

17. B 

_a_ = cos 46.000 

29.7 

a = 29.7 cos 46.00° 

~ • ~ 1.498426374 =20.6 m 

22° 

A C .s: = sin 46.00° 

29.7 

18. A 

b = 29.7sin46.00° 

=21.4 m 

.: 

25. B = 73.00°, b = 128 in. 

B 

5 ~a ~ 

A 128 in . C 

C B "" 56 .44269024° 

A = 90° -73.00° = 17.00° 

0 _ 128 

tan 73.00 - 

a 

128 

a=--- 

tan 73.00° 

20. If we are given an acute angle and a side in a right a =39.1 in. 

triangle, the unknown part of the triangle 

sin 73.000 = 128 . 

requiring the least work to find is the other acute 

c 

angle . It may be found by subtracting the given 128 

acute angle from 90°. 

23. A = 28.00°, c = 17.4 ft 

B 

a 

90· 28 .00 

C b A 

A+B=90° 

B=900-A 

B = 90° - 28.00° 

B=62.00° 

sinA=~ 

c=--- 

sin 73.00° 

c=134 in. 

26. A = 61°00', b =39.2 em 

B 

61.00· 90 · 

c A 39 .2m C 

a =csinA 

B = 90° - 61°00' = 29°00' 

a = 17.4sin 28.00° 

~= tan 61°00' 

a=8.17 ft 39.2 

b 

a 

cosA = 

= 39.2 tan 61°00' 

c 

=70.7 em 

b=ccosA 

_c_ = sec 61°00' 

b = 17.4 cos 28.00° 39.2 

b = 15.4 ft 

c = 39.2 sec 61°00' 

=80.9 em 

a 

64



27. a = 76.4 yd, b = 39.3 yd 29. a = 18.9 em, c = 46.3 em 

B 

76 .4 yd 

~.18.9 em 

90· 

A b C 

b=~c2_a2 

b = ~(46 .3)2 -(18.9)2 

90· 

C 39.3 yd A b =42.3 em 

c 2 =a 2 +b 2 . A 18.9 

sm =- 

46.3 

c=~a2+b2 

A =24°10' 

c = ~(76.4)2 + (39.3)2 B 18.9 

cos =- 

c=85.9 yd 46.3 

B= 65°50' 

tan A = 76.4 

39.3 

A = 62°50' 

tanB= 39.3 

76.4 

B = 27°10' 

28. a = 958 m, b = 489 m 

B 

30. b = 219 m, C = 647 m 

B 

a 

958m 

a=~6472-2192 

a = .J4 I 8,609 - 47,961 

a = .J370,648 

90· 

a=609m 

A 489m C 

958 219 

tanA= 

cosA= 

489 

647 

A = 63.0° = 63°00' 

A = 70.2° = 70° 10' 

489 . B 219 

tanB=- sm = 

958 

647 

B = 27.0° = 27°00' 

B = 19.8° = 19°50' 

Check: A + B = 90° Check that A + B = 90°. 

To find c from the given information, use the 

Pythagorean theorem. 

c 2 =a 2 +b 2 

c=~a2 +b 2 

c = ~9582 + 489 2 

c = .Jl, 156,885 

c= 1080 m 

Another way to find c is to use the given 

information together with the result for one of the 

acute angles. 

~=csc63.0° 

958 

c = 958 esc 63.0° 

c= 1080 m 

65



31. A = 53°24', C = 387.1 ft 

B =90° - 53°24' 

B = 89°60' - 53°24' 

B= 36°36' 

sin 53°24' = _a_ 

387.1 

a = 387.1sin53°24' 

a = 310.8 ft 

cos 53°24' = _b_ 

387.1 

b = 387. 1cos 53°24' 

b = 230.8 ft 

32. A = 13° 47', C = 1285 m 

B 

13°47' 90· 

Abe 

B= 90°-13°47' == 76°13 ' 

b = 1285 cos 13°47' 

= 1248 m 

a = 1285sin 13°47' 

=306.2 m 

33. B=39° 9', c= .6231 m 

B 

a 

90° 

c b A 

A =90° - 39°9' 

A = 50°51' 

cos 3909' = _a_ 

.6231 

a = (.6231) cos 39°9' 

a = .4832 m 

sin 3909' =_b_ 

.6231 

b = (.6231)sin39~' 

b=.3934 m 

a 

34. B = 82° 51', C = 4.825 em 

B 

~a 

Abe 

A = 90° - 82°51' = 7°9' 

b =4.825sin 82°51' 

=4.787 em 

a = 4.825 cos 82°51' 

=.6006 em 

35. The angle of elevation from X to Y is 90° 

whenever Y is directly above X. 

36. The angle of elevation from X to Y is the acute 

angle formed by ray XY and a horizontal ray with 

endpoint at X. Therefore, the angle of elevation 

cannot be more than 90° . 

39. Let h =the distance the ladder goes up the wall. 

sin43 050' = _h_ 

13.5 

h = 13.5 sin 43°50' 

h = 9.3496000 

The ladder goes up the wall 9.35 m. 

40. Let ()= the angle the guy wire makes with the 

ground. . 

. () 71.3 

S10 =- 

77.4 

() = 67.100475° 

() = 67° + (.100475)(60') 

() =67°10' 

The guy wire makes an angle of 67° 10' with the 

ground. 

41. Let x =the length of the guy wire. 

S10 

. 45030' =- 63.0 

x 

xsin 45°30' = 63.0 

63.0 

x =- 

""7 

sin 45°30' 

x =88.328020 

The length of the guy wire is 88.3m. 

66



42. Since angle T= 32° 10' and angle S = 57° 50', 

S + T =90°, and triangle RST is a right triangle. 

tan 32010' =- RS 

53.1 

RS =53.1 tan 32°10' 

RS"" 33.395727 

The distance across the lake is 33.4 m. 

43. The two right triangles are congruent. 

So corresponding sides are congruent. Since the 

sides that compose the base of the isosceles 

tnang 

. I 

e are congruent, eac 

. h id . 42.36 

si e IS --,or 

2 

21.18 in. Let x =the length of each of the two 

equal sides of the isosceles triangle. 

45. tan 30.0° =-y 

1000 

y =1000 · tan 30.0° 

=577 

However, the observer's eye-height is 6 feet from 

the ground, so the cloud ceiling 

is 577 + 6 = 583 feet. 

46. Let x = the length of the shadow. 

0 _ 5.75 

tan 23 . 4 -- 

x 

x tan 23.4° =5.75 

5.75 

x=-- 

tan 23.4° 

x ""13.287466 

The length ofthe shadow is 13.3 ft. 

cos38.120=21.18 

x 

xcos38.12° =21.18 

21.18 

cos38.12° 

X"" 26.921918 

The length of each of the two equal sides of the 

triangle is 26.92 in. 

44. The altitude of an isosceles triangle bisects the 

base as well as the angle opposite the base. Let 

h =the altitude. The base is 184.2 em so the 

altitude forms two congruent right triangles each 

with a leg of 184.2 =92.1 cm and with an 

2 

opposite angle of .!.(68°44') =34°22'. 

2 

A 

47. Let h =height of the tower. 

~h 

42.36-- 

A 40 .6m C 

In triangle ABC, 

h 

tan 34.6° =- 

40.6 

h =40.6 tan 34.6° 

x=--- 

h "" 28.0. 

The height of the tower is 28.0 m. 

48. Let 0 = the angle of elevation of the sun. 

J> 

./ 

./ 

63 .1 

tan 0 = 48.6 

63.1 

0"" 37.603681 

"" 37°+(.603681)(60') 

0",,37°40' 

The angle of elevation of the sun is 37° 40'. 

D 92.1 em C 92.1 em B 


tan 34 

022' 

-- 

_ 92.1 

h 

h tan 34~2' = 92.1 

92.1 

h=-- 

tan 34°22' 

h =134.67667. 

The altitude of the triangle is 134.7 em. 

67



49. Let d =the distance from the top B of the building 

to the point on the ground A. 

B 

A 


C 

sm . 32030' =- 252 

d 

252 

sin 32°30' 

d==469. 

The distance from the top of the building to the 

point on the ground is 469 m. 

tan () =12.02 

d=---5.93 

50. Let x =the horizontal distance that the plan must 

fly to be directly over the tree. 

tan 13°50' = 10,500 

x 

x tan 13°50' =10,500 

10,500 

tan 13°50' 

x == 42,600 

The horizontal distance that the plan must fly to 

be directly over the tree is 42,600 ft. 

51. Let x = the height of the taller building, 

h = the difference in height between the 

shorter and taller buildings, and 

d =the distance between the buildings along 

the ground. 

52. Let a = the angle of depression that the lens 

makes with the horizontal and 

() =the angle that the lens makes with the 

vertical. 

camera 

5.93~ 

ft~ 

12.02ft 

head 

teller 

() =63°40' 

a=900-() 

=90° - 63°40' 

a =26°20' 

The lens should make an angle of 26° 20' with 

the horizontal. 

53. Let ()=the angle of depression. 

tan () = 39.82 

51.74 

x=---() = 37.58° =37°35' 

54. Let d = the diameter of the sun. 

..!.(l°4') =32' =.533° 

2 

Ld 

2 =tan .533° 

92,919,800 

d =2(92,919, 8oo)(tan .533°) 

== 1,730,000 mi 

The diameter of the sun is about 1,730,000 mi. 

55. The triangle with 60° as vertex angle and x as a 

base is equilateral. d is the perpendicular bisector 

of x and forms a 30°-60° right triangle. 

i x i x 

60· 60· 

d 

--= cot 14°10' 

28.0 

d =28.0 cot 14°10' 

=111m 

h 

- =tan 46°40' 

d 

h = d tan 46°40' 

=111tan 46°40' 

=118m 

x=h+28.0 

= 118+28.0 

==146m 

The height of the taller building is 146 m. 

LX 

_2_ = tan 30° 

d 

x = 2dtan30° 

=2(2.894)(tan 30°) 

==3.342 mm 

68



56. Let x = the distance from the assigned target. 

A 

' .')"",",J; 


x 

tan 0°0'30" =-- 

234,000 

x = 234,000 tan 0°0'30" 

x'" 34.0 

The distance from the assigned target is 34.0 mi. 

57. (a) The height h of the peak above 14,545 is 

sin5.82 0 = h 

27.0134(5280) 

h ",14,463 ft. 

The total height is about 

14,545 + 14,463 = 29,008 ft. 

(b) The curvature of the earth would make the 

peak appear shorter than it actually is. 

Initially the surveyors did not think 

Mt. Everest was the tallest peak in the 

Himalayas. It did not look like the tallest 

peak because it was farther away than the 

other large peaks. 

58. (a) Let d be the distance between P and Q. 

Triangle PQC is a right triangle with 

() d 

tan -= 

2 R 

e 

d= R tan 

2 

37° 

=965 tan 

2 

= 965 tan 18.5° 

'" 323 ft. 

(b) From right triangle CPN, 

() NC 

cos-= 

2 R 

() 

NC=Rcos- . 

2 

Now, 

MN+NC=R 

MN=R-NC 

() 

= R-Rcos 2 

='I -COS~} 

59. (a) 

Section 2.5 

Exercises 

f3 ,., 57.3S = 57.3(336) ,., 320 

R 600 

d =R(I-COS ~ ) 

'" 600(1- cos 16°) 

= 23.4 ft 

(b) f3'" 57.3S = 57.3(485) = 46 .3° 

R 600 

d= 'I-COS~J 

00( 

,., 48.3 ft 

46 30) 

=6 I-COST 

(c) The faster the speed, the more land needs to 

be cleared on the inside of the curve. 

1. It should be shown as an angle measured from due 

north. 

2. It should be shown measured from north 

(or south) in the east (or west) direction. 

3. A sketch is important to show the relationships 

among the given data and the unknowns. 

4. The angle of elevation (or depression from X to Y 

is measured from the horizontal line through X to 

the ray XY. 

5. N 

(-4.0) 

W'''-+-.-l~+-t-+-+-i-t-+--r.-E 

s 

The bearing of the airplane measured in a 

clockwise direction from due north is 270°. The 

bearing can also be expressed as N 90° W, or 

S 90° W . 

69



6. N 

9. y 

-E-"T'T-=""'-t--- x 

The reference angle is 30°. A point on the ray is 

(-.J3, -1). Using the equation y = mx + b, the 



bearing can also be expressed '3S S 45° W. 

7. N 

equation of the ray is y = .J3 x, x ~ 0 since the 

3 

ray lies in quadrant 111. 

10. y 

wr.r-+-+-+-++-I~+-t-t--t-H-;-- E 

-----+=n---x 

(1,-..[3) 

s 



bearing can also be expressed as N 45° W. 

8. N 

1800 

The reference angle is 60°. A point on the ray is 

(1, -.J3). Using the equation y =mx + b, the 

equation of the ray is y =-.J3x, x~ 0 

since the ray lies in quadrant IV. 

11. Let x =the distance the plane is from its starting 

point. In the figure, the measure of angle ACB is 

40° + (180° - 130°) or 90°. Therefore, triangle 

ACB is a right triangle. 

N 

N 

s 



bearing can also be expressed as S 0° E or 

s o-w, 

A x B 

Distance traveled in 1.5 hr: 

(1.5 hr)(110 mph) = 165 mi 

Distance traveled in 1.3 hr: 

(1.3 hr)(110 mph) = 143 mi 

70



Using the Pythagorean theorem. we have 

x 2 :;:: 165 2 + 143 2 

x 2 :;::47.674 

x =220. 

The plane is 220 mi from its starting point. 

12. Let x =the distance from the starting point. 

In the figure, the measure of angle ACB is 

27° + (180° - 117°) or 90° . Therefore, triangle 

ACB is a right triangle. 

Applying the Pythagorean theorem, 

x 2 =50 2 + 140 2 

2 

x = 22,100 

x =..)22,100 

x=150. 

The distance of the end of the trip from the 

starting point is 150 krn. 

13. Let x =distance the ships are apart. 

(1.5 hr)(l8 knotsj e 27 nautical mi 

(1.5 hr)(26 knots) =39 nautical mi 

Since 130° - 40° =90°, we have a right triangle. 

Applying the Pythagorean theorem, 

x 2 =27 2 +39 2 

x 2 =2250 

x=47. 

The ships are 47 nautical mi apart. 

14. Let C =the location of the ship 

and c = the distance between the lighthouses. 

N 

129 043' 

B 

Since 50° 17' + 39° 43' :;:: 90°. C =90°. and we 

have a right triangle. 

. 39043' 3742 

Sin =- c 

3742 

c= sin 39°43' 

c = 5856 

The distance between the lighthouses is 5856 m. 

15. Let x =distance between the two ships. 

The angle between the bearings of the ships is 

180° - (28° 10' + 61° 50') or 90°. The triangle 

formed is a right triangle. 

Distance traveled at 24.0 mph: 

(4 hr) (24.0 mph) =96 mi 

Distance traveled at 28.0 mph: 

(4 hr)(28.0 mph) =112 mi 

Applying the Pythagorean theorem gives 

x 2 = 96 2 + 112 2 

2 

x = 21,760 

x = 148. 

The ships are 148 mi apart. 

16. Let C =the location of the transmitter 

and Q = the distance of the transmitter from B. 

angle CBA =90° - 53°40' 

=36°20' 

angle CAB =90 - 36°20' 

= 53°40' 

A + B = 90°, so C = 90° .. 

_Q- = sin A =sin 53°40' 

2.50 

Q =2.50 sin 53°40' 

=2.01 

The distance of the transmitter from B is 2.01 mi. 

17. Draw triangle WDG with Wrepresenting 

Winston-Salem, D representing Danville, and 

G representing Goldsboro. Name any point X on 

the line due south from D. 

N 

D 

S 

Angle BAC =180° - 129° 43' 

=50° 17' 

B 

.L.. -+ ~G 

x 

s 

Since the bearing from W to Dis 42°, 

angle WDX =42°. 

Angle XDG =48° 

So angle D =42° + 48° =90°, and triangle WDG 

is a right triangle. Using d =rt and the 

Pythagorean theorem, we have 

71



WG = ~(WD)2 +(DG)2 

23. Let x = the side adjacent to 49.2° in the smaller 

triangle. 

= ~[60(1)f + [60(1.8)]2 In the larger right triangle: 

h 

= ..)15,264 tan 29.5° = -- 

392+x 

'" 120. h = (392 + x) tan 29.5°. 

The distance from Winston-Salem to Goldsboro 

is 120 mi. 

tan49.2°=~ 

18. Let x = the distance from Atlanta to Augusta. x 

In the smaller right triangle: 

Atlanta x Augusta h = x tan 49.2°. 

N 

Substitute the first expression for h in this 

I• equation. and solve for x . 

! 63° (392 + x) tan 29.5° = x tan 49.2° 

392 tan 29.5° + x tan 29.5° = x tan 49.2° 

Macon 

392 tan 29.5° = x tan 49.2° - x tan 29.5° 

Note that the line from Atlanta to Macon makes 392 tan 29.5 = x(tan 49.2° - tan 29.5°) 

an angle of 27° + 63°. or 90°. with the line from 392 tan 29.5° 

Macon to Augusta. 

-------=x 

Distance from Atlanta to Macon: 

tan 49 .2° - tan 29.5° 

Then substitute this value for x in the equation for 

60(1~)=75 mi 

the smaller triangle. 

h = xtan 49.2° 

h = 392 tan 29.5°(tan 49.2°) 

Distance from Macon to Augusta: 

tan 49.2° - tan 29.5° 

h"'433 

60(Ii) = 105 rni 

The height of the triangle is 433 ft. 

Use the Pythagorean theorem to find x. 

24. Let x = the side adjacent to 52 .5° in the smaller 

x 2 = 75 2 + 105 2 

triangle. 

= 16,650 

In the larger triangle: 

x'" 130 mi 

The distance from Atlanta to Augusta is 130 mi. tan 41.2° = h 

168+x 

h = (168+ x) tan 41.2°. 

19. Solve the equation ax = b + ex for x in terms of 

a, b. and e. 

In the smaller triangle: 

ax =b+ex 

tan52.5°=~ 

ax-ex=b 

x 

x(a-e) =b 

h = x tan52.5°. 

b Substitute for h in this equation and solve for x. 

x=- 

a-e 

(168 + x) tan 41.2° = x tan 52S 

168tan 41.2° + x tan 41.2° = x tan 52.5° 

21. Using the equation y = (tan 9)(x - a) where (a, 0) 168 tan 41.2° = x tan52.5°-x tan 41.2° 

is a point on the line and (J>is the angle the line 168 tan 41.2° = x(tan52.5° - tan 41.2°) 

makes with the x-axis, y = (tan 35°)(x - 25). 168 tan 41.2° 

-------=x 

22. Using the equation y = (tan 9)(x - a) where (a. 0) 

tan 52.5° - tan 41.2° 

is a point on the line and 

Then substitute for x in the equation for the 

(J> is the angle the line 

makes with the x-axis, 

smaller triangle. 

y = (tan 15°)(x - 5). 

h = x tan 52.5° 

h = 168 tan 41.2°(tan 52.5°) 

tan 52.5° - tan 41.2° 

h",448 

The height of the triangle is 448 m. 

72



25. Let x =the distance from the closer point on the 27. Let x = the height of the antenna 

ground to the base of height h of the pyramid. 

and h =the height of the house. 

135 ft 

In the larger right triangle: 

h 

tan 21°10' =-- 


135+x 

h =(135 + x) tan 21°10'. h 

tan 18°10' = 


28 

h 

h = 28(tanI8°1O'). 

tan 35°30' = 

x 

In the larger right triangle: 

h = x tan 35°30'. 

tan 27 

010 ' 

-- 

_ x+ h 

Substitute for h in this equation, and solve for x. 

28 

(135 + x) tan 21°10' = x tan 35°30' x + h =28(tan 27° 10') 

135 tan 21°10' + x tan21 °10' =x tan 35°30' x =28(tan27°IO') 

135 tan 21°10' =x tan 35°30' - x tan 21°10' -28(tan 18°10') 

135 tan 21°10' =x(tan 35°30' - tan 21°10') 

X"" 5.18. 

135 tan 21°10' 

-----:-----=x 

The height of the antenna is 5.18 m. 

(tan 35°30' -tan21°1O') 

Then substitute for x in the equation for the 

smaller triangle. 

h =135 tan 21° 10'(tan35°30') 

(tan 35°30' - tan 21°10') 

~. 

• D 

h"" 114 ~B 

The height of the pyramid is 114 ft. 

In triangle ADC, 

28. Let x be the height of Mt. Whitney above the level 

of the road and y be the distance shown in the 

figure below. 

A 

26. Let x = the distance traveled by the whale as it tan 22°40' = ~ 

approaches the tower and 

y 

y = the distance from the tower to the 

Ytan 22°40' = x 

whale as it turns. 

x 

15·50' Y = tan 22040' . (1) 


x 

tan 10° 50' =-- 

y 

. y+7.00 

68.7 =tan 35°40' 

(y + 7.(0)tan 10°50' =x 

y 

y tan 10° 50' +7.00 tan 10°50' =x 

68.7 x - 7.00 tan 10°50' 

y =tan 35°40' Y = tan 10050' (2) 

68.7 =tan 15050' 

x+y 

Substituting (l)into (2), we have 

x 

x -7.00tinlOo50'~ 

----=------ 

68.7 tan 22°40' tan 10°50' 

x+Y=--- x tan 10°50' =x tan 22°40' 

tan 15°50' 

68.7 7.00 tan 10 050'(tan 22°40') 

x = tan 15050' y 

7.00tan 10 050'(tan 22°40') 

68.7 68.7 =x tan 22°40' - x tan 10°50' 

x=--- tan 15°50' tan 35°40' 

7.00 tan 10 

050'(tan 22°40') 

X"" 147 

=x(tan 22°40' -tan 10°50') 

The whale traveled 147 m as it approached the 

x ::::2.47 . 

lighthouse. 

The height of the top of Mt. Whitney above road 

level is 2.47 km. 

73



29. (a) From the figure in the text, 

b a b /3 

d = -cot - + - cot 

222 2 

d = !!'(cot a + cot /3). 

2 2 2 

(b) d = !!.(cot a + cot /3) 

2 2 2 

Let a = 37'48", /3 = 42'3", and b = 2. 

2 ( 37'48" 42'3") 

d=- cot---+cot- 

2 2 2 

d == 345.395 I m 

30. Let h =the minimum height above the surface of 

the earth so a pilot at A can see an object on the 

horizon at C. 

Using the Pythagorean theorem, 

(4.00 x 10 3 +h)2 = (4.00 x 10 3 )2 + 125 2 

(4000+h)2 = 16,015,625 

4000+ h = 4001.952648 

h = 1.95. 

The minimum height above the surface of the 

earth would be 1.95 mi. 

31. Let x =the minimum distance that a plant needing 

full sun can be placed from the fence . 

tan 23OZ0' = 4.65 

x 

x tan 23°20' = 4.65 

4.65 

tan 23°20' 

x == 10.8 

The minimum distance is 10.8 ft. 

x=--- 

2. 

sin A = side opposite = 40 = 20 

hypotenuse 58 29 

32. tan A =1.0837 

1.4923 

A = 35.987° or 35°59'10" 

1.4923 

tan B =- 

1.0837 

B = 54.013° or 54°00'50" 

33. Let y =the common hypotenuse of the two right 

triangles. 

x 

90° 52°20' 

cos 30050' =- 198.4 

y 

198.4 

y=--- 

cos 30°50' 

y = 231.06 

To find x, first find the angle opposite x in the 

right triangle by subtracting. 

52°20' - 30°50' = 51°80' - 30°50' 

=21°30' 

sin 21030' = .::. = __x_ 

y 231.06 

x = 231.06(sin 21°30') 

x == 84.7 m 

34. Since angle C is 90° and the angles of a triangle 

must sum to 180°, angles A and B must sum to 

90°, or be complementary, in order to ensure that 

AB is a straight line (and therefore the two ends 

meet). 


. side opposite 60 

1. smA= = 

hypotenuse 61 

cos A = side adjacent =.!..!. 

hypotenuse 61 

tan A = side opposite = 60 



cot A = = 


sec A = hypotenuse = ~ 


esc A = hypotenuse = ~ 


cos A = side adjacent = 42 = ~ 


tan A = side opposite = 40 = 20 

side adjacent 42 21 

side adjacent 42 21 

cot A = =-=side 

opposite 40 20 


sec A = =-=side 

adjacent 42 21 

esc A = hypotenuse = 58 = 29 

side opposite 40 20 

3. sin 4/3 = cos 5/3 

Since sine and cosine are cofunctions, 

4/3+ 5/3 = 90° 

9/3 = 90° 

/3 =10°. 

74



4. sec(2 y + 10°) = csc( 4y + 20°) 

Since secant and cosecant are cofunctions, 

(2y + 10°)+ (4y + 20°) = 90° 

6y + 30° = 90° 

6y =60° 

Y =10° 

5. tan (5x + 11°) = cot (6x + 2°) 

Since tangent and cotangent are cofunctions, 

5x + 11°+ 6x + 2° = 90° 

l l,r = 90° - 11° - 2° 

l l,r = 77° 

x=7°. 

6. co{~ +l1o)=sinG~ +40 0) 

Since cosine and sine are cofunctions, 

(~ +l1°)+G~ +40 0) =90° 

Q9+51°=9O° 

10 

7. sin 46° < sin 58° 

sin 9 increases as 9 increases from 0 to 90°. 

Since 

58° > 46° 

sin 58° > sin 46° 

True 

8. cos 47° < cos 58° 

False because cosine decreases from 0° to 90°. 

9. sec 48° ~ cos 42° 

sec 9 ~ 1 for all 9 and cos 9 S 1 for all 9. 

Therefore, 

cos9S1Ssec9. 

True 

10. sin 22° ~ esc 68° 

For all angles in quadrant I, 

sin 9 < 1. 

For all angles in quadrant I, 

esc 9 > 1. 

Therefore, 

esc 68° > sin 22°, 

and the given statement is false. 

12. 120° 

This angle is in quadrant II, so the reference angle 

is 180° - 120° = 60. 

Since 120° is in quadrant II, the cosine, tangent, 

cotangent, and secant are negative. 

sin 120° = sin 600 = ...[3 

2 

1 

cos 120° = -cos6O° =- 

2 

tan 120° = - tan 60° = -...[3 

...[3 

cot 120° = -cot60°=- 

3 

sec 120° = -sec60° =-2 

2...[3 

esc 120° = csc60°=- 

3 

13. 300° 

This angle is in quadrant IV, so the reference 

angle is 360° - 300° = 60°. 

Since 300° is in quadrant IV, the sine, tangent, 

cotangent and cosecant are negative. 

. 3000 . 600 ...[3 

sIn =-SIO =- 2 

1 

cos 300° =cos 60° = 

2 

tan 300° =- tan 60° = -...[3 

...[3 

cot 300° =-cot 60° =- 

3 

sec 300° =sec 60° = 2 

2...[3 

csc 300° =- csc 60° =-- 

3 

14. -225° 

-225° is coterminal with -225° + 360° = 135°. 




sin(-225 0) = sin 45° = .,fi 

2 

.,fi 

cos(-225°) = -cos45°=- 

2 

tan(-225°) = -tan 45° =-1 

cot(-225°) = -cot 45° =-1 

sec(-225°) =-sec 45° =-J2 

csc(-225°) =csc45° =.,fi 

75



15. -390° is an angle in quad rant IV with a reference 

angle £)'= 30°. 

sin(-390°) = - sin 30° = _! 

2 

..[j 

cos(- 390°) = cos 30° = 

2 

tan(-390°) = - tan 30° = - ..[j 

3 

cot(-390°) = - cot 30° = -..[j 

2..[j 

sec(-390°) = sec 30° = - 

3 

csc(-3900) = -csc30° =-2 

16. sin£)=_! 

2 

Since sin £) < 0, £) is in quadrant III or IV. Since 

sin £) = _!, the reference angle is 30°. 

2 

In quadrant III, £) = 180° + 30° = 210°. 

In quadrant IV, £) = 360° - 30° = 330°. 

1 

17. cos£)=- 

2 

Since the cosine is negative, £) is in quadrant II or 

quadrant Ill, Since cos 60° = !, £)' = 60°. 

2 

In quadrant II, 

e =180° - £)'= 180° - 60° = 120°. 

In quadrant III, 

e = 180° + £)'= 180° + 60° = 240°. 

18. cot £) =1 

Since cot £) < 0, £) is in quadrant 11 or IV. Since 

cot e = -1, the reference angle is 45°. 

In quadrant II, £) = 180° - 45° = 135°. 

In quadrant IV, £) = 360° - 45° = 315°. 

2..[j 

19. sec£)=-- 

3 

Since the secant is negative, £) is in quadrant II or 

quadrant III. Since sec 30° = 2..[j £)' = 30°. 

3, 

In quadrant II, 

£) = 180° - £)' = 180° - 30° = 150°. 


£) = 180° + £)'= 180° + 30° = 210°. 

2 

20. cos600+2sin 30°= ~+ 2(~r 

=~+2(±) 

1 1 

=-+ 

2 2 

=1 

2 

21. tan 120 0-2cot2400 = (_..[j)2 -2(~) 

7 

= 

2 

=3- 2..[j 

3 



the type andlor model of calculator being used. 

24. sin 72° 30' 

72030' = (72+ ~)0 "= 72.5 

Enter: 72.5 ~ 

or ~ 72.5 ~ 

Display: .95371695 

25. sec 222° 30' 

222°30' =(222 + ~~) 0 

= 222.50 

Enter: 222.5 ~ (jill 

or CD ~ 222.5 CD IE:) ~ 

Display: -1.3563417 

26. cot 305.6° 

Enter: 305.6 ~ (jill 

or CD ~ 305.6 CD IE:) ~ 

Display: -.71592968 

27. esc 78°21' 

78° 21'=(78+ ~~y =78.35°2 

Enter: 78.35 8 (jill 

or CD ~ 78.35 CD IE:) ~ 

Display: 1.0210339 

76



28. sec 58.9041° 34. sec 8 = 1.2637891 

Enter: 58.9041 @[) @I 

or OJ @[) 58.9041 CD IE:) E!§ 

Display : 1.9362132 

29. tan 11.7689° 

Enter: 11.7689 (E) 

or ~ 11.7689 E!§ 

Display: .20834446 

Enter : 1.2637891 @I {INV AR~~OS 

or 

ICOII"1 

or ~ @[) 1.2637891 IE:) E!§ 

Display : 37.695528 

8 = 37.695528° 

30. If 8 = 135°,8 = 45° . 

If8=45°, 8=45°. 

If 8 = 300°, 8 = 60°. 

If 8 = 140°, 8 = 40°. 

Of these reference angles 40° is the only one 

which is not a special angle, so (d) tan 140° is the 

only one which cannot be determined exactly. 




31. sin 8 = .8254121 

AR~SIN 

Enter: .82584121{INV 8 

or 

1·..·'1 

or 0 8 .82584121 ~ 

Display: 55.673870 

8 = 55.673870° 

32. cot 8 = 1.1249386 

Enter' 1.1249386 

or 

Gill {INV ARCTAN 

or 

IT.....J 

or IE) (E) 1.1249386 IE:) E!§ 

Display: 41.635092 

8 = 41.635092° 

33. cos 8 = .97540415 

Enter: '97540415{INV AR~~OS 

or 

[cos"J 

or 0 @[) .97540415 ~ 

Display: 12.733938° 

8 = 12.733938° 

35. tan 8 = 1.9633124 

Enter: 1.9633124{INV AR~~AN 

or 

IT.....I 

or 0 (E) 1.9633124 ~ 

Display: 63.008286 

8 = 63.008286° 

36. esc 8 = 9.5670466 

Enter :9.5670466 @I {INV AR~SIN 

or 

1"'~ 1 . 

or ~ 8 9.5670466 IE:) E!§ 

Display: 5.9998273 

8 = 5.9998273 0 

37. sin 8 = .73254290 

Since sin 8 is positive, there will be one angle in 

quadrant I and one angle in quadrant IT. If 8', is 

the reference angle, then the two angles are 8' 

and 180° - 8'. 

Enter: .73254290 INV 8 . 

or ~ 8 .73254290 ~ 

Display: 47.100000 

8'=47.1 0 

180° - 8' = 180° - 47.1° = 132.9° 

38. Since tan 8 = 1.3865342, 8 is in quadrant I or 

quadrant III. If 8' is the reference angle, the two 

angles will be 8' and 180° + 8'. 

Enter: l.3865342{INV AR~~AN ~ 

or 

[T.....I 

or IE) (E) 1.3865342 ~ 

Display: 54.200000 

8' = 54.2° 

180 0 + 8' = 180 0 + 54.2° = 234.2° 

77



39. sin 50° + sin 40° 1. sin 90° 

sin 50° + sin 40° = 1.4088321 and 

sin 90° = 1 

False 

40. cos21O° 1. cos 180° · cos 30° -sin 180° ·sin30° 

cos 210° = _ .J3 

2 

cos 180° .cos 30° - sin 180° .sin 30° 

=(-1)( ~J-(o{-}) 

.J3 

=- 

2 

True 

41. sin 240° 1. 2sin 120° cos 120° 

sin 240° = - sin 600 = _ .J3 

2 

2sin 120°· cos 120° = 2(sin600)(-cos600) 

=2( ~J(--}) 

True 

.J3 

=- 2 

42. sin 42° + sin 42 1. sin 84 

Using a calculator. we have 

sin 42° + sin 42° = 1.3382612. 

sin 84° = .99452190 

False 

1 

43. No. cot 25° = -- 

tan 25° 

Enter: 25 ~ om 

or OJ ~ 25 CIJ 0 ~ 

Tan- 125 is calculating the angle whose tangent 

is 25. 

46. e = 4000° 

cos40000= .766044443 

sin 4000° = .642787610 

Since cosine and sine are both positive. the 

angle (} is in quadrant I. 

47. A =58° 30'. c= 748 

A+B=90° 

B=900-A 

B = 90° - 58°30' 

B= 31°30' 

sin A =!: 

c 

a = csinA 

a = 748sin58°30' 

a=638 

b 

cos A = 

c 

b=ccosA 

b =748 cos 58°30' 

b=391 

48. a =129.70. b = 368.10 

tan A = 129.70 

368.10 

A = 19.41° = 19°25' 

tan B = 368.10 

129.70 

B = 70.59° = 70°35' 

Check that A + B =90°. 

c=~a2 +b 2 . 

=~129.702 +368.10 2 

c=390.28 

49. A =39.72°. b = 38.97 m 

A 

44. (} = 2976° 

cos 2976° = - .104528463 

sin 2976° = .994521895 

Since cosine is negative and sine is positive. the 

angle (} is in quadrant II. 

45. (} = 1997° 

cos 1997° = -.956304756 

sin 1997° =-.292371705 

Since sine and cosine are both negative. the 

angle (} is in quadrant III. . 

78



A+B=900=C 53. The diagonal d of the rectangle is the hypotenuse 

B=900-A 

of a right triangle whose legs are the adjacent 

B = 90° - 39.72° 

sides of the rectan Ie. 

= 50.28° 

tanA=!: 

b 

a=btanA 

10 .93cm 

a = 38.97 tan 39.72° 35.65° 90° 

a =32.38 m 

15.24cm 

b 

By the Pythagorean theorem, the length d of the 

cosA=c 

diagonal is 

b 

c=-d = ~r(I-5-.2-4)--:2-+-(-10-.-93-)7"2 "" 18.75 em, 

cos A 

38.97 

An alternative method gives the same answer. 

c=--- 

cos 39.72° Sin . 35 

. 650 =- 10.93 

c=50.66 m 

d 

d = 10.93 

SO. B = 47° 53', b = 298.6 m sin35.65° 

"" 18.75 em 

90· 

A 298.6m C 

C=9O° 

A = 90° - 47°53' 

= 42°7' 

B 

54. Let s = the length of one of the equal sides of an 

isosceles triangle. Divide the isosceles triangle 

into two congruent right triangles. 

49.28m 

....--..-- 

-Q- = tan 42°7' 

d 

298.6 1 

a = 298.6(tan 42°7') 

d = -(49.28 m) = 24.64 m 

2 

a =270.0 m 

(J =.!..(58.746°) = 29.373° 

_. _c_ = esc 47°53° 2 

298.6 

• £I d 

c = 298.6(csc47°53') Sin 17 = 

S 

c=402.5 m 

Sin. 

. 29 3730 =- 24.64 

51. Let x = height of the tower. s 

24.64 

tan 38020' = _x_ 

93.2 

sin 29.373° 

s =50.24 

x = 93.2 tan 38°20' 

Each side is 50.24 m long. 

x "" 73.693005 

.The height of the tower is 73.7 ft. 

52. Let h = height of the tower . 

h 

tan 29.5° =- 

36.0 

h = 36.0 tan 29.5° 

h ""20.4 

The height of the tower is 20.4 m. 

s=--- 

79



55. Draw triangle ABC and extend the north-south 

lines to a point X south of A and S to a point Y, 

north of C. 

N 

57. Suppose A is the car heading south at 55 mph, 

B is the car heading west, and point C is the 

intersection from which they start. 

After two hours, by d = rt, 

AC= 55(2) =110 

Since angle ACB is a right angle, triangle ACB is 

a right triangle. 

Since the bearing of A from B is 324°, 

angle CAB = 360° - 324° = 36°. 

w B C 

S 

AngleACB = 344° - 254°:;:; 90°, so ABC is a 

right triangle. 

Angle BAX = 32° since it is an alternate interior 

angle to 32°. 

Angle YCA = 360° - 344° = 16° 

Angle XAC = 16° since it is an alternate interior 

angle to angle YCA. 

Angle BAC = 32° + 16° = 48° 


AC 

cosA= 

AB 

AB= AC 

cos A 

AB=~ 

cos 48° 

= 1200. 

The distance from A to B is 1200 m. 

56. Draw triangle ABC and extend north-south lines 

from points A and B. Angle ABX is 55° (alternate 

interior angle 3 of parallel lines cut by a 

transversal are congruent) so Angle ABC is 

55° + 35° =90° . 

N N 

A 

x 

c 

6'o~ 

B 

Since angle ABC is a right angle, use the 

Pythagorean theorem to find the distance 

from A to C. 

(AC)2 =80 2 +74 2 

(AC)2 =6400 + 5476 

AC = ~ll,876 

= 110 km 

It is 110 km from A to C. 

s 

A 

AC 

cosCAB= 

AB 

110 

AB= AC =__ 

cos CAB 

::: 140 mi 

58. Let x =the leg opposite angle A. 

x h+x 

tan A =- and tan B :;:;- 

k 

k 

x=ktanAandx=ktanB- h 

Therefore, 

k tan A =k tan B - h 

h = k tan B - k tan A 

h =k(tan B - tan A) . 

side adjacent AB 

61. cot () = = 

side opposite OA 

Since OA =r =1, cot() =AlJ 

62. Label the point (0, 1) as P. 

sec () = hypotenuse =OC 

side adjacent OP 

Since OP =r =I , sec () =OC 

63. esc () = hypotenuse = OB 

side opposite OA 

Since OA =r = I, esc () = OB. 

64. (a) Let R:;:; 3955, T= 25, P = 140. 

h = {.os(~) 1) 

= 395{ co,(~) -I) 

=716 mi 

80



(b) Now let T~ 30. 

h-J 1 -1] -"l cos(~) 

~395{ cos (:Vl 1] 

'" 1104 mi 

In order to increase the communication 

time T, the satellite must have a higher orbit. 

65. (a) From the figure in the text, 

xQ-xp 

sin 8 =~'----'- 

d 

xQ = x p +dsin8. 

Similarly, 

YQ -YP 

cos 8 =-=-- 

d 

YQ =Yp +dcos8. 

(b) Let (x p ' Yp)= (123.62,337.95), 

8 = 17° 19' 22", and d = 193.86. 

xQ = x p +dsin8 

=123.62 + 193.86sin 17°19'22" 

=123.62 + 193.86sinI7.3228° 

'" 181.34 

YQ =Yp +dcos8 

=337.95 + 193.86cos 17°19'22" 

=337.95 +193.86 cos 17.3228° 

'" 523.02 

The coordinates of Q are 

(181.34,523.02). 


1. sin A = side opposite =12 

hypotenuse 13 

cos A =side adjacent =~ 

hypotenuse 13 

tan A = side opposite =.!3. 



cot A = = 


sec A = hypotenuse =~ 


esc A = hypotenuse =~ 


2. To find w: 

sin 30° =.i 

w 

4 

w=- 

sin 30° 

4 

w= 

(!..) 

2 

w=8 

To find x: 

4 

tan45°=x 

4 

x=- 

tan 45° 

4 

x= I 

x=4 

To findy: 

cos 30° =1:. 

w 

cos 30° = 1'. 

8 

8cos30° =Y 

/3 

8·-=y 

2 

4/3 =y 

To find z: 

sin 45° =.i 

z 

4 

z=- 

sin 45° 

2 

z=- 

(..[2) 

2 

z=4...[i 

3. sin (B + 15°)=cos (2B + 30°) 

Since sine and cosine are cofunctions, the 

equation is true when 

(B + 15°) + (2B + 30°) =90° 

3B+45 =90° 

3B=45° 

B=15° 

This is one solution; others are possible. 

4. sin 8 =.27843196 

INV 8 

or 

Enter: .27843196 ARCSIN 

81



5. 

or lEJ ~ .27843196 ~ 

Display: 16.16664145 

(J=16.16664145° 

-.,fi 

cos(J=- 

2 

Cosine is negative in quadrants II and III. The 

reference angle (J' is 45°. In quadrant II (J is 

180° - 45° = 135°. In quadrant III (J is 

180° + 45° = 225°. 

6. tan (J =1.6778490 

To find cot (J, 

Enter: 1.6778490 (ill) 

or 1.6778490 0 ~ 

Display: .5960011896 

cot (J =.5960011896 

7. (a) sin24° J: sin 48° 

Sin (Jincreases from 0 to 1in the interval 

0°::';(J::';90°. Therefore sin 24° is less 

than sin 48° . 

True 

(b) cos24°



Angle ABX is 55° (alternate interior angles of 

parallel lines cut by a transversal are congruent). 

Angle ABC is 55° + 35° = 90°, therefore triangle 

ABC is a right triangle. 

AC 2 =S02 +74 2 

AC=~S02 +74 2 

AC=llOkm 

k 

k.Jh 

t =-- or -- seconds 

4h 

4.Jh 

14. v=32tsin9 

v =32(k;';)(I) 

v = s.Jh ft per sec 

The velocity depends only on the height h; 

the length k does not affect the velocity . 

.. 

83


84 Chapter 3 Radian Measure and the Circular Functions 

CHAPTER 3 RADIAN MEASURE AND THE 

CIRCULAR FUNCTIONS 

Section 3.1 

Exercises 

1. Since 0 is in quadrant I, 

n 

0



34. 

lltr = IlJrC800) =660 

30 30 tr 

35. 39° = 39(...!!..- radian) = .68 radian 

180 

36. 74° = 74(~ radian) = 1.29 radians 

180 

37. 42°30' = 42.5(~ radian) = .742 radians 

180 

38. 53°40' = 53.6667(...!!..- radian) = .937 radians 

180 

39. 139°10' = 139"!'(...!!..- radian) = 2.43 radians 

6 180 

40. 174°50' = 174.833\...!!..- radian) = 3.05 radians 

180 

41. 64.29° = 64.29(.s: radian) 

180 

= 1.122 radians 

42. 85.04° = 85.04(...!!..- radian) = 1.484 radians 

180 

43. 56025' = 56 25 ( .s: radian) 

60 180 

= .9847 radian 

44. 122°37' = 122.61~...!!..- radian) = 2.140 radians 

. 180 

45. 47.6925° = 47.6925(.s. radian) 

180 

= .832391 radian 

46. 23.0143° = 23.0143(.s. radian) 

180 

= .401675 radian 

0) 

47. 2 radians = 2 C80 K 

= 114.5916° 

= 114° + (.5916)(60') 

= 114°35' 

48. 5 radians = 5 C 

K 8OO) 

= 286.4789° 

= 286° + (.4789)(60') 

= 286°29' 

0) 

49. C80 

1.74 radians = 1.74 K 

=99.6947° 

= 99° + (.6947)(60') 

=99°42' 

0) 

50. o{180 

3.06 radians = 3. K 

= 175.3251° 

= 175° + (.3251)(60') 

= 175°20' 

51. .3417 radians = .34 17 (180°) K 

= 19.5780° 

= 19° + (.5780)(60') 

= 19°35' 

52. 9.84763 radians = 9.8476 \180°) K 

= 564.228° 

= 564° +



58. cos : = cos(i,180°) Substitute 180° for TC. 67. sin 5: =sin(~'1800) 

= cos30° = sin 150° 

..[3 1 

= 

= 

2 

2 

59. tan~ = tan( ~ . 1800) 68. tan 5: = tan( ~ . 1800) 

= tan45° 

= tan 150° 

=1 ..[3 

=- 

3 

60. cot ~ = cot( ~ .1800) 

69. cos 3TC = cos(3. 180°) 

= cot 60° = cos540° 

=-1 

..[3 

= 

3 70. see TC = sec180° 

=-1 

61. sec: = sec(i,180°) 

71. sin 4 TC ~ sin( i .1800) 

= sec30° . 3 3 

2..[3 = sin 240° 

3 ..[3 

=- 

2 

62. csc~ = esc( ~ '180°) 

= esc 45° 72. cot(- 2;)= co{-~'1800) 

=../2 = cot(-120°) 

63. sin ~ = sin(± , 180°) 

..[3 

= 

3 

= sin90° 

=1 73. Sin(- 7:) = Sin(-i ,180°) 

= sin(-2100) 

64. esc ~ = esc(± ,180°) 1 

= 

2 

= esc90° 

=1 

74. cos(- :)= cos(-i, 1800) 

65. tan 2;= tan(~ ,180°) = cos(-300) 

= tan120° ..[3 

= 

=-..[3 2 

66. cot3 

2TC 

= cot (2 "3' 180° ) 

= cot120° 

..[3 

=- 

3 

86



75. Answers start at 30° and proceed 

counterclockwise around circle. 

30° =30 (..!!.... radian) =!E. radian 

180 6 

x . tr (1800) 

- radian =- -- =45° 

4 4 tr 

60° =60 (..!!.... radian) =tr radians 

180 3 

2tr diai 2tr (1800) 1200 

- ra lans = - -- = 

3 3 tr 

3tr. 3tr (1800) 

4 4 tr 

- radians =- -- =135° 

150° = 150(..!!.... radian) = 5tr radians 

180 6 

180°=180(..!!.... radian) =zrradians 

180 

210° =21O(..!!.... radian) =7tr radians 

180 6 

225° =225 (..!!.... radian) =5tr radians 

180 4 

4tr. 4tr (1800) 

3 3 tr 

- radians =- -- =240° 

- 

5tr. 

radians = - 

5tr (1800) 

-- = 300° 

3 3 tr 

315° =315(..!!.... radian) =7n radians 

180 4 

330° = 330(..!!.... radian) = 1m radians 

180 6 

76. In degree mode, using tan 9 =!..., 

x 

tan9 

6.3018678 

= 

9 

(} =35°. 

In radian mode, using tan 9 =!..., 

x 

tan(} =6.3018678 

9 

(} = .611 radian. 

9 n tr 

77. (a) _ ._= 

10 180 200 

(b) Degrees: 3.5 x .9° =3.15° 

Radians: 3.5 x ..!!.... ::; .055 radian 

200 

79. In each rotation, the pulley would rotate 2tr 

radians. 

(a) In 8 rotations, the pulley would tum 

8 . 2tr =2tr radians. 

(b) In 30 rotations, the pulley would tum 

30 . 2tr =60tr radians. 

80. (a) The hour hand completes 2 revolutions in 

24 hr. 

One revolution is 2tr. 

(} =2(2tr) = 4tr 

(b) In 4 hr it rotates through 

(} =C~)2tr) =2; . 

12 

81. In each rotation around the earth, the space 

vehicle would rotate 2tr radians. 

(a) In 2.5 orbits, the space vehicle travels 

(2.5)' 2tr= 5trradian~. 

(b) In ±of an orbit, the space vehicle travels 

3 

(~) . 2tr =8: radians. 

82. (a) S =1367, N = 80 

.1S =.034Ssin [2tr(82.5 - N)J 

365.25 

=.034(1367)sin [2tr(82.5 - 80)J 

365.25 

::; 1.998 watts per square m 

(b) S =1367, N = 1268 

.1S=.034S sin [2tr(82.5 - N) J 

. 365.25 

= .034(1367) sin [2tr(82.5 -1268)J 

365.25 

::; -46.461watts per square m 

78. The values of sin tr and cos tt are 0 and -1, 

respectively, when zrrepresents a measurement in 

radians, not degrees. Therefore, the calculator was 

in radian mode 

87



(c) The maximum value of!!.S would occur 3n 

5. 5=6n (}=when 

, 4 

. [2n(82.5 - N)J 1 5 = r(} 

sm = . 

365.25 5 6n 

In this case !!.S would be equal to 

r=(j=Jir 

T 

.034S =.034(1367) 4 

=46.478 watts per square m =6n· 

3n 

=8 

(d) !!.S= 0, so 

N)J 0 

sm 

. [2n(82.5 - 

= . 

n 

6. 5=3n (}= 

365.25 , 2 

This occurs when N = 82.5,265.125,447.75, 

5 =r(} 

630.375,813,995.625,1178.25, or 5 3n 

r=-= 

1360.875. Since N represents a day number, 

(} t 

which should be a natural number, we might 

interpret day 82.5 as noon on the 82nd day. 

Section 3.2 =6 

Connections (page 112) 7. r=3,5=3 

Answers will vary. The longitude at Greenwich is 0°. 

Exercises 

1. 5 =r(} 

1 

2 

=3n· 

5 =r(} 

x 

(}=~=~=1 

r 3 

8.5=6,r=4 

5 = r(} 

1=-(} 5 6 3 

2 

(} = - = - = - or 1.5 

r 4 2 

(}=2 

The gas gauge rotates through 2 radians. 

·2 (} 2n di 

9• r ='1 .3 em, =- 'ra tans 

. . 3 · 

2. Since the seesaw is 12 ft long, the radius of the 

seesaw is 6 ft. 

s =r(} 

3 =6(} =8.2n em 

1 

(}= 

::::25.8 em 

2 

The board rotates through '2 1 radian. 

5 =r(} =12.3(2;) em 

10. r =.892 em, (} =lIn radians 

10 

n 

3. r=4 (}= 

, 2 

5 =r(} =.892(\1;) cm 

=.9812n cm 

s = r(} = { ; ) = 2n 

:::: 3.08 em 

11. r =4.82 m, (} =60° 

Convert (} to radians. 

4. r =12 

, 

(} = n 

3 

n 

(} =60°=_ 

3 

s = re = 12( ~ ) = 4n 

s =r(} =4.82( ~ ) 

= 1.61n 

::::5.05m 

88



12. r= 71.9 em, 0= 135° 

Convert 0 to radians. 

135° = 135(-.!!....- radian) = 31t' radians 

180 4 

s =rO =71.ge: )cm =53.9251t'cm 

== 169 em 

13. r = 2 in., s = 5 in. 

s=rO 

o= ~ =~ or 2.5 radians 

r 2 

14. 0 =2 radians, s =3 ft 

s=rO 

s 3 

r =- =- or 1.5 ft 

o 2 

For Exercises 19-24, note that since 6400 has two 

significant digits and the angles are given to the nearest 

degree, we can have only two significant digits in 

the answers. 

19. 9° N, 40° N 

o= 40° - 9° = 31° 

s=rO 

nt1 311t') 

=64 \180 

== 3500 km 

= 31 (-.!!....- radian) 

180 

=- 

311t' 

ra 

di 

Ian 

180 

20: 36° N, 49°N 

o= 49° - 36° = 13° 

15. 0 =1t' radian, s =4 in. 

5 

s=rO 

s 4 

r=-=o 

1 

=4 '~ 

1t' 

20 . 

=-In. 

1t' 

s=rO 

s =~AntI 131t') 

~180 

== 1500 km 

= 13 (~ radian) 

180 

=- 131t' ra diIan 

180 

16. s = 30 em, r = 5 em 

s=rO 

o=-=-= 

s 30 6 

ra 

d"' 

ians 

r 5 

17. The formula for arc length is s = rO. Substituting 

2r for r we obtain s = aO. The length of the arc is 

doubled. 

18. Recall that z radians =180°. If 0 is measured in 

degrees, then the formula becomes 

91r 1t'r0 

s=r·-=-. 

180 180 

22. 45° N, 34° S 

34° S =-34° N 

0= 45°- (-34°) =79° 

s=rO 

s = 64nt1 79Jt) 

\180 

== 8800 krn 

=-- 

79Jt 

ra 

d' 

lans 

180 

89



23. r= 6400 km, s =1200 km 

s= r8 

1200 =64008 

8=1. 

16 

Convert l.- radian to degrees. 

16 

0) 

8 = 1.-(180 '" 11° 

16 1r 

The north-south distance between the two cities 

is 11°. 

Let x =the latitude of Madison. 

x-33°=II° 

x=44° N 

24. r= 6400 km, s =1100 Ian 

s =rO 

1100 =64000 

.!..!. =0 

64 

Convert .!..!. radian to degrees. 

64 

e= .!..!..180° '" 100 

64 1r 

The north-south distance between the two cities 

is 10°. 

Let x =the latitude of Toronto. 

x-33° = 10° 

x=43° N 

25. (a) The number of inches lifted is the arc length 

in a circle with 

r = 9.27 in. and 0 = 71° 50'. 

71050' = (71 ° + 50 

60 180° 

s =rO 

0 

)(...!E-) 

= 9.27(71 ° + 50·X...!E-) 

60 180° 

= 11.6 

The weight will rise 11.6 in. 

(b) When the weight is raised 6 in., 

s =rO 

O=!.. 

r 

=_6(1800) 

9.27 1r 

= 37.085° = 37° + (.085)(60') 

= 37°5'. 

The pulley must be rotated through 37° 5'. 

26. To find the radius of the pulley, first convert 51.6° 

to radians. 

o= 51.6° = 51.6 (...!!-) radians 

180 

Now substitute this value of 0 and s = 11.4 em 

into the equation s = rO and solve for r. 

s =rO 

11.4 = r(51.6)(~) 

180 . 

11.4 

r =-"""7"----, 

51.6(rn,) 

r = 12.7 

The radius of the pulley is 12.7 em. 

27. A rotation of 

o= 60.00 = 1r 

3 

on the smaller wheel moves through an arc length 

of 

s = rO =5 .2'~) =5.48 em. 

Since both wheels move together, the larger 

wheel moves 5.48 em, which rotates it through an 

angle 0, where 

5.48=8.160 

0= 5.48 = .671 radian 

8.16 

= .67 { 1 ~ 0 ) ':" 38.50. 

The larger wheel rotates through 38.5°. 

28. The arc length s represents the distance traveled 

by a point on the rim of a wheel. Since the two 

wheels rotate together, s w~1I be the same for both 

wheels. 

For the smaller wheel, 

o= 80° = 8rI...!!-) = 41r radians. 

\180 9 

s= -o = 11.\~) "" 16.336 em. 

For the larger wheel, 

o= 50° = 50 (...!!-) =51r radian. 

180 18 

s =rO 

16.336 = {~;) 

16.336 

r =---s;r- 

IB"" 

r = 18.7 

The radius of the larger wheel is 18.7 em. 

90


29. The chain moves a distance equal to the arc length 

on the larger gear. So, for the large gear and 

pedal, 

s = r() 

= 4.72(180)...!:. 

180 

= 4.72". 

Thus, the chain moves 4.72" in. The small gear 

rotates through an angle 

()=!.. 

r 

4.72" 

=- 

1.38 

=3.42". 

()for the wheel and ()for the small gear are the 

same, or 3.42". So, for the wheel, 

s =r() 

= 13.6(3.42,,) 

= 146 in. 

The bicycle will move 146 in. 

30. (a) In one hour, the truck travels 55 mi. Since 

the radius is given in in., convert 55 mi 

to in. 

s=55 mi 

= 55(5280) ft 

= 55(5280)(12) in. 

= 3,484,800 in. 

s =r() 

3,484,800 in. = (14 in.)() 

248,914 .29 radians = () 

Each rotation is 2" radians. Thus, the number 

of rotations is 

9 ' 248,914.29 

-=---'-- 

2" 2" 

= 39,616 rotations. 

(b) Find s for the 16-in. wheel. 

s=r9 

= (16 in.)(248,914.29) 

= 3,982, 628.64 in. 

=62.9 mi 

The truck with the 16-in. tires has gone 

62.9 mi in one hour, so its speed is 62.9 mph. 

Yes, the driver deserves a ticket. 


31. Let t = the length of the train. t is approximately 

the arc length subtended by 3° 20'. 

9=3°20'=3,( 

3 

() =(3~)C:0) 

= 

54 

" 

t = r() = 3.5(~) 

=.20 km 

The tra~n is about .20 km long. 

32. Let L = the length of the ship. 

L is approximately the arc length subtended 

by 1.5°. 

L= r() 

= (2.3)(1.5)(...!!-) 

180 

=.06 km 

The length of the ship is about.06 km (or 60 m), 

33. Let d = the diameter of the moon. d is 

approximately the arc length subtended by .!.0. 

2 

tt 

= 

360 

d=r9 

= 240,000(3:0) 

= 2100 mi 

The diameter of the moon is about 2100 mi. 

34. Let r = the distance of the boat. The height of the 

mast, 32 ft, is approximately the arc length 

subtended by 2° 10'. 

9 = 2° 10' = 2.!. (...!!-) 

6 180 

s 

r= 9 

32 

r=- 

13" 

=- 

1080 

1311" 

1080 

= 850 ft 

The boat is about 850 ft away. 

91



x 

3rr 

35. r=6 

, 

()=- 

3 

41. r=52 em, ()= 

10 

1 2 

A=-r() 

rr) 

A = .!..(52)2e 

2 2 10 

= 405.6rr 

=~(6)2(~) 2 

'" 1300 em 

= 6rr 

x 

42. r = 25 mm, () =!!.... radian 

36. r=S, ()=- 15 

2 

A = '!"r 2 A = '!'(25)2(!!"") 

() 2 15 

2 

125rr 

=- 

= ~(S)2(~) 6 

=16rr 

37. A = 3 units 2, r= 2 

2 

"'65 mm 

43. r =:: 12.7 em, () =:: SI° 

2 

A =.!. r 2 () 

The formula A = .!. r () requires that () be 

2 2 

measured in radians. Convert SI ° to radians by 

3 = '!"(2)2() 

2 multiplying by ...!E...-. 

3=2() ISO 

() = ~ or 1.5 radians () = S{ ...!E...-) =:: 9rr 

2 ISO 20 

38. A = S units-, r = 4 A = .!.(l2.7)2(9rr) 

2 20 

1 2 

A=-r() 

= 36.29rr 

2 

'" 114 em 2 

S=.!..(4)2() 

2 44. r = IS.3 m, () =125°· 

S = S() 

() = 1 radian The formula A = .!.r 2 () requires that () be 

5rr 

39. r=29.2 m,()= 

2 

measured in radians. Convert 125° to radians by 

. 6 multiplying by ..!!-. 

ISO 

A =.!. r 2 () 

2 A = .!..(lS.3)2(l25{...!E...-) 

2 180 

= ~(29.2)2( 5:) = 116.2Srr 

= 355.271t = 365 m 2 

'" 1120 m 2 

45. A = 16 in. 2, r;;;:: 3.0 in. 

40. r;;;:: 59.8 km, () = 2rr radians A =.!..r 2 () 

3 2 

A =.!. r 2 () 

16 = '!'(3)2() 

2 2 

9 

16 =-() 

;;;::~(59 .8)2(2:) 

2 

() '" 3.6 radians 

;;;:: 1192.01rr 

2 

'" 3740 lan 

92



46. A =36 ft2, () = lC radian 

4 

A =.!. r 2 () 

2 

36= ~r2(:) 

36 = lC r2 

8 

288 2 

-=r 

lC 

r == 9.6 ft 

47, A =64 m 2 , () = lC radian 

6 

A =.!. r 2 () 

2 

64=~r2(:) 

64 =.!!-r 2 

12 

768 2 

-=r 

lC 

r==16 m 

48. A =25 in. 2 , r= 10 in. 

A =.!. r 2 () 

2 

25=.!.(l0)2() 

2 

25 =500 

() =.50 radian 

2 

49. A =.!. r () 

2 

Substituting () =2lC, 

2(2lC) 

=1CT 2 

A =.!. r . 

2 

2 

The area of a circle of radius r is tcr . 

50. The formula for the area of a sector is 

A =.!. r 2(). Substitute 2r for r. 

2 

A =.!.(2r)20 

2 

=4(~r20 ) 

The area of the sector is quadrupled. 

52. x 2 +i =4 is the equation of a circle of radius 2. 

y = ( "'7} is the equation ofa line with slope 

.fj 

- . Therefore, 

3 

lan() =.fj 

3 

lC 

()=-. 

6 

Substitute r =2 and () = lC in the formula for the 

6 

area of a sector. 

1 2 

A=-r() 

2 

=~(2)2(:) 

x 

= 3 

53. (a) Central angle in degrees: 

360° =13.!.° 

27 3 

Central angle in radians: 

2lC 

27 

(b) Circumference =2tcr; 

radius =76 ft, so . 

circumference = 2(76)lC 

"" 480 ft 

2lC 

(c) r =76 ft ()= 

, 27 

s =r()= 7{~~) 

"= 160 "" 17.8 ft 

9 

(d) Area of sector with r = 76 ft and 0 = 2lC is 

27 

A =.!. r 2 0 

2 

='!'(76 2 ) 2lC 

2 27 

""672 ft2. 

51. Recall that zrradians = 180°. If 0 is measured in 

degrees, then the formula becomes 

1 2 81C tcr 2 () 

A=-r -=- 

2 180 360 

93



54. (a) 

55. The area cleaned is the area of the sector "wiped" 

by the total area and blade minus the area 

"wiped" by the arm only . 

10 - 7 =3, so the arm was 3 in. long. 

1 2(951r) . 2 

Aarm only =2" (3) 180 == 7.5 m, 

The triangle formed by the central angle and 

the chord is isoceles . Therefore, the bisector 

of the central angle is also the perpendicular 

bisector of the chord. 

sin Zl = 50 

r 

50 

r=- 

sin 21° 

= 140 ft 

(b) r=~' 0=42° 

sin21° ' 

Convert 0 to radians: 

(c) 

42(-!!..... radian) = 71r radians 

180 30 

s =rO 

50 71r 

=-_. 

sin 21° 30 

351r 

= 3sin 21° 

= 102 ft 

1 ( 50 )2(71r) 

Asector = 2" sin 21° 30 

== 7135 ft 2 

Atriangle =2" 1 bh 

= .!.(IOO)(~) 

2 tan 21° 

= 6513 ft 2 

The area bounded by the arc and the chord is 

7135 - 6513 = 622 ft2. 

1 2(951r) . 2 

A arm and blade =-(10) -- =S2.9tn. 

2 180 

The area of the region cleaned was about 

82.9 -7.5 = 75.4 in. 2 . 

56. Use the Pythagorean Theorem to find the 

hypotenuse of the triangle, which is also the 

radius of the sector of the circle. 

r 2 = 30 2 +40 2 

r = ../900+ 1600 

= ../2500 

=50 

The total area of the lot is the sum of the areas of 

the triangle and the sector. 

Atriangle = 2" 1 bh =2" 1 (30)(40) 

=600 yd 2 

.tl = .!.r 20 = .!..(50)2(60 .-!!.....) = 12501r yd 2 

'"Sector 2 2 ISO 3 

12501r 2 

Total area = --+600 = 1900 yd 

3 

57. The angle of inclination of the sun's rays at 

Alexandria is equal measure to the central angle 

that subtends the arc from Alexandria to Syene 

(5000 stades) because these angles are 

corresponding angles formed by a transversal that 

intercepts the parallel rays of the sun. To find the 

radius r of the earth, use 

s = rO, with s = 5000, and 0 = l...(21r) =.!!.... . 

50 25 

5ooo=r(~) 

125,000 

r = 

== 39,78S.736 stades 

1r 

C = 2Trr = 21r(125~000) = 250,000 stades 

d=2r 

= 2( 125~000) stades 

= 2e25~000}516.7) ft 

__ 2e25~OOO)(516.7) 

......:..__.!...-_- mi 

5280 

== 7800 mi 

94



58. Use Volume V =Area of base X height. e h e 

62. cos- = -, so h = rcos-. 

Area of base = .!.r 2 e 2 r 2 

2 

height =h 

63. d=r-h 

V=(~r2e )h) 

e 

=r- rcos- 2 

r 2 eh 

=-- 

2 

=r(I-COS~) 

(e is in radians.) 

. L 

64. Smcer=-, 

59. The area of the sector using the outside radius is (} 

1 2 

A, =-rl e. d =r(1-COS~) 

2 

The area of the sector using the inside radius is 

1 2 

d = ~ (1- cos ~) 

A2 =-r2 e. 

2 

To find the area of the base, take the larger area 65. (a) MQP is similar to 6.RMO, because 

minus the smaller area. angle R = angle R and angle Q = angle M. 

A = A,-A 2 

1 2 1 2 (b) OR RM 

=-r, e--r2 e -=- 

2 2 RP RQ 

1 (2 2) r £ 

="2e r, -r2 

_=.1.. 

To find the volume, multiply the area of the base 

c b 

2 

by the height. 

c 

rb= 

V=A ·h 

2 

= ~eh2 - r2 2)h. 

r=- 

c 2 

2b 

Note that these calculations assume that (} is 

measured in radians. 2 

(c) By the Pythagorean theorem, + b 2 2 

a = c . 

60. (a) Substitute 950,000 for A in the formula for 

c 2 a 2 + b 2 

r=-= 

area of a circle. 2b 2b 

2 

A =rrr 

2 (d) Since a = 1.4 and b =.2, 

950,000 = rrr 

1.4 2 + .2 2 1.96 +.4 2 

r =~950;rOOO r= = 

2(.2) .4 .4 

",,550m 

= 5 in. 

(b) 35° = 35(~) =71C radian 

180 36 

Section 3.3 

Exercises 

Substitute 950,000 for A and 71C for (} in 

36 

the formula for area of a sector. 1. Since (} = 1C , 

.... 2 

A =.!.r 2 (} 

2 sin 1C is a circular function. 

950 000 =.!. r 2 2 

( 71C ) 

, 2 36 

2. Since e= -150°, 

_ ~ 68.400.000 

r - 

71C 

cos(-150)0 is a trigonometric function. 

""1800 m 

3. Since e= 45°, 

tan 45° is a trigonometric function. 

61. Since L is the arc length, 

L 

L=r9 andr=e 

95



3Jr 

4 

II 

4• S·mce U;;- , 

3Jr) . . If ' 

cot 4 IS a circu ar unction. 

( 

II 

5 S· 5Jr 

• mce u;;-3' 

sec - 3 5Jr) IS . a circu . I ar fu nction. . 

( 

6. Since ();; 240°, 

esc 240° is a trigonometric function . 

7. Since ();; 3, 

tan 3 is a circular function. 

8. Since ();; -5, 

cos (-5) is a circular function. 

9. cos.8 "".7 

10. cos 4 "" - .65 

11. sin.2 "" .2 

12. sin 4 "" - .75 

13. x;; -.65 

when ();; 4 radians 

14. y=-.95 

when () "" 4.4 radians 

15. (a) Since 1.57 < 2 < 3.14, Jr < 2 < Jr, 

2 

2 radians is in quadrant II. Therefore, cos 2 

is negative. 

(b) Since -1.57 < -1 < 0, - Jr < -1 < 0, 

2 

-1 radian is in quadrant IV. Therefore, 

tan -1 is negative. 

(c) Since 4.71 < 5 < 6.28, 3Jr < 5 < 2Jr, 

2 

5 is in quadrant IV. Therefore, sin 5 is 

negative. 

(d) Since 4.71 < 6 < 6.28, 3Jr < 5 < 2Jr, 

2 

6 is in quadrant IV. Therefore, cos 6 is 

positive. 

· d . 7Jr 

16 • F 10 SIO-. 

6 

In radians, the reference angle for 7Jr is 

6 

7Jr Jr 

--Jr;;-. 

6 6 

7Jr .. dIll h he si . 

- IS 10 qua rant ,were t e sme IS negative. 

6 

. 7Jr . Jr 1 

SIO-=-Sln-=--. 

6 6 2 

Or, convert to degrees. 

7Jr = 2(1800) = 2100 

6 6 

SIO-=SIO . 7Jr . 2100 ;;-1 

6 2 

F 'nd 

5Jr 

3 

17• I COS- . 


3 

5Jr n 

2Jr--=-. 

3 3 

- 5Jr .. IS 10 qua d rant IV , were h t h e cosine .. IS 

3 

positive. 

5Jr Jr 1 

cos- = cos-= 

332 


.5Jr = ~(1800) = 3000 

3 3 

5Jr 

1 

cos - ;; cos 300° ;; 

3 2 

· d 3Jr 

18• F10 tan-. 

4 


4 

3Jr Jr 

Jr--=-. 

4 4 

3Jr is in quadrant II, where the tangent is 

4 

negative. 

3Jr Jr 

tan - ;; - tan - ;; -1 

4 4 

Or, convert to degrees . 

3Jr ;; ~(1800);; 1350 

4 4 

tan 3Jr ;; tan 135 0 =-1 

4 

96



· d . 5Jr 

19• F10 SIO-, 

3 

20• F10 


3 

5Jr Jr 

2Jr--=-. 

3 3 

- 5Jr IS . . 

10 qua d rant IV , were h t h e sine . IS , negative . . 

3 

. 5Jr ,Jr .J3 

s1o- = -s1o-=- 

3 3 2 


5Jr = ~(1800) = 3000 

3 3 

. 5Jr . 3000 .J3 

s1O-=s1o =- 

3 2 

" d 4Jr 

tan-. 

3 


3 

4Jr tt 

--Jr=-. 

3 3 

-4Jr IS .. 

10 qua d rant ill , were h th e tangent IS . 

3 

positive. 

4Jr Jr t: 

tan-=tan-=",3 

3 3 


4Jr = ~(1800) = 2400 

3 3 

4Jr 

h 

tan- = tan 240 0 = ",3. 

3 

21 F· d 2Jr 

• 10 sec-. 

3 

2Jr , 2Jr Jr 

The reference angle for - IS Jr - - = -, 

. 3 3 3 

2Jr is in quadrant II, where the secant is 

3 

negative. 

2Jr x 

sec-=-sec-=-2 

3 3 


2Jr = ~(1800) = 120 0 

3 3 

sec 2Jr = sec 120 0 =-2 

3 

' d IlJr 

22• F10 CSC-. 

6 

llJr , 11Jr Jr 

The reference angle for - IS 2Jr - - = - . 

6 6 6 

- 

11Jr 

IS 

. . d IV 

10 qua rant , where the cosecant is 

6 

negative. 

IlJr Jr 

csc- = -csc- =-2 

6 6 


IlJr =!.!.(1800) = 3300 

6 6 

IlJr 

csc- = csc330° =-2 

6 

23• Fm 

' d 

cot-. 

5Jr 

6 

5Jr . 5Jr Jr 

The reference angle for - IS Jr - - = -. 

6 6 6 

5Jr is in quadrant II, where the cotangent is 

6 

negative. 

5Jr Jr h 

cot - =- cot - =-V3 

6 6 


51t = ~(1800) = 150 0 

6 6 

cot 51t =cot 150 0 = -J3 

6 

24. Find cos(- 4;J 

4Jr " . I . h 2Jr . 

- - IS cotermma WIt -. 

3 3 

2Jr . 2Jr Jr 

The reference angle for - IS Jr - -. - =-. 

333 

2Jr " d II h h ." 

- IS m qua rant , were t e cosine IS 

3 

negative. 

cos(- 4;) = cos 2: =-cos~ =-~ 

Or, convert to degrees . 

2Jr = ~(1800 ) =1200 

3 3 

cos(- 4Jr) =cos 2Jr =cos 120 0 =_! 

332 

97



d 231C 

25. FindSin(_ 5:)- 27• Fm ' sec--. 

6 

51C . . I . h 71C 231C . bOd 

- - IS cotermma Wit -. 6 IS not etween an 21C. Subtract 21C. 

6 6 

71C. 71C 1C 231C _ 21C = 231C _ 121C = ll1C 

The reference angle for - IS - - 1t =-. 

6 6 6 6 6 6 6 

51C . . dr III h th ' . 

231C . . I . ll1C 

-- IS m qua ant • W ere e sine IS -- IS cotermma With--. 

6 

6 6 

negative. ll1C . 1m 1C 

The reference angle for -- IS 21C - - = - . 

. (51C) . 71C . 1C 1 

6 6 6 

sm -6 =sm 6=-sm'6=-2' 2~1C is in quadrant IV. where the secant is 

Or. convert to degrees. 

positive. 

_ 51C = -~(1800) = -1500 

6 6 231C ll1C 1C 2-./3 

sec--=sec-=sec- =- 

6 6 6 3 

sin(_ 5:) = sin(-150 0) =_~ 

Or. converting to degrees. 

1m =.!..!(1800) = 3300 

F· d 171C 6 6 

26• m tan--. 

3 231C ll1C ° 2-./3 

sec--=sec-=sec330 =--. 

171C is not between 0 and 21C. Subtract 21t twice. 663 

3 

that is. 41C. 

28• Fm · d csc-. 131C 

171C -41C= 171C _121C = 51C 

3 

3 3 3 3 131C _ 41C =131C _ 121C = 1C 

171C . . I ith 51C 3 3 3 3 

- IS cotermma WI -. 

3 3 131C . . I . h 1C 

-- IS cotermma Wit -. 

51C . 51C 1C 3 3 

The reference angle for - IS 21C - - = -. 

333 The reference angle for 1C is 1C . 

51C is in quadrant IV. where the tangent is 

· 3 3 

3 ~ is in quadrant I. where the cosecant is positive. 

negative. 3 

tan 171C = tan 51C = _ tan 1C = -.Ji 131C 1C 2-./3 

3 3 3 

csc-- = csc-=- 

333 



51C = ~ (1800)=3000 

1C =.!. (1800) =600 

3 3 

3 3 

51C 

r;:; 

tan- = tan 300° = -v3. 1C 2-./3 

3 

csc- = csc6O° =--. 

3 3 

98



29• P· m d cos-. 13K 

4 

13K _ 2K = 13n _ 8K = 5K 

4 4 4 4 

13K . . I . h 5K 

- IS cotermma wit -. 

4 4 

5K. 5K K 

The reference angle for - IS - - K = -. 

4 4 4 

13K .. dr ill h h . . 

- IS m qua ant ,were t e cosine IS 

4 

negative. 

13K 5n K..fi 

cos- = cos- = -cos-=- 

4 4 4 2 

Converting to degrees, 

5K = ~(l800) = 2250 

4 4 

13K 5n o..fi 

cos- = cos- = cos225 =--. 

4 4 2 

For Exercises 30-53, your calculator must be set in 

radian mode. Keystroke sequences may vary based on 


30. sin .8203 

Enter: .8203 ~ 

or ~ .8203 ~ 

sin .8203 = .73135046 

31. cos .6429 

Enter: .6429 ~ 

or §J .6429 ~ 

cos .6429 = .80036052 

32. tan .9047 

Enter .9047 ~ 

or ~ .9047 ~ 

tan .9047 =1.2723944 

33. sin 1.5097 

Enter: 1.5097 ~ 

or ~ 1.5097 ~ 

sin 1.5097 = .99813420 

34. cot .0465 

Enter: .0465 ~ 8 

or ITJ ~ .0465 CIJ [2] l§l 

cot .0465 = 21.489874 

35. esc 1.3875 

Enter: 1.3875 ~ . ~ 

or ITJ E.J 1.3875 CIJ [2] 

§3 

esc 1.3875 = 1.0170372 

36. sec 7.4526 

Enter: 7.4526 §J EJ 

or ITJ ~ 7.4526 CIJ [2] 

~ 

cot 7.4526 =2.5595706 

37. cos 6.6701 

Enter: 6.6701 ~ 

or §] 6.6701 ~ 

cos 6.6701 = .92607765 

38. tan 4.0230 

Enter: 4.0230 §) 

or §) 4.0230 ~ 

tan 4.0230 = 1.2131367 

39. cos 4.2528 

Enter: 4.2528 §J 

or ~ 4.2528 ~ 

cos 4.2528 = -.44357977 

40. sin 3.4645 

Enter: 3.4645 E.J 

or ~ 3.4645 ~ 

sin 3.4645 = -.31732498 

41. sin (-2.2864) 

Enter: 2.2864 EJ EJ 

or ~ CEJ 2.2864 ~ 

sin(-2.2864) = -.75469733 

42. cot(-2.4871) 

Enter: 2.4871 EJ ~ ~ 

or CD ~ (BJ 2.4871 CD [2] 

~ 

cot(-2.4871) = 1.3032411 

43. cos(-3.0602) 

Enter: 3.0602 EJ §J 

or §J CEJ 3.0602 ~ 

cos(-3.0602) = -.99668945 

44. tan s = .21264138 

Enter: .21264138 EJ ~ 

or 8 ~ .21264138 l§l 

Display: .20952066 

s = .20952066 

45. cos s =.78269876 

Enter: .78269876 ~ §J 

or El §J .78269876 ~ 

Display: .67180620 

s = .67180620 

46. sin s =.99184065 

Enter: .99184065 EJ EJ 

or El ~ .99184065 ~ 

Display: 1.4429646 

s = 1.4429646 

99



47. cot S =29949853 

Enter: .29949853 

or 8 EJ .29 

~ 

Display: 1.2797997 

S = 1.2797997 

~ 

949853 

8 EJ 

~ 

48. cot S = .62084613 

Enter: .62084613 ~ 8 EJ 

or 8 EJ .62084613 ~ 

~ 

Display: 1.0151896 

S = 1.0151896 

49. tan S = 2.6058440 

Enter: 2.6058440 8 

EJ 

or 8 EJ 2.6058440 ~ 

Display: 1.2043741 

S = 1.2043741 

50. cos S = .57834328 

Enter: .57834328 8 EJ 

or 8 EJ .57834328 ~ 

Display: .95409991 

S =.95409991 

51. sin S = .98771924 

Enter: .98771924 8 EJ 

or 8 ~ .98771924 ~ 

Display: 1.4139143 

S = 1.4139143 

52. cot s = .0963704 1 

Enter: .09637041 ~ 8 EJ 

or 8 EJ .09637041 ~ 

~ 

Display: 1.4747226 

s = 1.4747226 

53. esc s =1.0219553 

Enter: 1.0219553 ~ B ~ 

or EJ ~ 1.0219553 IE] 

~ 

Display: 1.3631380 

s =1.3631380 

54. 

1C ]. 1 

[2'1C ; Slns='2 

Because sin s = .!... the reference angle for s must 

2 

be -1C. since Sill- . 1C = - 1 . F or s to be 

In 

' t he e i mterva I 

6 6 2 

[~ , 1CJ. we must subtract the reference angle 

from 1C. 

Therefore, 

1C 51C 

S=1C--=-. 

6 6 

55. 

56. 

57. 

58. 

[~.1Cl coss =-~ 

1 

Because coss = --, the reference angle for s 

2 

1C . 1C 1 F be' th 

must be - since cos - = -. or s to In e 

3 3 2 

interval 

[~ , 1CJ. we must subtract the reference angle 

from 1C. 

Therefore, 

1C 21C 

S=1C--=-. 

3 3 

[1C' 3;l tans =~ 

Because tan s =~. the reference angle for s 

must be 1C since tan 1C =~. For s to be in the 

3 3 

interval [1C' 3;J. we must add the reference 

angle to 1C. Therefore, 

1C 41C 

S=1C+-=--. 

3 3 

31C] . 1 

[1C'T ; Slns=-'2 

1 . 

Because sin s = - -, the reference angle for s 

2 

1C • .1C IF be in th 

must be - since Sin- = -. or s to III e 

662 

interval [1C' 3;J. we must add the reference angle 

to 1C. Therefore. 

1C 71C 

s=1C+-=-. 

6 6 

[3;, 21Cl tans =-1 

Because tan s = -1, the reference angle for s must 

be 1C since tan 1C = 1. For s to be in the interval 

4 4 

[3; , 21CJ. we must subtract the reference angle 

from 21C. Therefore, 

1C 71C 

S=21C--=-. 

4 4 

100



59. [3;, 21rl coss = ~ 

Jj 

Because cos s =-, the reference angle for s 

2 

1r . 1t Jj . 

must be - since cos- =-. For s to be In the 

662 

interval [3; , 27rJ. we must subtract the 

reference angle from 21r. Therefore, 

. 1r 111r 

s=21r--=-. 

6 6 

sinx . sinx 

60. x -- (radians) -- (degrees) 

x 

x 

.1 .998334 17 .0174532837 

.01 .99998333 .0174532924 

.001 .99999983 .0174532925 

61. Since X =cos s, 

coss = .55319149 

Use radian mode. 

Enter.55319149 ~ ~ 

or ~ ~ .55319149 ~ 

Display : .984605869 

s



(c) May: x=4 73. 

4n 

t =60-30cos 

6 

2n 

t =60 - 30cos 

3 

t =60-3~-~) = 60+15 

t = 75° 

(d) June: x=5 

5n 

t = 60-30cos 

6 

1= 60-3{ "7) = 60+ 15.J3 

1= 86° 

(e) August: x = 7 

7n 

1= 6O-30cos 74. 

6 

1=60-3{- "7)=6O+15.J3 

t "" 86° 

(f) October: x = 9 

9n 

1= 60-30cos 

6 

3n 

1= 6O-30cos 

2 

1= 60-30(0) = 60-0 

1=60° 

72. T(x) = 37sin[ 2n (x -101)]+ 25 

365 

(a) March 1 (day 60) 

T(6O) = 37sin[ 2n (60 -101)]+ 25 

365 

""1°F 

(b) April 1 (day 91) 

T(91) = 37sin[2n (91-101)]+25 

365 

...19°F 

(c) Day 150 

T(150) se 53°F 

(d) June 15 is day 166. 

T(166) '" 58°F 

(e) September 1 is day 244. 

T(244) '" 48°F 

(f) October 31 is day 304. 

T(304) = 12°F. 

75. 

76. 

sin x =sin (x + 2) 

Ifx is in quadrant I, then x + 2 is in quadrant II. 

Therefore, 

x+2=n-x 

2x =n-2 

n 

x = --1. 

2 

On the interval 0 :s; x :s; 2n, this would also be true 

if we use an angle of 3n- x which is coterminal 

to n - x. Therefore, 

2+x=3n-x 

2x = 3n-2 

3n 

x=--1. 

2 

For the statement sin x =sin (x + 2) to be true 

n 3n 

x = - - l or x = - - 1. 

2 2 

(a) New Orleans has a latitude of L =30° or 

.5236 radian. Since the day and time have 

not changed, 

D = -.1425 and CO = .7854. 

sin (J = cos Dcos Lcosco + sin Dsin L 

= cos(-.1425)cos(.5236) .cos(.7854) 

+ sin(-.1425) ·sin(.5236) 

= .5352 

Thus, 

(J= .5647 radian or 32.4°. 

To solve this problem make sure your calculator 

is in radian mode. 23.44° "" .4091 radian and 

44.88°= .7833 radian. 

Shortest Day: 

cos (.1309H) = -tanDtan L 

= -tim (-.4091) tan (.7833) 

"" .4317 

.1309H "" 1.124 

H ""8.6 hi' 

Longest Day: 

cos (.1309H)=-tanDtanL 

cos (.1309H)=-tan (.4091) tan (.7833) 

"" -.4317 

.1309H "" 2.017 

H ...15.4 hr 

Hartford's latitude Lis 41.93° ....7318 radian and 

D= .26724. 

cos (.1309H) = -tan Dtan L 

= - tan (.26724) tan (.7318) 

"" -.2459 

.1309H "" 1.819 

H "" 13.9 hr 

102


3.4 Linear and Angular Velocity 103 

Section 3.4 

6• 6 2n di 5n d" . 

=""9 ra ran, co = 27 ra Ian per mm 

Exercises 

6 

co= 

1. The circumference of the unit circle is 21t 

t 

co= 1 radian per sec, 6= 2nradians 

5n =If 

6 

27 t 

co=t 

2n 27 6 ° 

t=_·_=- mm 

6 

t=co 

9 5n 5 

2n 3n di n dO ° 

t = - =2nsec 7• 6 = - ra ians.m = - ra Ian per mm 

1 8 24 

6 

co= 

2• eo = -2n ra d" lans per sec, t = 3 sec 

t 

3 6 

t= 

6 

eo= 

eo 

t 

2n 6 

t= ~ 

.1L 

-= 24 

3 3 

=_0 

3n 24 

6 = 2n radians 

8 n 

=9min 

3· co = 4" n ra d' 

° 

Ian per mm, t = 5' mm 

8. 6=3.871142 radians, t =21.4693 sec 

6 

eo= 

6 

t 

eo= 

t 

n 6 

-= 3.871142 

4 5 eo=-- 

21.4693 

= .180311 radian per sec 

6=\:) 

9. eo = .90674 radian per min, t = 11.876 min 

=- 5n ra dilans 

4 

6 

eo=t 

4• 

.6 

= - 

3n 

ra 

d' 

lans, 

• 

t = 

8 

sec 

.90674 =_6_ 

4. 11.876 

6 3zr 6 = 10.768 radians 

eo=-=-.i.. 

t 8 10. The circumference of the unit circle is 21t. 

3n di eo = 1 unit per sec, 6 = 2n radians 

= 32 ra I3n per sec 

6 

eo=t 

5. 6 = 2n radians, t = 10 sec 6 

t= 

5 eo 

6 2n 

eo=- 

=-=2nsec 

t 

1 -------- - 

llL 

eo =..2 

10 13. r = 12 m, co = 2n radians per sec 

2n 3 

=- v = reo 

50 

= -n ra di Ian per sec 

v = 12(2;) = 8n m per sec 

25 

103



8 91r di 

121r 3 21r . 

5 5 ' 2' 5 

v = rOJ 

s = rOJt 

14. r = em, OJ =- ra tans per see 22. s =-- m r =- m OJ =- radians per sec 

v = 8(9;) 

721r 

=-- em per see 

121r = ~(21r} 

5 2 5 

5 t= 1~1r(~)(;1t) 

t=4see 

15. v = 9 m per sec, r = 5 m 

v = rOJ 

31r 

9=5OJ 23. s = - km, r =2 km, t = 4 sec 

4 

OJ = -9 raditans per see 

s = rost 

5 31r 

-=2(OJ)4 

16. v = 18 ft per sec, r =3 ft 

4 

v= rOJ .k 

18 = 30J 

OJ =....!... 

8 

18 

OJ=- =- 

31r 

ra 

d' 

Ian per sec 

3 32 

= 6 radians per see 

81r 4 

17. v = 107.692 m per sec , r= 58.7413 m 24. s=- m r=- m t= 12 sec 

v=rOJ 

9 ' 3 ' 

107.692 =58.7413OJ 

s =riot 

OJ == 1.83333 radians per sec 

81r=i OJ(12) 

9 3 

18. r = 24.93215 em, (0= .372914 radian per sec 

81r ( 3 ) 

v=rOJ OJ =9 4(12) 

v = (24.93215)(.372914) 

== 9.29755 em per see OJ =-1r radiJan per sec 

18 . 

19. r = 6 em, OJ =1r radians per sec, t = 9 sec 27. The hour hand of a clock moves through an angle 

3 of 21rradians in 12 hours, so 

s = r(Ot 

(J 21r 

(0=-=t 

12 

S=6(~)9 

= 181r em =!!.- radian per hr 

6 

20. r = 9 yd, (0 = 21r radians per sec, t = 12 sec 

28. The minute hand makes one revolution per hour. 

5 Eaeh revolution is 21rradians, so 

s =rtat 

(0= 21rradians per hr 

s=,2;}12) 

_1_ (21r) =.!!.... radian per min, 

60 30 

= 2161r yd 

29. The second hand of a clock moves through an 

5 

angle of 21rradians in 60 sec, so 

(J 21r 

(0=-= 

21. s = 61tem, r = 2 em, OJ = 1r radian per sec t 60 

4 

s =rOJt =-1r 

ra d' Jan per see. 

30 

61r= 2(:} 

30. The line makes 300 revolutions per minute. Eaeh 

61r 

revolution is 21rradians, so 

t=-=12 sec 

1 

OJ =21r(300) 

=6oo1r radians per min. 

104


3.4 Linear and Angular Velocity 105 

31. The minute hand of a clock moves through an 

angle of 2rr radians in 60 min, and at the tip of the 

minute hand, r = 7 em, so 

r(} 7(2rr) 

bicycle is 

v=-=- 

t 60 

7rr . 

=- cm per 10m. 

30 

32. The second hand makes one revolution per 

minute. Each revolution is 2rr radians, so 

ro =2rr radians per min. 

v =rro, r =28 mm 

v=28(2rr) =56rr mm per min 

56rr 

= 60 mm per sec 

14rr 

=15 10m per sec 

37. At 200 revolutions per minute, the bicycle tire is 

moving 200 (2rr) = 400rr radians. This is the 

angular velocity roo The linear velocity of the 

v=rro 

= 13(400rr) 

= 5200rr in. per min 

To convert this to miles per hour: 

v =(52007t in'X 60 minutesXJ..!!....)( I mi ) 

minute hour 12 in. 5280 ft 

"" 15.5 mph 

38. Mars will make one full rotation (of 2rr radians) 

during the course of one day. 

2rr radians ( I hr ) "" 24.62 hr 

0.2552 radians 

33. The flywheel making 42 rotations per min turns 

through an angle 42(2rr) or 84rrradians in I min 

with r=2m. 

r(} 2(84rr) . 

v=-=---= 168rrm permm 

t I 

34. The point on the tread of the tire is rotating 

35 times per min . Each rotation is 2rr radians. 

ro =35(2rr) =70rr radians per min 

v =rm, r =18 cm 

v = 18(70rr) = I 260rr ern per min 

35. At 500 rotations per min, the propeller turns 

through an angle of 

()= 500(2rr) = lOOOrrradians in I min with 

3 

r=-= 1.5 m 

2 

r(} 1.5(1000rr) 

v=- =-...:...----' 

t . I 

=1500rr m per min 

36. The point on the edge of the gyroscope is rotating 

680 times per min. Each rotation is 2rrradians. 

ro =680(21t) =1360rr radians per min 

v = rca. r = 83 cm 

v =83(1360rr) 

=I 12,880rrcm per min 

39. (a) () =_1_ (2rr) = 2rr radian 

365 365 

(b) ro = -2rr ra di Ian per day 

365 

=_.- 2rr I ra d' Ian per hr 

365 24 

=-..!!..- radian per hr 

4380 

(c) 

v =rro 

=(93,000,()()()(-..!!..-) 

4380 

=66,700 mph 

40. Since s =56 cm of belt go around in t = 18 sec, 

the linear velocity is 

s 

v=t 

56 

= 

18 

28 

="9 em per sec. 

v = rro, r = 12.96 cm 

28 =(l2.96)ro 

9 

18.. 

ro=_9_ 

12.96 

"".24 radian per sec. 

105



41. The larger pulley rotates 25 times in 36 sec or (c) At the equator, r= 6400 km. 

v = rW 

- 25. times per sec. Th· us, Its angu I ar ve I ocity ·· IS 

= 64oo(21l") 

36 

= 12,8001l"km per day 

W = 25 (21l") = 251l" radians per sec. The linear 

36 18 

12,8oo1l" km r hr 

24 pe 

velocity of the belt is 

v =rto 

es 5331l" km per hr 

251l" ) 1251l" 

= 15 -- =-- cm per sec. 

( 18 6 

To find the angular velocity ofthe smaller pulley, 

1251l" 

usev =-- em per sec and r= 8 cm . 

6 

v = rW 

1251l" =8ro 

6 

W = 12:1l" (i) 

1251l" . 

=-- radians per sec 

48 

42. v = 1.46 m per sec and 46 revolutions per min is 

46(21l") 231l" . 

W =---=-- radians per sec. 

60 15 

So, 

v =rW 

1.46={2:;) 

r= ~ "".303m. 

15 

43. W = (l52)(21l") = 3041l" radians per min 

3041l" d. 

= ~ ra tans per sec 

v =rW 

59.4 ={7:;) 

r z3.73cm 

761l" d. ' 

=15 ra ians per sec 

44. (a) The earth completes one revolution per day, 

so itturns through (} = 21l" radians in time 

t = 1 day =24 hr. 

(} 21l" 

w=-=t 

1 

=21l" radians per day 

21l" 

W= 

24 

=!!- radian per hr 

12 

(b) At the poles, r = 0, so 

v =rto =O. 

" 

(d) Salem rotates about the axis in a circle of 

radius r at an angular velocity W = 21l"radians 

per day . 

South Pole 

sin45o=_r 

6400 

r = 6400 sin 45° 

=64'~) 

r =3200-.12 km 

v=rW 

=3200-.12(21l") 

z 90501l"-km per day 

= 28,000 Ion per day 

=90501l" km per hr 

24 

= 3771l" km per hr 

z 1200 km per hr 

45. Let s = the length of the track on the arc. 

400 =4rf...!!.-) = 21l" 

s 

\.180 9 

The length of track, s, is the arc length when 

21l" 

(} = - and r= 1800 ft. 

9 

s =r(} 

s =1800(2;) =4001l" ft 

;:.'. . 

106



Express the velocity v =30 mph in ft per sec. 

v=30 mph 

30 . 

=-- ml per sec 

3600 

(30)5280 ft 

= per sec 

3600 

=44 ft per sec 

s 

v=t 

s 

t=v 

400;rr 100;rr 

44 11 

== 29 sec 

180 9 

t=--=- 

46. In one minute, the propeller makes 5000 

revolutions. Each revolution is 2;rrradians, so 

5000(2;rr)= lO,OOO;rr radians per min. 

Since there are 60 sec in a minute, 

IO,OOO;rr 

ro=---'- 

60 

500;rr 

=- 

. 3 

== 523.6 radians per sec. 


2. (a) ;rr < 3 < n , so the terminal side is in 

2 

quadrant·ll. 

3;rr h . al 'd . . 

(b) ;rr < 4



20. 9= 1O.5(21r) = 211r 27. 2soN, 12° S 

5 =r9= 2(211r) =421r in. 12° S = _12° N 

9 = 2So- (-12°) = 40° 

21. r=15.2 em, 9=4 

5 =r9 =15.2e:) 

31r =4~1:0)= 2; 

s= r9 

= 11.41r =6400(2;) 

"" 35.S em ""4500 km 

22. r = 11.4, 9 = .769 28. Let 5 = distance between cities 

5=r9 at 72° E and 35° W. 

= (11.4)(.769) em 35° W =-35° E 

=S.77 em 9= 72° - (-35°) = 107° 

23. 

1071r . 

r=S.973 em, 107° = -- radians 

ISO 

9=49.06° s=r9 

= 49.06(.!!.-) = 6400( 1071r) 

ISO 

ISO 

5 =r() = 12,OOOkm 

= S.973(49·06{1:0) 29. 5 = 1.5, r = 2 

"" 7.6S3 em 5=r9 

5 1.5 3 

71r 

9=-=-= 

24. 9=-, r=2S.69 r 2 4 

4 

1 2 

A=-r9 

2 

A =.!. r 2 () 

2 

=~(2)2(~) 

= ~ (2S.69)2(7;) 

=- 3 or 15 " umts "2" 

2 

"" 2263 in. 2 

25. 4 = 3S.0 m, 9 = 21°40' 30. 5=4, r= S 

5= r9 

9=21°40'=21,°= 6:C:0) 

541 

9=-=-=r 

S 2 

A =.!. r 2 () 

2 A =.!.r 2 9 

65X.!!.-) 

= .!.(38.0)2( 

2 3 ISO =~(S)2(~) 

2 

""273 m 

2 

= 16 units 2 

26. Because the central angle is very small, the arc 

31. (a) The hour hand of a clock moves through an 

length is approximately equal to the length of the 

angle of 21rradians in 12 hr, so 

inscribed chord. (See directions for 

Exercises 31-34 in Section 3.2) 

co = - 21r = - 1r ra di Ian per h r. 

Let h = height of tree, r = 2000, 9 = 1° 10'. 12 6 

h"" r9 

In 2 hr the angle would be 

rvv11 =2 +.!Q.X.!!.-) 2(:) or ~ radians. 

vvv\ 60 ISO 

",,41 yd 

108



(b) The distance s the tip of the hour hand travels 

during the time period from I o'clock to 

3 o'clock. is the arc length when ()= nand 

3 

r= 6 in. 

s =r(J = ~ ~ ) = 2n in. 

33. tan n = tan 60° = .J3 

3 

34. cos 2n = cos 120° = _.!. 

3 2 

35.' Si{- 5:) = sin(-~ ,1800) 

= sin(-150°) 

= sin(-150° + 360°) 

= sin 210° 

I 

=- 

2 

36. tan(- 7;) = tan(-~ ,180°) 

= tan(-420°) 

= tan[-420° + 2(360°)] 

= tan 300° 

=-.J3 

37. ~j lIn) (11 0) 

c""~-6 =csc -(;'180 

= csc(-3300) 

= csc(-330°+ 360°) 

= esc 30° 

=2 

38. cot (--3- 17n) =cot (17 -"3,1800) 

= cot(-10200) 

= cot(-1020° + 3 .360°) 

= cot 60° 

.J3 

= 

3 

39. tan I> tan 2. 0 

Enter: 7.3159 ~ 56. 

or ~ 7.3159 ~ 

tan 7.3159:: 1.6755332 

48. sin 4.8386 

Enter: 4.8386 ~ 

or ~ 4.8386 ~ 

sin 4.8386 :: -.99204596 

49. cos s:: .92500448 

Enter: .92500448 EJ [§ 

or ~ ~ .92500448 ~ 

Display: .38974894 

s :: .38974894 

50. tan s:: 4.0112357 

Enter: 4.0112357 EJ ~ 

or ~ ~ 4.0112357 ~ 

Display: 1.3264768 

s :: 1.3264768 

51. sin s :: .49244294 

Enter: .49244294 EJ ~ 

or ~ ~ .49244294 ~ 

Display: .51489440 

s:: .51489440 

52. esc s:: 1.2361343 58. 

Enter: 1.2361343 ~ EJ ~ 

or ~ ~ 1.2361343 ~ ~ 

Display : .94240395 

s :: .94240395 

53. cot s:: .50221761 

Enter: .50221761 ~ 8 ~ 

or ~ ~ .5022176 1 ~ §) 

Display: 1.1053762 

s:: 1.1053762 

54. sec s :: 4.5600039 

Enter: 4.5600039 ~ EJ ~ 59. 

or ~ §] 4.5600039 ~ §) 

Display: 1.3497014 

s:: 1.3497014 

55. [0,;J. cos s:: ~ 

60. 

$ 

Because cos s > -, the reference angle for s 

2 

must be -TC smcecos-::-. . TC $F or s to be emi the 

4 4 2 

interval [0, ~ Js must be the reference angle. 

TC 

Therefore, s :: -. 

4 

57. 

[;, TCJ. tans::-J3 

Because tan s :: -J3, the reference angle for s 

must be TC since tan TC ::.J3. For s to be in the 

3 3 

interval [~, TCJwe must subtract the reference 

angle from TC. Therefore, 

TC 2TC 

S::TC--=-. 

3 3 

n 3TC] sec s = _ 2.J3 

[ , 2 ' 3 

2.J3 

Because sec s :: - --, the reference angle for 

3 

s must be - 

TC. n 2..[3 

6 6 3 

smce sec- =--. For s to 

be in the interval [TC, 3;Jwe must add the 

reference angle to TC. Therefore, 

TC 7TC 

s=TC+-::-. 

6 6 

3 TC 2] ' 1 

- TC sms=- 

[ 2" 2 

. 1 

Because sin s :: - -, the reference angle for s 

2 

TC. . TC 1 F be' th 

must be - smce sm - = -. or s to 10 e 

6 6 2 

interval [3;, 2nJwe must subtract the 

reference angle from 2TC. Therefore, 

TC lIn 

s=2TC--::-. 

6 6 

2 radians is in quadrant II so cosine must be 

negative. In quadrant II cosine decreases from 

oto -1. Therefore, cos 2 is closest to -.4. 

2 radians is in quadrant II so sine must be 

positive. In quadrant II sine decreases from 1 to O. 

Therefore, sin 2 is closest to .9. 

110



12tr 3 

61. tan 8= Y 66. s =-- ft, r = - ft t = 15sec 

X 25 5' 

5.1961524 s 12n 12tr 

= 

3 

v=-=~=-- 

t 15 375 

"" 1.73205 v = rro 

Use degree mode. 

12tr 3 

Enter: 1.73205 ~ §] 

--=-ro 

or ~ ~ 1.73205 ~ 375 5 

Display: 60° 

ro=~(12tr) 

8=60° 3 375 

8 = 6O(.!!...-) =!!. radian ro = 75 ra Ian per sec 

180 3 

62. Let s = the length of the arc. 67. r = 11.46 cm, ro= 4.283 radians per sec, 

y 

t = 5.813 sec 

tans =- 

s = r8 = r(Ot 

X 

-.5250622 

= (11.46)(4.283)(5.813) 

= 

.85106383 

=285.3 em 

= -.6170. 

Use radian mode. 

68. The flywheel is rotating 90 times per sec or 

Enter: .6170 

90 . 2tr radians per sec, 

~ §] 

or 

r=7m 

~ ~ .6170 ~ 

Display: -.5528258 

v = rro 

s = .5528 

= 7(90 ·2tr) 

Note that s is positive since it represents a length. 

= 1260tr m per sec 

5tr 8tr . 69. (b) csc8 =2 

63. 8 = -, (0= - radians per sec 

12 9 . 8 1 

SID = 

8 

(0=tr 

2 

t 8= 

6 

-= 8tr "* 

d is double h when the sun is 30° above the 

9 t 

5tr 9 

horizon. 

t=_· 

12 8tr 

tr _ tr 5 

45 15 () c csc-=1 and csc- "" 1.1 

=-=-sec 2 3 

96 32 When the sun is lower in the sky 

64. t = 12 sec, ro= 9 radians per sec (8 = ~). sunlight is filtered by more 

8 

ro= 

atmosphere. There is less ultraviolet light 

t 

8 

reaching the earth's surface, and therefore, 

9= there is less likelihood of becoming 

12 

sunburned. In this case, sunlight passes 

8 = 108 radians 

through 15% more atmosphere. 

65. t = 8 sec, 8 = 2tr radians 

5 

8 

ro=t 

l2r. 

=.2.... 

8 

2tr tr d' 

=-=- ra tans per sec 

40 20 

4tr 

di 

111




2. 120 0 = 12"/....!!..- radian) = 21r radians 

\180 3 

0) 

3. 91r =91r(180 =1620 

10 10 1r 

4. r = 150 em, s = 200 em 

s =r() 

()=!.. 

r 

200 

= 

150 

4 

= 

3 

4 

r = 150 em, s = 200 em, (J = 

5. 3 

A =.!. r 2 (J 

2 

=~(150)2(~) 

71r . 

6. s=6 

=15,000 em 2 

. 71r 1 

SIO-=- 

6 2 

71r .,fj 

eos-=- 

6 2 

71r .,fj 

tan-= 

6 3 

7. {s Is '¢ ~ +n1r} are the domains of the tangent and secant circular functions. 

8. sin s = .82584121 

Enter: .82584121 8 ~ 

or B ~ .82584 12 1 ~ 

s « .97169234 

112



9. In 11.64 years, Jupiter travels 2n- radians. 

r = 483,600,000 

o 2n 

m=-=- 

t 11.64 

v=rm 

= 483,600, rvvI~) 

vv'\I1.64 

= 261,043,678.2 miles per year 

To convert this to miles per second: 

)( 1minute) 

year 365 days }.24 hours 60 minutes 60 seconds 

= 8.3 miles per second 

v = (261,043,678.2 milesX 1 year Y 1day X 1 hour 

10. If the person travels n- radians, then he or she is at the top of the Ferris wheel, so he or she is 50 ft off the 

ground. The person is !!.. radians short of n- radians if he or she travels 5n- radians. 

6 6 

distance = 50 ft - ( 25 - 25sin ~ ) 

:::: 46.65ft 

11. t = 30 sec, 0 =5n- radians 

6 

e 

m= t 

=~ 

30 

n- di = - ra Jan per sec 

36 

12. (a) sin 0 = l. 

r 

. 0 y 

Sin =- 

2.625 

y = 2.625 sin 0 

(b) cosO =!!. 

r 

cosO=_u 

2.625 

u = 2.625 cos 0 

s 

(c) cosa= 

L 

s 

cosa=- 

10.5 

s= 10.5cos a 

(d) x=u+s 

x = 2.625cos 0 + 10.5cos a 

(e) v=-2.625sin11+ ~ cosO }505,168.1'( 1 ) 

vl 15 + cos 2 0 \12 ·5280 

The maximum velocity of 2 L757 mph occurs when 0 = 4.94415 radians. 

113


114 Chapter 4 Graphs of the Circular Functions 

CHAPTER 4 GRAPHS OF THE 

CIRCULAR FUNCTIONS 

Section 4.1 


2 

1. One example is f(x) =x . 

2. One example is f(x) =x 3 . 

3. (a) f(x) =1x 1 

(b) 

(c) 

f(-x) =1- xl= I(-I)x 1=1xl 

even 

g(x) =V; 

g(-x)=~ =~(-I)x =~ 

odd 

h(x) =x 

2 

+x 

h(-x) =(_x)2 +(-x) =x 

2 

- x 

neither 


1. X = -0.4161468, Y= 0.90929743, X is cos 2 and 

Y is sin 2. 

2. X = 1.9, Y= 0.94630009; sin 1.9 = 0.9463009 

3. X = 1.9,Y= -0.3232896; cos 1.9 = -0.3232896 

Exercises 

1. y= sin x 

The graph is a sinusoidal curve with an amplitude 

of 1 and a period of 21r. Since sin 0 =0, the point 

(0. 0) is on the graph . This matches with graph G. 

2. y= cos x 


of 1 and a period of 21r. Since cos 0 = 1, the point 

(0, I) is on the graph. This matches with graph A. 

3. y = -sin x 


of 1 and a period of 21r. Because a = -1, the graph 

is a reflection of y =sin x in the x-axis. This 

matches with graph E. 

4. y =-cos x 


of 1 and a period of 2n: Because a = -1, the graph 

is a reflection of y = cos x in the x-axis. This 

matches with graph D. 

5. y= sin 2x 

The graph is a sinusoidal curve with a period of 

1'C and an amplitude of 1. Since sin 2(0) = 0, the 

point (0, 0) is on the graph . This matches with 

graph B. 

6. y=cos 2x 


1rand an amplitude of 1. Since cos 2(0) =1, the 

point (0, 1) is on the graph. This matches with 

graph H. 

7. y = 2 sin x 


21rand an amplitude of 2. Since 2 sin 0 = 0, the 


graph F. 

8. y=2cosx 


21rand an amplitude of 2. Since 2 cos 0 = 2, the 


graph C. 

9. y = 2 cos x 

Amplitude: 121 =2 

o 1r 21r 

2 ~x 1----------- 

2 o -2 o 2 

This table gives five values for graphing one 

period of the function. 

Repeat this cycle for the interval 

[-2n; 0] . 

y 

y=2cosx 

10. y =3 sin x 

Amplitude: 131 =3 

xj_o_ 

1'C 21r 

3 sin x 0 3 0 -3 0 

This table gives five values for graphing one 

period of y =3 sin x. Repeat this cycle for the 

interval [-21r, 0]. 

114



y 

y 

y=3sinx 

y=-cosx 

115


116 Chapter4 Graphs of the CircularFunctions 

y 

y 

y=-3 cos x 

-+-I-Hk-::+-I--i-HI--x 

116



21. y = sin 3x 

P iod 2n en : 

3 

Amplitude: 111 = 1 

y 

y=2sin!x 4 

x 

X 0 

t t 

3x 0 

f n Jzr. 

2 

sin 3x 0 

y 

y=sin3x 

1r 

"2 If 

2n 

0 -1 0 24. y = 3 sin 2x 

P en iod : -=1r 21r 

2 

Amplitude: 3 

x 

3':2X I 

y 

0 

0 

i 

3 

f 

0 

-f 

-3 

1r 

0 

22. y=cos2x 

P en iod : -=1r 21r 

2 

Amplitude: 111 = 1 

x 

x 

2x 

cos2x 

y 

0 

0 

i 

1r 

"2 

0 

f 

-1 

1r 

Jzr. 

4 

¥ 

0 

1r 

21r 

25. y=-2 cos 3x 

P en iod : 2n 

3 

Amplitude: 2 

y=cos 2x x 0 t t f 1f 

3x 0 f 

1r 

-l 

21r 

x 

-2 cos 3x -2 0 

y 

y=-2cos 3x 

2 

0 

-2 

23. y= 2 sm-x ' 1 

4 x 

Period: 2; = 8n 

Amplitude: 2 

x 

l.x 

4 

2sintx 

0 

0 

0 

21r 

f 

2 

41r 

1r 

0 

61r 

¥ 

-2 

81r 

21r 

0 

26. Y=-5 cos 2x 

P 

eno 

. d 

: -=1r 

21r 

2 

Amplitude: 1- 51=5 

-5~'2x I 

0 

-5 

i 

0 

f 

5 

-t 

0 

1r 

-5 

117



-5 

y 

y=-5 cos 2x 

---+-f+-F+,t-HH--- x 

1l rr 

2 

27. y = cos 1C x 

Period: 21C =2 

1C 

Amplitude: 1 

30. The graph has an amplitude of .!.. and a period of 

4 

4. Then y = a sin bx where a =.!.. and b = 1C , so 

4 2 

1 . 1C 

y=-sm-x. 

4 2 

31. The graph has an amplitude of 4 and period of 41C. 

Then y = a cos(bx - ~) where a = 4 and 

b=21C=.!.. SOY=4COJ.!..x- 1C) . 

41C 2' \2 2 

y 

o I 1 3 2 

2" 2" 

1 o -1 o 1 

y = COSlrX 

32. The graph has an amplitude of .!.. and period of 4. 

4 

Then y =a cos(bx :... ~) where a =±and 

1C). 

b= 21C = 1C so y=.!..coJtr x_ 

4 2' 4 \.2 2 

-t-+-t-I"'*"+-t-I-t-~ x 

33. The graph repeats each day, so the period is 

24 hours. 

. I 2.6-0.2 2 

34• Approximate y =1. 

2 

28. y = -sin 1C x 


1C 

Amplitude: I-~ =1 

x o 1 

2 

o 

1 1. 

2 

-sin 1rX o -1 o o 

y 

Y =-SiD 10: 

I 

In Exercises 29-32, there are other correct answers. 

2 

21C 

29. The graph has an amplitude of 4 and a period of 

. 1 

41C. Then y = a sin bxwhere a = 4 and b = -, so 

2 

. 1 

y= 4sm-x. 

2 

35. Approximately 6 P.M., approximately 0.2 feet. 

36. The low tide occurred at 6 P.M. + 1:19, or 

7:19 P.M., the height was approximately 

0.2 - 0.2 =0 feet. 

37. Approximately 2 A.M.; approximately 2.6 feet. 

38. The high tide occurred at 2 A.M. + 1:18, or 

3:18 A.M.; the height was approximately 

2.6 - 0.2 =2.4 feet. 

39. (a) The amplitude is 120 - 80 = 40 =20. 

2 2 

(b) Since the period is .8 sec, there are 

- 

1 

=1.25 beats per sec and the pu 

I 

se rate 

. 

IS 

.8 

60(1.25) =75 beats per min. 

40. (a) The latest time that the animals begin their 

evening activity is 8:00 P.M., the earliest 

time is 4:00 P.M. 

4:00 s y s 8:00 

Since there is a difference of 4 hr in these 

times, the amplitude is .!..(4) =2 hr. 

2 

(b) The length of this period is 1 yr. 

118

\ ••+ 



41. (a) The highest temperature is 80°; the lowest is (d) E 

50°. 

5 

(b) The amplitude is .!.(800-500) = 15°. 

2 

.0083 .025 

(c) The period is about 35,000 yr. o 

.0 167 .033 

(d) The trend of the temperature now is 

downward. 

- 5 

42. (a) The amplitude of the graph is .!.; the period 

3 

. 3 S· 2n 3 k 4n Th . 

=3· e equanon 

IS 2'. mce 1:=2" 

. 1 . 4n t 

IS y=-sm-. 

3 3 

E=5 cos 120m 

46. E = 3.8 cos 40n t 

(a) Amplitude: 3.8 

Period: 2n =_1 

40n 20 

3 

(b) It takes - sec for a complete movement of (b) Frequency =number of cycles per second 

2 . 

1 

the arm. =--=20 

period 

43. -1:5y:5 1 

Amplitude: 1 

(c) t= .02 

Period: 8 squares = 8(30°) 

E = 3.8 cos 401l(.02) "" -3.074 

4n 

t= .04 

=240° or E = 3.8 cos 40n(.04) = 1.174 

3 

t= .08 

44. -1:5y:51 

E = 3.8 cos 40n(.08) "" -3.074 

Amplitude: 1 

1= .12 

E = 3.8 cos 40n(.12):;: -3.074 

Period: 4 squares = 4(300) 

t =.14 

21r 

=1200 or E = 3.8 cos 40n(.14) "" 1.174 

3 

(d) E 

45. E = 5 cos 120m 

(a) Amplitude: 151 =5 

Period: ~ =_1_ sec 

120n 60 

(b) Number of cycles per second -3 .8 

=_1_=_1 =60 

period 

io 

(c) t = 0 

E =5 cos 12011(0) = 5 

3.8 

E = 3.8 cos 40m 

47. (a) The graph has a general upward trend along 

with small annual oscillations. 

1= .03 L (x ) = .022x 2 + .55x + 316 

+ 3.5 sin(21Tx) 

E = 5 cos 120Jr(.03) "" 1.545 

'------"'"" 

t= .06 365 

E =5 cos 120n(.06) "" -4.045 

t= .09 

E =5 cos 120n(.09) "" -4.045 

t= .12 

E =5 cos 120Jr(.12) "" 1.545 

15 . ". . 35 

325 

119



(b) The seasonal variations are caused by the 

term 3.5 sin (2tr x). The maximums will 

occur when 

159 

x=4"' 4"' 4"' 

S· mce x IS 

. 

in 

. 

years, x = 4" 

1 

correspon 

d 

s to 

April when the seasonal carbon dioxide 

levels are maximum. The minimum levels 

will occur when 

3 7 11 

x=4"' 4"' 4' 

This is .!. yr later which corresponds to 

2 

October. 

48. (a) The graph of C has a general upward trend 

similar to L (in Exercise 47) except that both 

the carbon dioxide levels and the seasonal 

oscillations are lar er for C than L. 

C(x) = .04x 2 + .6x + 330 

+ 7.5 sin(21Tx) 

380 

(c) To solve this problem, horizontally translate 

the graph of C a distance of 1970 units to the 

right. The new C function can be written 

C(x) = .04(x -1970)2 + .6(x -1970) 

+ 330+ 7.5sin[21t(x -1970)], 

where x is the actual year. This function 

would now be valid for 1970 Sx S 1995. 

53. Since cos (x) S 1 for all x, the calculator will 

return 1. 

54. Since sin (x) S 1 ~ trfor all x, the calculator will 

return 1. 

57. (a) The maximum heights will occur when 

135 

t=4"' 4"' 4' .... 

The maximum height above equilibrium is 

s(±)= -5COS'±) = -5(-1) =5 in. 

1 1 

(b) Frequency = --.-= ..1 = 2 cycles per sec 

penod ~ 

Pen 

iod =-=- 2Jr 1 sec 

4Jr 2 

(c) The weight first reaches its maximum height 

1 

at t = - sec. 

4 

(d) s(1.3) = -5 cos 41r(1.3) ""4 

After t =1.3 sec, the weight is about 4 in. 

above the equilibrium position. 

58. (a) The maximum heights occur when 

Jr 3Jr 5Jr Jr 

t=IO'1O'1O=2',···· 

The maximum height above equilibrium is 

s(~) =-4COS1O(~) 

=-4cos Jr =-4(-1) = 4 in. 

1 1 5 

(b) Frequency: -'--=- =- cycles per sec 

period f Jr 

(c) 

Period: 2Jr = Jr sec 

10 5 

The weight first reaches its maximum height 

Jr 

when t =-sec. 

10 

(d) s(1.466) =-4 cos 10(1.466) "" 2 

After 1.466 sec, the weight is about 2 in. 

above the equilibrium position. 

55. sin ~ = 1 and sin 3Jr = -1, so Isin 3Jr I= 1-II= 1. 

222 

Jr 3Jr 

Thus, x=- andx=-. 

2 2 

56. cos 0 =1 and cos Jr=-1, so Icos JrI= 1-II= 1. 

Thus, x = 0 and x =Jr. 

120



59. (a) Use the model y =a cos bx. 

Since the weight starts at 3 in. below the 

equilibrium position, a =-3 and the cosine 

function can be used without shifting it to the 

right or left. 

1 

Frequency = --. 

penod 

6 1 

-=-- 

1C period 

1C 'od 

6"= pen 

Period = 21C = 1C . b =12 

b 

y =-3cosI2t 

(b) Period = 1C sec 

6 

6' 

60. (a) Use the model y = a cos bx. 

Since the weight starts at 2 in. below the 

equilibrium position, a =-2 and the cosine 

function can be used without shifting it to the 

right or left. 

Section 4.2 

Exercises 

Period = 21C =1.. b = 61C 

b 3' 

y =-2cos 61C t 

1 1 

(b) Frequency =--.- =- = 3 cycles per sec 

penod 

1. B; amplitude =3, period = 21C = 1C, 

2 

phase shift = .± =2 

2 

2. D; amplitude =2, period = 21C, phase shift = .± 

3 3 

3. C; amplitude =4, period = 21C, phase shift = 3 

3 3 

4. A; amplitude = 2, period = 21C =1C , 

4 2 

phase shift = ~ 

4 

t 

5. y =sin(x - :) 

The graph is a sinusoidal curve y = sin x shifted 

1C units to the right. This matches graph D. 

4 

6. y = sin ( x + : ) 

The graph is a sinusoidal curve y =sin x shifted 

1C units to the left. This matches graph G. 

4 

7. y=cos(x- =) 

The graph is a sinusoidal curve y = cos x shifted 

1C units to the right. This matches graph H. 

4 

8. y=cos(x+:) 

The graph is a sinusoidal curve y =cos x shifted 

1C units to the left. This matches graph A. 

4 

9. y =1 + sin x 

The graph is a sinusoidal curve y =sin x 

translated vertically 1 unit up. This matches 

graph B. 

10. y=-1 +sinx 

The graph is a sinusoidal curve y = sin x 

translated vertically 1 unit down. This matches 

graphE. 

11. y =1 + cos x 

The graph is a sinusoidal curve y =cos x 

translated vertically 1 unit up. This matches graph 

F. . 

12. y =-1 + cos x 

The graph is a sinusoidal curve y =cos x 

translated vertically 1 unit down. This matches 

graph C. 

13. right 

14. left 

15. y =2 sin (x - 1C) 

The amplitude is 121,which is 2. 

The period is 21C , which is 21C. 

1 

There is no vertical translation. 

The phase shift is 1C units to the right. 

121



16. 2 . ( n) 

Y="3 sm x+2" 21. Y = 2 - sin(3x - ~) 

17. 

The amplitude is I~ I. which is ~. 

The period is 2n. which is 2n. 

I 


The phase shift is n units to the left. 

2 

Y=4co{f+ ;) 

I 

y = 4cos-(x +n) 

2 

The amplitude is 141. which is 4. 

22. 

Y=-Isin{x- ~)+2 

The amplitude is 1-II. which is 1. 

Th epen iod i IS-. 2n 

3 

The vertical translation is 2 units up. The phase 

shift is !!.... units to the right. 

15 

1 

Y =-1+ -cos(2x - 31t} 

2 

y=-I+2"cos2 1 ( x-T 31t) 

The period is 2;. which is 4n. The amplitude is I~I. which is ~. 


The phase shift is n units to the left. 

Th e pen iod i IS -2n W hi IC h iIS 1t. 

2 

The vertical translation is 1 unit down. 

18. y=-cos~(x- ~) The phase shift is 3n units to the right. 

2 

Y=-ICOS~(X- ~) 

The amplitude is I-II. which is 1. 

The period is 2;. which is 3n. 


The phase shift is n units to the right. 

3 

23. Y=Co{x- ;) 

The amplitude is 1. 

The period is 2n. 

There is no vertical-translation, 

The phase shift is n units to the right. 

2 

y 

19. Y=3COS2(X-:) 

The amplitude is 131. which is 3. 

Th e pen iod i IS 2n w hi IC h iIS n. 


The phase shift is ~ units to the right. 

4 

... 20. 1 . (x ) 

Y=2"sm 2"+n 

IO~·X 

-I -!! ~ 2!!' 

222 

Y =.!.sin.!.(x + 2m 

2 2 

The amplitude is I~ I. which is ~. The period is 

2n h' h i 4 -)- W IC IS n: 

'2 


The phase shift is 2n units to the left. 

122



24. Y =sin(x - :) 


The period is 21r. 


The phase shift is 1r units to the right. 

4 

y 

27. Y=2CO{X- ~) 





3 

y 

25. Y =sin(x +:) 


The period is 2It 


The phase shift is 1r units to the left. 

4 

y 

28. Y =3sin(x _ 3;) 





2 

y 

-~-+-Hrr+t-f--J( 

-'> 

4 

y = sin (x + ~ 

y =3 sin (x - ¥) 

26. y=co{x- ~) 




The phase shift is 1r units to the right . 

3 

y 

29. Y =~sin2( x+ :) 

The amplitude is l. 

2 

Thepen 'ad 

IS 

' 2' 21r 

W 

hiIC h iIS x. 


The phase shift is 1r units to the left. 

4 

y 

2 

y=cos(x-f) 

-2 y= ~sin 2(x+~) 

123



30. y = -±COS4(x+ ~) 

The amplitude is 1-±I, which is ~ , 

IS 

,2TC 

-, W 

hi 

IC 

h i 

IS-, 

TC 

4 2 

There is no vertical translation, 

The phase shift is TC units to the left. 

2 

y 

The period 

32. y = 3 cos (4x + 1t) 

Y=3COS4(X+~) 


Th iod i 2TC hi h i TC 

e pen IS "4' W IC IS "2' 


The phase shift is TC units to the left. 

4 

y 

--+--;---,f-+--\o-+--+-~ x 

2" 

-I 

-+-~H-++----l-_ x 

_!J 0 !J 

4 4 

31. 

y = -4 sin (2x - TC) 

Y =-4Sin2(x- ~) 

The amplitude is 1- 41, which is 4, 

Th e pen iod i IS -, 2TC W hi IC h iIS TC. 

2 


The phase shift is TC units to the right. 

2 

4 

y 

33. 

-3 

y=3cos(4x+ n) 

y =..!.cos(.!.x _ TC) 

2 2 4 

y =.!.cos.!.(X _ TC) 

2 2 :2 

The amplitude is .!., 

2 

--1-.....-+---+--'-+---i~X 

o ' ~ n 3" 

'2 

The period is 2 1 

1t , which is 41r. 

'2 

This is no vertical translation, 

-4 

y =-4 sin (2x 11") 

The phase shift is TC 

2 

y 

units to rhe right. 

5" 

"2 

y=!cosHx-~) 

124


34. Y=-'41. (3 sm '4 

1r) x +g 

1 " 3 ( 

Y=-'4 sm'4 1r) 

x+"6 


2 . 3 

37. y= 1 --sm-x 

3 4 

The amplitude is 1-±I, which is ±. Th iod i 

Th iod i 21r hi h i 81r 

epen IS T'w IC IS 

4 3 

This is no vertical translation. 

The phase shift if 1r units to the left. 

6 

y 

The amplitude is 1-%I, which is %. 

epen IS T' 21t w hi 

IC 

h 

IS-. 

i 8n 

'4 3 

The vertical translation is 1 unit up. 

There is no phase shift. 

y 

y = I_l sin Jx 

3 4 

-~ -~ 

3 3 

38. y = -1 - 2 cos 5x 

The amplitude is 1- 21. which is 2. 

35. y =-3 + 2 sinx 



The vertical translation is 3 units down. 


y y = -3 + 2 sin x 

-211" -II" 0 II" 2Il" 

-I 

36. y =2 - 3 cos X 

The amplitude is 1- 31, which is 3. 


The vertical translation is 2 units up. 


y 

y=2-3cosx 

-t-Jt--t-~-Jt--t-t\-t- .. 

• 

39. 

Th epen iod IS-. i 2n 

5 

The vertical translation is 1 unit down. 


y 

y =- 1- 2 cos 5x 

2Jr 

5" 

--+fIt+iHH---X 

I 

y= 1-2cos-x 

2 


The period is 2 11t, 

which is 4n. 

Z 



y 

y= 1-2 cos!x 2 

125



40. y=-3+3 sin.!..x 

2 


Th "00' 2rr hi h i 4 

e pen IS T' w IC IS n. 



y 

y = -3 + 3 sin ~x 

43. Y=-3+2sin(x+ ~) 


The period is 2rr. 


The phase shift is rr units to the left . 

2 

y 

-+--=+--+~f--+-. x 

-411' 411' 

41. y=-2+.!..sin3x 

2 

The amplitude is .!... 

2 

Th epen 

iod i 2rr 

IS 3' 

The vertical translation is 2 units down . 


y 

44. y = 4 - 3 cos (x rr) 


The period is 2rr. 

The vertical translation is 4 units up. 

The phase shift is rr units to the right. 

y 

7 

y =4- 3 cos (x-Ir) 

4 

1 

o 

IC 21r 3Jr 

42. 

2 1 

y=l+-cos-x 

3 2 

The amplitude is %. 

The period is 2 1 

lt , which is 4rr. 

"2 


There is no phase shift . 

y 

45. y =~ + sin 2(x +:) 


. Th . e pen iod i IS - 2rr . w hi Ie h iIS rr. 

". 2 

The vertical translation is .!.. unit up. 

2 

The phase shift is !!.. units to the left. 

4 

y 

-II' 

" 

2 

Jr 

y= I +~COS!X 

3 2 

126


46. y =-%+ cos 3( x - : ) 


51. Converting to a new cosine equation, the 

amplitude a =3 and period p = 21C =1C remain 

The amplitude is I. 2 

the same. We must shift the graph an additional 

Thepeno . d i 21r 

IS-. 

3 - 1C11C'th . - = - Units to eng 'ht so 

2 2 4 ' 

The vertical translation is ~ units down . 

2 y =3cos 2(x - : - :)=3cos 2(x - ~ ) , 

The phase shift is 1C units to the right. 

6 

y 

If 

v- 

52. Converting to a cosine equation, the amplitude 

(; 

a =1C and period p = 21C = 2 remain the same. 

1r 

---=+-+-I-++-lH--- x 

o 

-~ 

We must shift the graph an additional 1C •..!.. = .!.. 

2 x 2 

units to the right, so, 

y =1C cos n(x -t-t) = 1C cos 1C(X-I), 

48. 

53. (a) Let January correspond to x = 1, February to 

x =2, ... , and December of the second year 

to x =24. Yes, the data appear to outline the 

graph of a translated sine graph. 

70 

Using the trace function, we see that the 

y-maximum occurs at I and y-minimum occurs at 

(b) The sine graph is vertically centered around 

1-(-1) 2 the line y =50. This line represents the 

-I, so the amplitude a = =- =1. average yearly temperature in Vancouver of 

2 2 

50°F. (This is also the actual .average yearly 

Changing the x-scl to ~, we find that the period temperature. ) 

3 

is 51C _ (_!!.) =61r =21C and the phase shift is 

333 

1C 

a=-. 

3 

In Exercises 49-52, there are other correct answers. 

,: :::L. .~ 

49. The graph has an amplitude of 3, a period of 1C, 

and a phase shift of 1r units to the right. In the 

4 

equation y =a sin btx - cl),a =3, b =2, and 

d = : ; therefore, y = 3sin 2(x - :). 

. Ḍ.. 

50. The graph has an amplitude of n, a period of 2, 

and a phase shift of .5 unit to the right. In the 

equation y = a sin b(x - cl), a =1C, b =n, and 

d =,5; therefore, y =zrsin 1l(x - .5). 

127



(c) The amplitude of the sine graph is 

approximately 14 since the average monthly 

high is 64, the average monthly low is 36. 

and 64 -36:= 14. The period is 12 since the 

2 

temperature cycles every twelve months. Let 

b =21r =1r. One way to determine the 

12 6 

phase shift is to use the following technique. 

The minimum temperature occurs in January. 

Thus, when x =I, b(x - d) must equal 

( - ~) + 2nn, where n is an integer, since the 

sine function is minimum at these values . 

Solving for d 

~ (l-d):=- ~ 

d=4. 

This can be used as a first approximation. 

(d) Letj{x) =a sin b(x -d) + c. 

Since the amplitude is 14, let a = 14. The 

period is equal to one yr or 12 mo, so 

b =1r. The average of the maximum and 

6 

. . . 64+36 50 Le 

mrrumum temperatures IS 

:=. t 

2 

the vertical translation be c = 50. 

Since the phase shift is approximately 4, it 

can be adjusted slightly to give a better 

visual fit. Try 4.2. Since the phase shift is 

4.2, let d =4.2. Thus, 

f(x) = 14sin[~(X-4 .2)J+50 . 

(e) Plotting the data with 

f(x) =14sin[~ (X-4.2)J+50 

on the same coordinate axes gives a good fit. 

70 

f(x) = 14 sin [~(x - 4.2)] + 50 

(f) 

From the sine regression we get 

y:= 13.21sin(.52x -2.18)+49.68 

:=13.21sin[.52(x - 4.19)] + 49.68. 

54. (a) The maximum average monthly temperature 

is 90°F, and the minimum average monthly 

temperature is 51°F. The average of these 

two values is 90 + 51 =70.5°F, which is 

2 

very close to the actual value of 70°F. 

(c) Let the amplitude a be 90-51 =19.5.Since 

2 

the period is 12, let b =1r. Let 

6 

c:= 90+51 :=.. 705 Th emmlmum . . 

2 

temperature occurs in January. Thus, when 

x =1, b(x - d) must equal an odd multiple of 

1rsince the cosine function is minimum at 

these values . Solving for d 

!:(1-d):=-1r 

6 

d:=7 

d can be adjusted slightly to give a better 

visual fit. Try d:= 7.2. Thus, 

f(x):= acosb(x -d)+ c 

:= 19 .5CO{~ (x -7.2)J+ 70.5. 

1 "-.;"""""'__-....;.._....;....'--'--~.25 

30 

128

' 


4.3 Graphs of the Other Circular Functions 129 

d. Plotting the data with 

30 

f(x) = 19.5CO{: (x -7.2)]+ 70.5 

on the same coordinate axes give a good fit. 

I(x) = 19.5 cos [*(x - 7.2)J + 70.5 

95 

-.1 13 

-2 

Section 4.3 

Exercises 

55. a. 

e. SinRe9 ":'" . 

· . ~=a"'sin ( bx+c ) .+d 

:· a= 1 9 ~ 72 '- , ' 

.'b=· i 2 '.r...:-./ .:.' 

c= .17 '.." . . 

~ · ~.=r ~.~~~ < ~ . ~~~ .: ~, > . 

~ . l ; " . ~. ~:: ~ . - '. ." 

TI-83 fixed to the 

nearest hundredth 


y = 19.72sin(.52x -2.17)+ 70.47 

= 19.72sin[.52(x- 4.17)]+ 70.47. 



1. True; since cos!£ = 0, tan lC is undefined. 

2 2 

2. False; the smallest such number is n. 

. sinx 1 

3. True; since tan x =-- and sec x =--, the 

cos x 

cosx 

tangent and secant functions will be undefined at 

the same values. 

4. False; secant values are undefined when 

lC . 

x = - + nlC, while cosecant values are undefined 

2 

when x = nn: 

sin(-x) -sinx 

5. False; tan(-x) = =--=-tanx 

cos(-x) cos x 

(since sin x is odd and cos x is even) for all x in 

the domain. 

I I 

6. True; sec(-x) = =--=sec(x) 

cos(-x) costx) 

(since cos x is even) for all x in the domain. 

7. y =csc x 

The graph is the reflection of the graph of 

y = esc x about the x-axis. This matches with 

graph B. 

-.1 13 

20 

b. SinRe9 . 

~=a*s i n ( bx+c )+d 

a=11.36 

b=.48 

c=1.22 

d~12.69 . 



8. y=-secx 


y = sec x about the x-axis, This matches with 

graph C. 

9. y =-tan x 


y = tan x about the x-axis, This matches with 

graph E. 

10. y =-eotx 


y = cot x about the x-axis. This matches with 

graph A. 

129



11. y =tan(x - :) 

The graph is the graph of y = tax x shifted 

1r units to the right. This matches with graph D. 

4 

12. y=co{x- :) 

13. 

The graph is the graph of y =cot x shifted 

1r units to the right. This matches with graph F. 

4 

y =csc(x =) 

We graph this function by first graphing the 

corresponding reciprocal function 

15. y =sec(x +:) 



y =cos(x + :). The period is 21r. 

0 

-I 

y 

1: 

3" ~ 

4" IV: 

n7" 9" 

4 4 4" 4" 

I I I 

I 

I 

y=sec(x+n 

x 

14. 

y =sin(x : ). The period is 21r. 

y 

1 

UI I 

0 

-I 

" 

Y1 

4 4 4 

I I I 

I 

I I I 

y=csc(x-~) 

y=sec(x+ 3:) 



y = co{x+ 3:). The period is 21r. 

y 

i"U 

-+-!.,+H--hH-+---'x 

SIr 7" 

4" 4" 

I 

I 

I 

y=sec(x+¥) 

x 

16. y =csc(x+ ~) 



y =sin(x + ~). The period is 21r. 

y 

7" 

6" 

'n 

,,2tr 

SIC 

-1-:o:J+--+-+--+-=-x 

-~-I 

I 

I 

I 

y=csc(x+~) 

130



17. y=sec(~x+ ~) 

1 ( 21l) 

y=sec'2 x+"'3 



y =cos~(x + 2:). The period is 41l. 

y 

I 

I 

I UI I 

1 I ¥ I I 

--,::++++l-++++l-++++Ir- x 

10K 13" 

-IOl



22. y =2 +±sec(~ x - Jr) 

1 1 

y =2 +-sec-(x-2Jr) 

4 2 



y = 2 + .!.cos.!.(x - 2Jr). The period is 4Jr. 

4 2 

y 

I 

U , 

I 

I 

I 

I I I 

--t: -t' t- 

-lC 0 lC 2lC 3lC 

I 

I 

I 

Y= 2 +i sec (!X-lC) 

23. y = 2 tan x 

Two adjacent vertical asymptotes are x =_ Jr 

2 

and x = Jr . The graph is "stretched" because 

2 

a =2and 121 > 1. 

y 

1 

25. y= -cotx 

2 

Two adjacent vertical asymptotes are 

x =0 and x =Jr. The graph is "compressed" 

because a =~ and I~I< 1. 

y 

--"7I--"'1o..----I---x 

y=!cotx 

1 

26. y=2tan-x 

4 


1 Jr 1 Jr 

-x=-- and -x= 

4 2 4 2 

or x =~2Jr and x =2Jr. 

y 

-/---+---:ioI'''''-+--+_x 

---+.....::,jL.-+--_x 

y = 2 laDix 

y=2tanx 

24. y= 2 cotx 

Two adjacent vertical asymptotes are x =0 

and x = Jr. The graph is "stretched" because 

a = 2and 121 > 1. 

y 

27. y =cot3x 


3x =0 and 3x =Jr 

Jr 

or x =0 and x =-. 

3 

y 

2 

---+~,.---/--+-_x 

y=2cotx 

--"7I----\----,I---x 

1< 

:3 

I 

-I I 

I 

I 

y= cot3x 

132



1 

28. Y =-cot-x 

2 


1 1 

-x=O and -x=tr 

2 2 

or x =0 and x =2tr, 


y =cot'!" x about the x-axis. 

2 

y 

30. y =tan(i + tr) 

1 

y =tan -(x + 2tr) 

2 


1 tr 1 tr 

-(x+2tr)=-- and -(x+21r)= 

2 2 2 2 

or x + 2tr =-tr and x + 2tr =tr 

or x =-3tr and x=-tr. 

y 

----j.,.--..".,,4



35. y = 1 - cot x 

32. y =cot(2x _ 3;) This is the graph ofy = cot x reflected about the 

x-axis and then translated vertically 1 unit up. 

y 

y = cot 2(x _ 3:) 


2(X- 3:)=0 and 2(X- 3:)=1r 

.1 

31r 31r 1r ---:~4--t"""'t-+---x 

or x--=O and x--= 

12K 

442 I 

31r 51r 

or x=- and X=-. 

I 

4 4 y=l-cotx 

y 

36. y = -2 - cot x 

This is the graph of y = cot x reflected about the 

x-axis and then translated vertically 2 units down. 

y 

I 

I 

o 

Ill' 

" 4 -+--+-+t=--t-f+-__ x 

-1 I 

I 

I 

I 

y= cot (2x-¥) 

33. y= 1 + tan x 

This is the graph ofy = tan x translated vertically 

1 unit up. 

y 

y =-2-cotx 

37. y =-1 + 2 tan x 

This is the graph of y = 2 tan x translated 

vertically 1 unit down. 

y 

34. y = -2 + tax x 

This is the graph of y = tan x translated vertically 

2 units down. 

-4--=t-I-t-t++-_x 

+- 

I 

I 

I 

I I 

y=-I +2tanx 

1 

38. y= 3+-tanx 

2 

y 1 

I I This is the graph of y = - tan x translated 

2 

"I 13" 

-21 IT vertically 3 units up. 

I 

-I- 

I 

I 

I 

I 

x 

y 

--++~---r+-t--+-_x 

I I I 

y=3+ ~ tan x 

134



39. 

1 

y =,...1 +-cot(2x - 

2 

y = -1+~cot2(x _ 3;) 

Th epen 

'00' 1r 

IS "2' 


2(X- 3;) = 0 

31t) 41. y = ~tan(~x -1C)-2 

and 2(X- 3;) = 1r 

31r 31r 1r 

or x--=o and x--= 

222 

31r 

or x = and x = 21r. 

2 

This is the graph of y = .!. cot 2x translated 

2 

vertically 1 unit down. 

y 

2 

I 

-+:-+t--++-+--+-~ x 

-l--- 

-2 I 

I 

I 

I I 

y=-I +! cot (2x-3n'> 

Y=-2+~tan~(x-41C) 

3 4 3 

Th '00' 1C 41C 

epen IS.l =-. 

4 3 


~(x- 4;)=_~ and ~(x- 4;)= ~ 

41C 21C 41C 21C 

or x--=-- and x--= 

3 3 3 3 

21r 

or x = and x = 21C;' 

3 

This is the graph of y = ~ tan(~ x 1C) translated 

vertically 2 units down. 

y 

I 

I 

2 ~ ~ 

o 1r 

---l-! 

-2 I 

I 

I 

y =~ tan (~x - n) - 2 

L_ 

I 

I 

I 

I 

40. y=-2+3tan(4x+1r) 

y=-2+3tan{x+ :) 42. y= 1-2cot2(X+ ~) 

Th epen '00' 

IS "4. 1r 

Th epen '00' 


{x+ :)= ~ and {x+ :)= ~ 

1C 1C 1C 1C 

or ·x+-=- and x+-= 

4 8 4 8 

31C 

1C 

or x=-- and X=- . 

8 8 

or 

or 

"2. 

IS 1C 

The adjacent vertical asymptotes are 

2(X+~)=O and 2(X-+~)=1C 

x+ 1C=O and x+ 1C=1C 

2 2 2 

1C 

x=- and x=O. 

2 

This is the graph of y = 3 tan { x + :) translated This is the graph of y = -2cot2(x + ~) 

vertically 2 units down. 

_~I 

81 

I 

1 

1 

---t 

1 

1 

1 

y 

1 

1 0 

I lr Ilr 

18 )8 

1-1 I 

I I 

I 

oF 

I 

-t-- 

I I 

I I 

I I 

I 

Y=-2 +3 tan (4x+ n) 

translated vertically 1 unit up. 

I 

I 

I 

I 

I 

--I 

y 

I 

I 

I 

Ị 

-- 

I" 

1 2 

I 

y=I-2cot2~+~) 

135



43. The domain of the tangent function is 

{XF¢ ~ +n1r, wherenisanYinteger}, 

and the range is (--00, 00). For the function 

!(x) =-4 tan(2x +1r) 

=-4tan2(x+ ~J 

the period is 1r . Therefore, the domain is 

2 

: + ~ n, wherenis any integer} . 

{+ ¢ 

This can also be written as 

{xr ¢ (2n + 1) : ' where n is any integer}. 

The range remains (-00, 00). 

44. The domain of the function y =esc x is 

{xix ¢ nn, where n is any integer} , 

and the range is (-00 , - 1]u [1, 00). 

For the function 

g(x) = - 2 csc(4x + rr) 

(d) d =4 tan 21r(1.2) 

=4 tan 2.41r 

'" 4(3.0774) =12.3 m 

(e) t =.25 leads to tan 1r, which is undefined. 

2 

48. a =4j sec 21rt I 

a. t = 0 

a = 41 sec 01 =4 m 

b. t = .86 

a = 41 sec 21r(.86) I= 6.3 m 

c. t =1.24 

a =41 sec 21r(1.24) 1= 63.7 m 

51. 3 

~t•.!2:. .; v,~_, 

=-2CSC4(X+ :) 

the period is 1r. Therefore, the domain is 

2 

{xr ¢ n;, where nis any integer}. 

The range becomes (-00, - 2] u [2, 00) since 

a=-2. 

45. The function tan x has a period of 1C, so it repeats 

four times over the interval (-21r, 21r). Since its 

range is (-00, 00), tan x =c has four solutions for 

every value of c. 

52. 

." • -t • 

X=.snS917i ' Y=l:3S&02:S~ . 

-3 

sin(1r) + sin(21r) =.!..+ $ = 1+ $ 

6 6 2 2 2 

46. None; cos x ~ 1 for all x, so 

1 

--~ 1 and sec x ~ 1. 

cos x 

Since sec x ~ 1, secx has no values in the domain 

-1



5n: 

55. y=-+nn: 

4 

56. approximately .3217505544 

57. .3217505544 + n « 3.463343208 

58. .3217505544 + nn: 


1. B; the amplitude is 4 and period is 2n: = x 

2 

2. D; all other statements are true, 

3. The range for sin x and cos x is [-I, I]. The range 

for tan x and cot x is (-00, 00). Since .!. falls in 

2 

these intervals, those trigonometric functions can 

attain the value .!.. 

2 

4. The range for esc x and sec x is 

(-00, -1]u[l, 00). The range for tan x and cot x 

is (-00, 00). Since 2 falls in these intervals, those 

trigonometric functions can attain the value 2. 

5. y = 2 sin x 

Amplitude: 2 

Period: 2n: 

Vertical translation: none 

Phase shift: none 

6. y= tan 3x 

Amplitude: not applicable 

Period: n: 

3 



1 

7. y = - - cos 3x 

2 

Amplitude: .!. 

2 

P iod 2n: en : 

3 



8. y= 2 sin 5x 

Amplitude: 2 

P iod 2n: en : 

5 



9. y = 1+ 2 sin.!. x 

4 

Amplitude: 2 

Period: 2~ = 8n: 

"4 

Vertical translation: up 1 unit 


1 2 

10. y=3--cos-x 

4 3 

Amplitude: 1-±1= ± 

Period: 2; = 3n: 

Vertical translation: up 3 units 


11. y=3CO{X+ ~) 

Amplitude: 3 

Period: 2n: 


Phase shift: n: units to the left 

2 

12. ' 3n:) 

y=-sm( x-4'" 

Amplitude: I 

Period:2n: 


Phase shift: 3n: units to the right 

4 

13. y=t cSc(2X- :) 


P en iod : -=n: 2n: 

2 


Phase shift: 1r units to the right 

8 

14. y = 2 sec (n:x - 2n:) 


Period: 2n: = 2 

n: 


Phase shift: 2n: = 2 · the rizh 

- urnts to e ng t 

n: 

137



15. 

I 

y =~ tan(3x - ~ ) 

26. y=-secx 

2 

y =~ tan,x - : ) 

Period: 2n 

n 3n 


2 2 

Period : n 

3 


Phase shift: !!.. units to the right 

9 

X 

( 

3Jr) 

16. y=cot '2+4" 


Asymptotes: x =-, x= 

y 

-+--+.:--+--+---+---;.- x 

Period: t=2n 

27. y =-tanx 

r 

Period:n 


Jr Jr 

Phase shift: = 3; units to the left 2 2 

Asymptotes: x =--, x= 

y 

17. The tangent function has a period of nand 

x-intercepts at integral multiples of Jr. 

18. The sine function has a period of 2n and passes 

through the origin. 

-+--+-~-+--.-_ x 

19. The cosine function has a period of 2Jr and has the 

n 

value 0 when x =-. 

2 28. y = -2 cos x 

Amplitude: 2 

20. The cosecant function has a period of 2n and is 

Period: 2n 

not defined at integral multiples of Jr. 

y 

y= -tan X 

21. The cotangent function has a period of nand y=-2cosx 

decreases on the interval 0 < x < n. 

2 

22. The secant function has a period of 2Jr and --=+--+--il-+--t--x 

vertical asymptotes at odd multiples of !!.., that is, 

2 

x =-n + nit, were 

at here n is IS an imteger. 

2 

-2 

25. y =3 sin x 

Period: 2Jr 

Amplitude: 3 

y 

y= 3 sinx 

3 

-f;:--t--+--+----i.....-.- x 

2Jr 

- 3 

138



29. y= 2+ cotx 32. y = tan 3x 

Period: n 

Vertical translation: up 2 units 

Period: ~ 

3 

Asymptotes: x = 0, x = 1C 

y Asymptotes: x =- -, x = 

6 6 

y 

. 1C 

1C 

----1-:+-+---. x 

y=2+cotx 

30. y=-l+cscx 33. y=3cos2x 

Period: 2n Amplitude: 3 

Vertical translation: down 1 unit 

31. 

iod 21C 

y 2 

y 

I 

2 . I 

Asymptotes: x = 0, x = n:, x = 2n 

3K I 

"I T 2,. 2 

~....30+o"E'-+-+--j-. x 

- __.... 4 1 

y= sin 2x 

nl 

I 

I I 

I I 

y=-I +cscx 

P en iod : -=1C 2n 

2 

Amplitude: 1 

y 

34. 

Pen : -=1C 

3 

--:+-Hr--f-t--. x 

y=3cos2x 

1 

y =-cot3x 

2 

Period: ~ 

Asymptotes: x = 0, 

y 

1C 

x = 

3 

y=sin 2x 

---+--~--+---_ x 

-1 

y=!cot3x 

139



35. 

Y=CO{X- :) 37. y =sec(2x+ ;) 

=See[2(X+ :)J 

Amplitude: 1 

Period: 21C 

Phase shift: 1C units to the right .00 21C 

4 Pen : -=1C 

y 2 

Phase shift: 1C units to the left 

6 

y 

'V' 

--f-+---"'k--+--#--+- x 

I 

I I I 

I I I 

36. y = tan(x - ~ ) 

1C 71C J31C 

12 12 12 

---::++----1f-+--+-.... x 

o I .s I ¥ 

-1 if\3: 

I 

I 

Y =sec (2x+ n 

Period: 1C 

Asymptotes: x =0, x = 1C 38. Y =Si{3x + ~) 

Phase shift: !E. units to the right 

2 

y 

=sin[{x+ :)J 

Amplitude: 1 

.00 21C 

Pen : 

3 

Phase shift: 1C units to the left 

6 

y 

y =sin (3x+f) 

-+-~"""",....--+-_x 

39. Y =1 + 2 cos 3x 

Amplitude: 2 

.00 21C 

Pen : 

3 

Vertical translation: 1 unit up 

y 

3 

2 

--:-H~H----x 

y=I+2cosh 

140


40. 

41. 

y =-1 - 3 sin 2x 

Amplitude: 3 

iod : -=1C 

21C 

Pen 

2 

Vertical translation : down 1 unit 

y 

y = - 1- 3 sin 2x 

2 

---::-t-+-lt-Ht---- x 

-4 

y= 2 sin 1CX 

Amplitude: 2 

Period: 2 

y 

2 

y = 2 sin xx 

--.--+- .--__ x 

o 

2 

-I 

-2 


(b) The maximum average monthly temperature 

is 75°F, and the minimum is 25°F. Let the 

amplitude a be 75 - 25 =25. Since the 

2 

period is 12, let b = 1C. The data are 

6 

centered vertically around the line 

75+25 

y = =50, so let c =50. 

2 

The minimum temperature occurs in January. 

Thus, when x =I, b(x - d) =- 1C since the 

2 

sine function is minimum at _ 1C • Solving 

2 

ford, 

1C 

1C 

-(l-d)=- 

6 2 

d=4. 

d can be adjusted slightly to give a better 

visual fit. Try d =4.2. 

Thus, !(x) =a sin b(x - d) + C 

(d) Plotting the data with 

=25sin[: (x -4.2)]+ 50. 

42. 

1 

Y =--COS(1C x -1C) 

2 

Amplitude: .!.. 

2 


1C 

Phase shift: 1C =1 unit to the right 

1C 

y 

!(x) =25sin[: (x - 4.2)] + 50 

on the same coordinate axes ive a good fit. 

f(x) = 25 sin [-ifx - 42)] + 50 

80 

y =-! cos (1« -tr) 

(e) 

45. 

(a) Let January correspond to x = 1, February to 

x =2, ... , and December ofthe second year 

to x = 24. The data appear to follow the 

pattern of a translated sine graph. 

10 

.. - ~-• . ~:;:. :'.. . ,!,. 

.... .. 

-- "... • ~ :• .~ .. -;~ . :. a 

:. ~ 

. . .. 

,.: . ..-- 

• 

1• ~...;.._.;..........;;;.......;.._;.,;,25 

" 

20 


y =25.77sin(.52x-2.19)+ 50.57 

=25.77sin[.52(x - 4.21)] + 50.57. 

141



46. The shorter leg of the right triangle has length 

b: -hI ' 

(b) When h 2 ::::;:; 55 and hI::::;:; 5. 

d = (55 -5)cot8 = 50cot8. 

The period is n, but the graph wanted is d for 

0


7. y = 3 csc itx 

4. y=tan(X- ~) Period: 2 

Period:rr 

Amplitude : not applicable 

Asymptotes : x = -x, x = 0, x = rr 


Phase shift: rr units to the right 

2 I I 

I I 

y Y=IaD(x-~) I 

I 

I 

I 

I 

I 

x =- n 

5. y =-co{~x) 



Asymptotes: x = -2, x = -I, x = 0, X= I, x = 2 



'V' 

n 

I 

I 

I 

8. (a) 

I f(x) = 175 sin[~(x - 4) ] + 67.5 

x=n 

90 

Period: 2rr 


Asymptotes: x = -2rr, x = 0, x = 2rr 


. Phase shift: none 

y _ 1 

y--COli (b) Amplitude: 17.5 

I iod 2rr 6 

I 

I 

I 

y 

Pen : -=2rr·- = 12 

f rr 

Phase shift: 4 units tothe right 

Vertical translation: 67.5 units up 

(c) Approximately 52°F 

x=o 

x= 2lr 

(d) 50°F January; 85°F in July 

(e) Approximately 67.5°; this is the vertical 

translation. 

x =¥ 

U 

I 

( 

I 

I 

I y=-sccx 

I 

I 

10. C; the graph shows this is an odd function, so 

either A or C, Since sin x> 0 from 0 < x < !!... 

2 ' 

the function in the graph must bey =-3 sin x, 

143


144 Chapter 5 Trigonometric Identities 

CHAPTER 5 TRIGONOMETRIC IDENTITIES 

Section 5.1 

Exercises 

1. By a negative-angle identity, tan (-x) =-tan x. 

Since tan x = 2.6, tan (-x) =-tan x = -2.6. 

2. By a negative angle identity, cos (-x) = cos x. 

Since cos x = -.65, cos (-x) =cos x =-.65. 

3• By a reciproca . l I identi entity, cot x =--. 1 

tan x 

Since tan x = 1.6, cot x = _1_=.625. 

1.6 

4• By a quotient I identi entity, tan x =--. sinx 

cosx 

Since cos x =.8 and sinx= .6, 

sinx .6 

tanx=--=-=.75. 

cosx .8 

By a negative angle identity, 

tan(-x)=-tanx, so tan(-x) =-.75. 

3.. dr I 

5· cos s = '4 ' S IS 10 qua ant . 

sin 2 s+cos 2 s = 1 

2 

sin s + (~r = 1 

sin 2 s =1-i.=.2 

16 16 

sins =± .fi 

4 

S· mce s IS 

. 

10 

. qua drant, I' sms=-. .fi 

4 

6. cots = _.!., s in quadrant IV 

3 

1+ cot 2 s = csc 2 s 

1+(-~r =csc 2s 

10 = csc 2 s 

9 

Since s is in quadrant IV, esc s < 0, so 

cscs=-N=- ~. 

7. coss = S' ../5 tans < o : 


sin' s+( ~r =1 

sin 2 s = 1_ 2. = 20 

25 25 

4 

= 5 

. 2 

SIOS=±../5 

Since s is in quadrant IV, 

. 2 2../5 

SIOS=- ../5=--5-. 

.fi 

8. tans =-- sees> 0 

2 ' 

tan 2 s+l=see 2 s 

(- ~r +1=see 

Imp I' res s .. IS 10 qua dr ant IV . 

2s 

2 

!!= sec s 

4 

.J11 

--=secs 

2 

2 

--r:-:" =cos s 

...,11 . 

sin 2 s + cos 2 s = 1 

2s+(lY 

sin =1 

. 2 7 

SIO 

s= 

11 

Since tan s < 0 and sec s > 0, s is in quadrant IV, 

so sin s


5.1 Fundamentalldentities 145 

14. This is the graph off(x) = cotx. 

9• sees =-, 11 tans < O unpI' ' res S IS " ID quadrant IV. 

4 

Since f(-x) = cot(-x) 

. 1 

cos(-x) 

SIDee secs=- 

= 

coss 

sin(-x) 

11 1 cosx 

-=-- =- 

4 coss -sinx 

4 =-cotx 

-=coss. 

11 =-f(x), 


f(-x) =-j(x). 

2 

sin s +(.~r = 1 15. sin 0 = ~. 0 in quadrant II 

S· 

3 

. 2 

SID 

16 1 

s+-= sin 2 0 +cos 2 0 = 1 

121 

. 105 cos 2 0 = I-sin 2 0 

SIDS= 

12 

. .;/105 

=I-(%r =i 

SIDS=±- 

11 

cosO=± ..J5 

.. ad IV' , SID ..}105 3 

S 

11 

IDee S IS ID qu rant =---. 

Since 0 is in quadrant II, cosO = _ "f5 . 

3 

10. cscs =-- 8 sinO t 2 2..J5 

5 tanO=--=--=--=-- 

. 1 cosO _:/l ..J5 5 

SIDee CSCs=--, 

sins 1 1 "f5 

8 1 cotO=--=--=- 

--=-- tan 0 2 

5 sins 

3 

-i 

-8sins =5 .. 1 1 3 3..J5 

secO=--=--=--=-- 

. 5 

SinS =--. 

cosO -4 ...J5 5 

8 

1 1 3 

cscO=--=-= 

12. This is the graph of f(x) = sec x. sinO t 2 

Since f(-x) = sec(-x) 

1 

=-- 

16. cosO =.!., 0 in quadrant I 

5 

cos(-x) 

1 sin 2 0 + cos 2 0 = 1 

=- 

cos(x) sin 2 0 = l-cos 2 0 

= sec(x) 

=f(x), 

=1-(~r 

f(-x) =f(x). 24 

13. This is the graph off(x) = esc x. 

Since f(-x) =csc(-x) 

1 

=-- 

sin(-x) 

1 

f(-x) =-j(x). 

= -sin(x) /) 

=-~ - CSGc '" 

=-f(x), 

= 

25 

145



17. 

Since (J is in quadrant I and sin (J > 0, 

sin (J= f24 = 2../6 " 

V"25 5 

. (J 2-/6 

SID - 

t 

tan (J = -= _5_ = 2../6 

cos(J 

1 1 ../6 

cot(J =-=-=tan(J 

2../6 12 

1 

sec(J=--=5 

cos(J 

1 5 5../6 

csc(J = sin(J = 2../6 = U 

tan (J= _.!., (J in quadrant IV 

4 

sec 2 (J= 1+ tan 2 (J 

=1+(-if 

17 

= 16 

S· (J.. dr IV (J..ffj 

mce IS ID qua ant , sec = 4 " 

1 

cos(J=- 

sec(J 

1 4 4~ 

= .ffi =..ffj =17 

4 

sin (J = tan (J"cos(J 

1 4 1 ~ 

=-4"~ = .,Jfi =-17 

csc(J =_1_ =_1_ =_~ 

sin(J --i7 

1 1 

cot(J=--=-=-4 

tan(J -i 

18. csc(J=-~, (J in quadrant m 

. 2 

sin (J= _1_ = _ 3 

csc(J 5 

cos 2 (J= 1-sin 2 (J 

=1-(-~r 

21 

= 

Since (J is in quadrant III, cos(J < 0, so 

..fij 

cos(J =---. 

5 

tan (J = sin (J = --=.L = _2_ = 2..fij 

cos(J -41 ..fij 21 

1 ..fij 

cot(J=--=- 

tan(J 2 

1 5 5..fij 

sec(J = -=--=-- 

cos (J ..fij 21 

19. cot(J = ±, sin(J > 0 

3 

Since cot(J > 0 and sin (J > 0, (J is in quadrant I 

and all the functions are positive. 

1 1 3 

tan=--=-=cot(J 

t 4 

( 

3)2 25 

sec 2 (J= 1+ tan 2 (J= 1+ - = 

4 16 

sec(J =~ 

4 

1 1 4 

cos(J = -= = 

sec(J i 5 

sin2(J=I-cos2(J=I-(~r = ~ 

. (J 3 

SID = 5 

1 5 

csc(J=--=sin(J 

3 

3 

cos(J =- 5 

25 

20. sin(J = _±, cos(J < 0 

5 . 

Since sin (J< 0 and cos(J < 0, (J is in quadrant ill. 

2 " 2 (4)2 9 

cos (J=I-sID (J=I- -'5 = 25 

sin(J 4 

tan(J =--= 

. cos(J 3 

1 3 

cot(J=--=tan(J 

4 

1 5 

sec(J=--=- 

cos(J 3 

1 5 

csc(J =-=- 

sin (J 4 

146



21. sec 0 = i, sinO < 0 

3 

Since sec0 > 0 and sin 0 < O. 0 is in 

quadrant IV. . 

1 1 3 

cosO=--=-=sec0 

t 4 

· 2 2 (3)2 7 

SID O=I-cos 0=1-"4 =16 

S· mce SID . 0 < 0 , SID . 0 =--. ../7 

4 

sinO · -:if ../7 

tanO=--=--=- 

cosO i 3 

4 

1 1 3../7 

cot 0 =--=--=-- 

tanO 7 

1 1 4../7 

-:if 

cscO=--=--=-- 

sinO 

7 

22. cosO = _.!.. sinO> 0 

4 

Since cos0 < 0 and sin 0 > 0, 0 is in quadrant n. 

· 2 2 (1)2 15 

SID O=I-cos 0=1- -"4 =16 

sinO= "ff5 

4 

sinO ~ 

tanO=--=-",15 

· cosO . 

1 .J15 

cotO=--=-- 

tan 0 15 

secO=_I_ =-4 

cosO 

1 4.J}5 

cscO=--.=- 

sinO 15 

cos x 

23. --=cotx 

sin x 

Choose expression B. 

sinx 

24. tanx=- 

cos x 

Choose expression D. 

25. cos(-x)=cosx 

Choose expression E. 

26. tan 2 x+l=sec 2 x 

Choose expression C. 

27. 1=sin 2 x+cos 2 .x 

Choose expression A. 

sinx . . ( ) 

28. -tanxcosx =--_·cosx =-SIDX=SID-X 

cosx 

Choose expression C. 

2 2 sin 2 x 

29. sec x-I = tan x =--2 

cos x 

Choose expression A. 

30. 

.....L . 

sec x cosx sm r 

csc x = sJ x = cos x = tan x 

Choose expression E. 

2 

31. l+sin 2 x=csc x-cot 2 x+sin 2 x 

Choose expression D. 

2 1 

32. cos x=--2 

sec x 

Choose expression B. 

35. Find sin 0 if cos 0 = _x_. 

(x+l) 

. x 

oosO=- 

x+l 

Since sin 20+cos20=1 , 

sin 2 0 = 1-cos 2 0 

sin2 0 = 1-(_x_)2 

x+l 

(x+ 1)2_x 2 

=-'---'--..,....- 

(x+ 1)2 

x 2 +2x+l-x 2 

= 

(x+ 1)2 

2x+l 

- (x+ 1)2 

. 0 ±..,J2x+l 

SID = . 

x+l 

147



43. (1- cos 0)(1+ sec 0) 

36. Find tana if sec a = p + 4 . 

p 

=1+secO-cosO-cosOsecO 

tan 2 a + 1 = sec 2 a 

= 1+ sec 0 - cos 0 - cos 0(_1_) 

cosO 

( +4)2 

tan 2 a+ 1= P 2 

=1+secO-cosO-l 

P 

=secO-cosO 

2 (p+4)2 

tana= 2 1 cosO+sinO cosO sinO .0 1 

p 44. =--+--=cotu+ 

sinO sinO sin O 

2 p2+ Sp+ 16 p2 

tan a = p2 - p2 

cos 2 0 -sin 2 0 

45. 

2 Sp+ 16 

tan a= 

sinOcosO 

p2 20 

cos sin 2 0 

2 4(2p+4) = 

tan a = 2 

sinOcosO sinOcosO 

P cosO sinO 

=---- 

±2.J2P+4 

sinO cosO 

tan a = 

= cot 0 -tanO 

p 

l-sin 2 0 cos 20 

.0 • .0 cosO . .0 .0 

37. cot uSIO u =--·SlOu = cOSu 46. 1+ cot 2 0 = csc2 0 

sinO 20 

cos 

. 1 cosO sinO 

=-. 

38. sec 0 cot 0 SIO 0 =-_.-_._ 

sio 

cosO 2 8 

sinO 1 

sinO cosO 

=sin 2 Ocos 2 0 

= cosOsinO sinO cosO 

=1 47. tan 0 + cot 0 =--+- 

.. cosO sinO 

1 sin 2 0.+cos 2 0 

39. cosOcscO =cosO·_ = 

sinO 

cosO sinO 

cosO 1 

=- =--- 

sinO 

cosOsinO 

= cot 0 

=-_._ 

1 1 

cosO sinO 

20 

2 2 cos 2 

40. cot 0(1+ tan 0) = --2-(sec 0) 

= secOcscO 

sin 0 

20( 48. (secO+ cscO)(cosO- sin 0) 

cos 1 ) 

= sin 2 0 cos 2 0 = (_1_ + _._1_)(cos 0 - sin 0) 

1 

cosO SIOO 

= sin 2 0 = -1-(cosO) - -1-(sinO) 

2 cosO cosO 

= csc 0 

+ -._I_(cosO) __._I _ (sin 0) 

SIOO SIOO 

2 2 

41. sin 2 0(csc 0-1)= sin 0(+-1) = 1- sin 0 + cos 0 _ 1 

SIO 0 

cosO sinO 

• 2.0 

SIO u . 2.0 = -tanO+cotO 

=---SIO u 

sin 20 :;::; cotO - tan 0 

=1-sin 20 

20 49. sinO(csc 0 - sinO) = sinOcscO -sin 2 0 

=cos 

= l-sin 2 0 

42. (sec 0 -1)(sec 0 + 1) = sec 2 0 -1 = cos 

2 0 

= tan 2 0 

148

0 


5.1 Fundamental Identities 

149 

1+ tan 2 0 see 2 0 

50. l+cot20 = CSC2 0 

51. 

1 

0 - 

29 

coo 

=-1 

sio 2 9 

I sin 20 

=COS 2 0 ' - 1 

sin 20 

= COS 2 0 

=~~.J1 0 

sin 20+tan20+cos20 

= (sin 2 0 + cos 2 0) + tan 2 0 

=1+tan 20 

=see 2 0 

tan(-0) 

- tan0 

52. =-1 

see 0 oos9 

sinO cosO 

=--_._ 

cosO 

=-sinO 

53. Express sin 0 in terms of cot 0 and in terms of 

see O. 

0 

I 1 

sinO = --= -;======= 

cscO rJI + cot 2 0 

_rJI+cot 20 

- l+cot 20 

sinO = cos 0 tan 0 

=_I_.±~see2 0-1 

see 0 

±.J'-sec--:2=-0---1 

= 

see 0 

54. Express cos 0 in terms of sin O. cot O. and 

csc O. 

sin 2 0+cos 2 0 =I 

cosO =±~I-sin2 0 

. cosO 

Since ootO=--, 

sinO 

cosO = sin 0 cot 0 

I 

=--cotO 

cscO 

I 

= cot 0 

±.JI + cot 2 0 

rl-+-c-o-t 

± cot 0.J 2:--0 

= 

l+cot 20 

cosO = ± It - _1 2 

 

" csc 9 

20-1 

=± csc 

20 

csc 

±~csc2 0-1 

= 

cscO 

55. Express tan 0 in terms of sin O.cos O. sec 

esc 0, 

sinO 

tan O=- 

cosO 

sinO 

=~;===~ 

±~I-sin2 0 

±sinO 

- ~1-sin20 

±sin 0~""'I---s-in""'2""'0- 

= 

l-sin 20 

sinO 

tan O=- 

cosO 

±.Jr- -0 

I-_-co-s-=-2 

= 

cosO 

tan 0 =±~see2 0-1 

I 

tanO=- 

cot 0 

I 

=--;=== 

±~csc2 0-1 

_ ±~csc2 0-1 

- csc 20-1 

O. and 

149



56. Express cot 0 in terms of sin O. cos O. and 58. Express esc 0 in terms of cos 0, tan 0, cot O. 

esc O. and sec O. 

1 

cotO=±~cscZ 0 -1 

cscO=- 

sinO 

=±~~-1 

1 

=---;==:== 

±~I-sinZO 

±~I-cosZ 0 

= 

sinO _±..Jl-cosZO 

- l-cosZO 

±~I-sinZ 0 

cot 0 =---'--- 

sinO 

±cosO 

=--;==== 

= 

esc 0 = ±~1 + cot Z 0 

~1-cosZO 

=±~1+ l z 

tan 0 

± cos O~""'I---c-o""sZ:-O- ±~I+tanZO 

l-cos Z 0 

= 

lanO 

cotO=±~cscZ 0-1 cscO= ±~1 + cot Z 0 

±~I+tanZO 

57. Express sec 0 in terms of sin 0, tan O. cot O. 

and esc O. 

1 

secO=- 

cosO 

1 

=---;==~ 

esc O = ---'--- 

tan 0 

±secO 

= ~secZO-l 

± sec o-J,....-sec----:Z:--O---l 

= 

sec Z 0-1 

±~I-sinZO 

±~I-sinZO 

= l-sin Z 0 59. Let cos x = ~. Then x is in quadrant I or quadrant 

sec 0 = ±..Jtan Z 0+ 1 

Since tan 0 = _1_. 

cotO 

sec 0 = ±~rta-n-=-Z-0-+-1 

=± (_l z )+1 

cot 0 

±~I+cotZ 0 

= 

cotO 

1 

secO=- 

cosO 

1 

=-p===== 

.s:»: 

1 

=---;=== 

+~1- cs:2 8 

~cscZO 

=---;=;:==== 

±~cscZ 0-1 

± esc O~""'c-sc-Z-O---l 

= 

csc Z 0-1 

IV. Then. 

sin x = ±~""'i---c-o-sZ-x-. 

=±FG) 

= ± f24 = ±2J6 . 

'/"25 5 

. + z../6 

smx --5 rz: 

tanx=--=-t-=±2",6 . 

cosx S 

1 

secx=--=5 

cosx 

150



Quadrant I: 

sinx 

Quadrant IV: 

secx-tanx 5-(-2..,[6) 

= 

---r 

25+10../6 

= -2..,[6 

-25..,[6 -60 

= 

12 

sin x 2../6 

60. Let esc x = -3. Then x is in quadrant ill or 

quadrant IV. 

Quadrant ill: 

. 1 1 

SIDX=--=- 

cscx 3 

cos x = -~1- sin 2 x 

=-FG) 

=_~= __2~_2 

1 3 

secx =cosx =-"M 

. l 2J2 

smx+cosx -~-3 

sec x =-~ 

=(-1-2../2X_2~) 

3 

2../2 +8 

= 

9 

Quadrant IV: 

. 1 

SIDX =- 

3 

cos x = ~1-sin2 x 

= f! = 2../2 

V9 3 

3 

secx=- 

2../2 

sinx+cosx _ -i+¥ 

secx - ~ 

61. y =sin(-2x) 

y = -sin(2x) 

2 

=(-1 +3 ../2X2~) 

-2../2 +8 

= 

9 

62. It is the negative of sin(2x). 

63. y = cos (-4x) 

y= cos (4x) 

64. It is the same function. 

65. (a) y = sin (-4x) 

y=-sin (4x) 

(b) y = cos(-2x) 

y= cos(2x) 

(c) y = -5 sin(-3x) 

y = -5[-5in(3x)] 

y = 5 sin(3x) 

1 1 

66. y=sec(-x)= =--=secx 

cos(-x) cos x 

Since sec(-x) = sec x, the graph of y = sec(-x) is 

exactly the same as the graph of y = secx. 

1 1 

67. y=csc(-x)= . =-.-=-cscx 

sm(-x) -smx 

Since csc(-x) = -esc x, the graph of y = csc(-x) is 

a reflection across the x-axis of the graph of 

y = cscx. 

sin(-x) -sinx 

68. y = tan(-x) = =--=-tan x 

cos(-x) cosx 

Since tan(-x) = -tan x, the graph of y = tan(-x) is 


y = tan x. 

151


152 

Chapter 5 Trigonometric Identities 

69. 

70. 

cos(-x) cosx 

y=cot(-x)= . =-.-=-cotx 

sm(-x) -smx 

Since cot(-x) = -cot x, the graph of y = cot(-x) is 


y =cot x, 

74. Does cos(x - y) = cos x - cosy 

If it does, then the graphs of Y\ = cos(x - y) and 

Yz =cos x - cos y for specific values of y will 

overlap. We will graph 3 cases: y = 1r, Y = 1r • 

4 2 

and y= n. 

71. 

identity 

-4 

72. 

not an identity 


73. 


YJ=cos2x 

Y 2 = cos 2x - 

4 

sin 2x 

75. Does sin(x + y) = sin x + sin y If it does. then the 

graphs of Y J =sin(x+y) and Y2 =sinx+siny 

for specific values of y will overlap. We will 

1r 

graph 3 cases: y = -. y = -. and y =1r . 

4 2 

,....---.....,....---...... 

1r 

-4 

-4 

identity 

152



4. cos p(sec fJ + esc 13) = cos J_l_ + _._1_) 

I-'l cos fJ smfJ 

=l+cotp 

5. _1_ 2-+_1-2-=sin 

20+cos20=1 

esc 0 sec 0 

-4 

6. 

1 ___ = (sina+ 1)-(sina -1) 

sina-l sina+l (sina-l)(sina+l) 

2 

- sin 2a-l 

= 2_ or -2sec2 a 

cos 

2a 

Section 5.2 

-4 


Connections (pageJ99) 

Exercises 

~(I_X2)3 =~(l-(cosO)2)3 

=~(sin28)3 

= ~sin6 0 

= sin 3 0 

cos 0 is an appropriate choice since 

cos 2 0 + sin 2 0 = 1. 

1 cosO sinO 

1. co t [7+--=--+- 

Ll 

cot 0 sin 0 cos0 

cos 2 0 + sin 2 0 

= sin8cosO 

1· 

= or esc 0 sec 0 

sin OcosO 

secx cscx sinx cos x 

2. --+--=--+- 

cscx secx cos x sinx 

sin 2 x+cos 2 x 

= sinxcosx 

1 

= or cscxsecs 

sinxcosx 

3. tan s(cots+cscs) =-- 

sins (coss 

-.-+-. 

1) 

coss sins sms 

1 

=1+- 

coss 

=1+secs 

cos x sinx . 

7. --+--=cosx(cosx)+smx(sinx) 

sec x cscx 

= cos 2 x +sin 2 x 

=1 

cosr + 

2 . 2 

sinr = cosr + cos r+sm r 

8. sin r 1+ cos r (sin r)(1 + cos r) 

cosr+1 

=----''--- 

(sin rXI + cos r) 

1 

=-.- or csc y 

sm j 

9. (l+sint)2 +cos 2 t = 1+2sint+sin 2 t+cos 2 t 

=2+2sint 

10. (1+ tans)2 -2tans = 1+2 tans+ tan 2 s-2 tans 

11. 

l+cosx 

1 

= l+tan 2 s 

2 

=sec s 

12. (sina -cosa)2 = sin 2 a-2sinacosa+cos 2 a 

= 1-2sinacosa 

13. sin 2 r -1 = (sin r -1)(sin r + 1) 

14. sec 2 0 -1 = (sec 0 -1)(sec 0 + 1) 

153



15. (sin x + 1)2 - (sin x _1)2 

Let a = sin x + 1 and b = sin x-I. 

(sin x + 1)2 - (sin x _1)2 

=a 2 _b 2 

=(a-b)(a+b) 

= [(sin x + I) - (sinx -I)][(sinx + 1)+(sinx-I)] 

= 2(2 sin x) 

= 4sinx 

16. (tanx+cotx)2 -(tanx-cotx)2 

Let a =tan x + cot x and b =tan x - cot x. 

(tanx+cotx)2 -(tanx-cotx)2 

=a 2 _b 2 

=(a+b)(a-b) 

= [(tan x +cotx) + (tan x - cot x)]· 

[(tan x + cot x) - (tan x - cot x)] 

=(2 tan x)(2cot x) 

=4tanxcotx 

=4 

17. 2sin 2 x+3sinx+1 

Let a = sin x. 

2sin 2 x+ 3sinx + 1 = 2a 2 + 3a+ 1 

= (2a+ l)(a+l) 

=(2 sin x + 'l)(sin x + I) 

18. 4 tan 2 fj + tan p- 3 

Let a =tanfJ. 

4 tan 2 p+tan fj - 3 = 4a 2 +a-3 

=(4a - 3)(a + 1) 

= (4 tan p- 3)(tan fj + 1) 

4x+2cos2x+l 

19. cos 

2 

Let cos x =a. 

cos 4 x+2cos 2 x+l =a 2 +2a+l 

=(a + 1)2 

=(cos 2 x + 1)2 

20. cot 4 x+3cot 2 x+2 = (cot 2 x+2)(cot 2 x+l) 

21. sin 3 x - cos 3 x 

Let sinx = a and cos x = b. 

sin 3 x-cos 3 x 

=a 3 _b 3 

= (a -b)(a 2 +ab +b 2 ) 

= csc 2 x(cot 2 x + 2) 

= (sin x - cosx)(sin 2 x + sin xcosx + cos 2 x) 

= (sin x - cos x)(1+ sin x cos x) 

22. sin 3 a+cos 3 a 

= (sin a + cos a)(sin 2 a - sin a cos a + cos 2 a) 

= (sin a + cos a)(I- sin a cos a) 

sinO . 

23. tan 0 cos 0 = --cosO = smO 

cosO 

. cosa. 

24. cot a sma =-_.sma = cos a 

sin a 

1 

25. secrcosr=--·cosr=1 

cosr 

cost sint 

26. cotr tanr =-_._-= 1 

sint cost 

27. sinptanfj = tanfjtanp= tan 2 p 

cosp 

28. 

srh-. cok 

cot 0 ~ 

=-_.__._ 1 sinO 

sinO cosO cosO 

1 

= cos 2 0 

=sec 

2 0 

esc 0 sec0 

2 1 

29. sec x-I=---d 

cos 

2 

x 

l-cos 2 x 

- cos 2 x 

2 

sin x 2 

=--=tan 2 x 

cos x 

2 2 

30. csc t -1 = cot t 

sin 2 x . 2 . 1 

31. --2-+smxcscx = tan X+SIDX-. 

cmx ~x 

=tan 2 x+l 

2 

=sec x 

32. _1_ 2 

- + cot atana =cot 2 a + 1=csc 2 a 

tan a 

. cotO 

33. Venfy --=cmO. 

cscO 

ros9 

cotO -;--9 

--= .1!!L- = cos 0 

esc 0 si~9 

154



ify tana , 

34. Ven --=sma. 

see a 

tana ~ sina . 

--=:L =-_·cosa=sma 

see a cosa cosa 

I-sin z f3 f3 

35. Verify = cos . 

cosf3 

Z Z 

1-sin f3 = cos f3 = cos f3 

cosf3 cosf3 

ify 

36• Ven 

tan r+1 

=seer. 

seer 

Z 

tan Z r + I see Z r 

-----''--- =--= see r 

seer seer 

Z 

37. Verify cos Z O(tan 0 + I) = 1. 

Z Z 

z{sinZO) 

cos O(tan O+I)=cos --z-+I 

cos 0 

Z Z 

Z 

= cos 

{sin O+cos 0) 

2 

cos 0 

=1 

38. Verify sin Z ,8(1 + cot Z fJ) = 1. 

sin Z f3(l + cot Z fJ) = sin Z f3 csc Z f3 

. Z f3 I 

=sm . _ 

sin Z f3 

=1 

39. Verify cot s + tan s = see s esc s. 

coss sins 

cot s + tan s =--+- 

sins coss 

cos Z s + sin Z s 

= 

sinscoss 

I 

=--- 

sinscoss 

= seescscs 

sin z r 

42. Verify --=seer-cosr. 

cosy 

sin z r l-cos z r 

--= 

cosr cosy 

cos 

z 

r 

=----- 

cosy cosy 

=secr-cosr 

43. Verify sin 4 0 - cos 4 0 = 2 sin Z 0-1. 

sin 4 0-cos 4 0 

= (sin Z 0 - cos Z O)(sin Z 0 + cos Z 0) 

44. 

= sin Z 0 -cos Z 0 

= sin Z 0 -(l-sin Z 0) 

= 2sin Z 0-1 

Verify cosO = 1. 

sin ocot 0 

cosO = cosO . sinO = sinOcosO = I 

sinOcotO sinO cosO sinOcosO 

45. Verify 

(l-cos Z a)(l+cos Z a) = 2sin z a-sin 4 a . 

(1-cos z a)(1 + cos z a) = sin z a(l + cos z a) 

= sin z a(2 - sin z a) 

= 2sin z a -sin 4 a 

2 

46. Verify tan Z r sin z r = tan r + cos z r-1. 

Work with the left side. 

tan Z rsinz r = tan Z r(l- cos z r) 

Z Z z 

= tan r - tan r cos r 

Z 

= tan r-sinZr 

Now work with the right side. 

tan Z r+cosZ r -I = tan Z r-(I-cos Z r) 

2-sinz 

= tan r 

40. Verify sin Z a+ tan Z a+cos Z a = sec 2 a . 

sin Z a+ tan Z a+cos Z a = 1+ tan Z a = see Z a 

. cosa sina Z Z 

41. Venfy --+--=sec a-tan a. 

seea csca 


cosa sina cosa sina 

--+--= ---l...- +-1 

sec a esc a cos a 5iiia 

= cos 

za + sin Z a 

=1 


2 

sec a - tan 

Za = I 

155



ify cosO+l cosO 

47. Ven =-- 

tan 2 0 sec 0 -1. 


cosO+ 1 cosO+1 

tan 20 - sec20-I 

eosO+1 

=--...,,..-- 

-1--1 

2 

cos o 

eosO+I 

- l-cos 20 

cos 2 o 

cos 2 O(eosO + 1) 

= (1- cosO)(I+cosO) 

20 

cos 

= I-cosO 


20 

cosO cosO cosO cos 

secO-I = -1--1 = I-cosO = I-cosO 

cosO cosO 

_ 

48. Verify 

....:. (s_ec_ 0_- _ta_n_0....:,. )_2 _+_1_ = 2 tan O. 

sec 0 esc 0 - tan 0 esc 0 

(secO-tanO)2 +1 

secOcsc0- tanOcsc 0 

sec 2 0-2secOtanO+ tan 2 0+ 1 

= 

esc 6(sec 0 - tan 0) 

2 sec 2 0-2secOtanO 

= esc O(sec 0 - tan 0) 

2 sec O(sec0 - tan 0) 

= esc 6(sec 0 - tan 0) 

2secO 

=-- 

cscO 

=2 sinO 

cosO 

=2tanO 

49. Verify 1 + 1 2 sec 2 O. 

I-sinO l+sinO 

1 1 (l+sinO)+(I-sinO) 

---+---= 

I-sinO 1+ sinO (1- sin 0)(1 + sinO) 

2 

- I-sin 2 0 

2 

= cos 20 

= 2sec 2 0 

50. Verify =seca+tana. 

seca-tana 

1 seca + tan a 

-----=---- 

seca-tana 

sec a - tan a seca+ tan a 

sec a + tan a 

- sec 2 a - tan 2 a 

seca+tana 

=----..,...... 

±_ sin 2a 

cos a cos 2 a 

sec a + tan a 

= 2 

C:2~ 

= seca+tana 

tans sins 

51. Verify + = cots+ secscscs. 

I+coss l-coss 

---+ tans sins =_--:.__---:._-;:-.....:....__....: 

tans(l-coss)+sins(l+coss) 

l+coss l-eoss I-eos 2 s 

tans -sins + sins +sinscoss 

= 

sin 2 s 

tans coss 

=--+- 

sin 2 s sins 

sins 1 

=--'--+cots 

coss sin 2 s 

1 1 

=--'--+cots 

coss sins 

= secscscs+ cots 

. l-cosx 2 

52. Venfy = (cot x -csex) . 

I+cosx 


l-cosx (1- cosx)(1-cosx) 

---=-=---_....:....:._---' 

l+cosx (l+cosx)(I-cosx) 

1-2cosx+cos 2 x 

= 

l-cos 2 x 

2 

1- 2 cos x + cos x . 

= 

sin 2 x 

Work with the right side. 

2 (cosx 1)2 

(cotx-cscx) = -.---. 

smx 

smx 

= (co~x _1)2 

sm,r 

cos 2 x-2cosx+l 

= 

sin 2 x 

'f eota+ 1 1+ tan 

53. 

a 

Ven y =-- 

cota-l I-tana 

cot a + 1 tarhi + 1 It~'aa 1+ tan a 

cot a -1 =tarhi-1 = I~~':.ra = 1- tan a 

156



54. Verify + =-2tana. 58. 

tana-seca tana+seca 

1 1 

-----+---- 

tan a -seca tan a +seca 

tana+seca+ tan a -seca 

= 2 2 

tan a - sec a 

2tana 

= . 2 

~~2~ - c~2a 

2tana 

= -(1-sin 2 a) 

cos 2a 

2tana 

= 2 

-:2~ 

= -2 tana 

55. Verify sin 2 a sec 2 a + sin 2 acsc 2 a = sec 2 a. 59. 

" 2 

. 2 2 . 2 2 sm a 1 

sm asec a+sm acsc a=--2-+ 

cos a 

= tan 2 a+I 

2 

= sec a 

. cscO+cotO 

56. Venfy = cotOcscO. 

tanO+sinO 

cscO+cotO srne+~ 

tanO+sinO = ~+sinO 

- ~ 

- sin6(1+cosB) 

cose 

(1 + cosO) cosO 

= sin 2 0(1+ cosO) 

cosO 

= sin 20 

=-_._- 

1 cosO 

sinO sinO 

=cscOcotO 

57. Verify sec 4 x-sec 2 x=tan 4 x + tan 2 x , 

Simplify left side. 

4 

sec x-sec 2 x= sec 2 x(sec 2 x-I) 

60. 

2 2 61. 

= sec x tan x 

Simplify right side. 

4 2 

tan x+tan x=tan 2 x(tan 2 x+I) 

2 2 

= tan x sec x 

. I-sinO 2 2 

Venfy =sec 0-2secOtanO+tan O. 

1+sinO 


sec 2 0 - 2secOtan 0 + tan 2 0 

1 2sinO sin 20 

=------+-- 

20 20 20 

cos cos cos 

1-2sinO+sin 2 0 

= 20 

cos 

(1 - sin 0)(1-sin 0) 

- I-sin 20 

_ (1 - sin 0)(1- sin 0) 

- (1 + sinOXI-sinO) 

I-sinO 

= 1+ sinO 

.. sinO cosO 

Venfy smO+cosO = ~~(J + -.Sin1! • 

I-~ I-~ 


sin 0 cosO sin 20 cos 20 

6+ " 6 = + 

I-~ 1- ~e sinO-cosO cosO-sinO 

sin 20 cos 

= 20 

+ 

sinO- cosO -{sinO ~ cosO) 

sin 2 0-cos 2 0 

= sinO-cosO 

(sin 0 + cosOXsin0 - cos 0) 

=-'--------:...;:.-_--~ 

(sin0 - cos 0) 

= sinO+cosO 

sinO sinOcosO 0( 2) 

Verify 

I-cosO l+cosO 

esc I + cos 0 . 

sinO sinOcosO 

I-cosO I+cosO 

sin 0 + sinOcosO - sin OcosO+ sin0 cos 2 0 

= 

sin 20 

20 

cos 

=cscO+-- 

sinO 

=cscO(I + cos 2 0) 

4 

sec s - tan 4 s 

Verify 

2s-tan2s 

2 2 =sec . 

sec s+ tan s 

2 2 

sec 4 s-tan 4 s _ (sec s-tan 2 sX sec2 s+tan s) 

sec 2 s+tan 2 s - sec 2 s+tan 2 s 

2 2 

= sec s - tan s 

157



62. Veriy f 

cot2 t -I I 2 ' 2 t 

2=-sm. 

1+ cot t 

2 

cot 2 t-I ~-I 

l+cot 2 t = I+~ 

SID t 

COS 2 t-sin 2 t 

_ sin 2 t 

- sio 2 t+cos 2 t 

sin 2 t 

cos 2 t - sin 2 t 

= 

sin 2 t 

2t-sin2t 

=cos 

=(l-sin 2 t) -sin 2 t 

=1-2sin 2 t 

tan 2 t -I tant - cott 

63. Verify sec2 t = tan t + cott 


tant -cott tant-~ 

= -ltant 

+ cott 

tan t + taot 

.1a1U:J. 

=--¥L- 

JilId.±l. 

tant 

tan 2 t-I 

- tan 2 t+1 

2 

tan t -I 

= sec 2 t 

64. Verify (l+sinx+cosx)2 =2(I+sinx)(l+cosx). 


2(1 + sin x)(1 + cosx) 

= 2(1 +sinx +cosx+sinxcosx) 

= 2 + 2sin x + 2 cos x + 2sin x cos x 


(l+sinx+cosx)2 

. . 2 • 

= I+sinx+cosx+smx+sm x+smxcosx 

+cosx+sinxcosx+cos 2 x 

= 2 + 2sinx + 2cosx + 2sin xcosx 

I+cosx I-cosx 

65. Verify - =4cotxcscx. 

I-cosx I+cosx 

I+cosx I-cosx 

I-cosx I+cosx 

l+cos 2 x+2cosx-I-cos 2 x+2cosx 

= 

l-cos 2 x 

4cosx 

= sin 2 x 

=4cotxcscx 

2 I-sina 

66. Verify (seca-tana) = . . 

I+sma 

(seca-tana)2 = sec 2 a- 2secatana+tan 2 a 

I 2sin a sin 2 a 

--------+-- 

2 2 2 a 

- cos a cos a cos 

1-2sina+sin 2 a 

2a 

cos 

(l-sina)2 

= l-sin 2 a 

(l-sina)2 

=-~----,..---"----:- 

(I-sin a)(1+sin a) 

I-sina 

= I+sina 

67. Verify (sec a + csca)(cosa -sina) =cow -tana. 

(seca + esc a)(cosa - sin a) 

= (_1_ + -._I_)(cos a - sin a) 

cosa smrz 

= I +cota - tana-I 

=cota-tana 

. 4 4 

sin a-cos a =I . 

68. Verify 2 2 

sin a-cos a 

sin 4 a - cos 4 a 

sin 2 a - cos 2 a 

71. 

(sin 2 a - cos 2 aXsin 2 a + cos 2 a) _ 

= 

sin 2 a-cos 

2 

a 

-I 

The graph of y = (secx + tanx)(l-sinx) 

appears to be the same as the graph of y = cos x. 

(sec 0 + tan 0)(1- sin0) =(_1_ + _si_n_0)1-sin 0) 

cosO cosO 

=(I + sin 0)(1- sin 0) 

cosO 

l-sin 20 

= 

cosO 

cos 

20 

=-- 

cosO 

=cosO 

158


72. 74. 

+ cotx)(secx - 1) y 1 = tan x sin x + cos x 


73. 

The graph of y = (cscx+cotxXsecx-l) 

appears to be the same as the graph of y = tan x . 

(cscO+ cotOXsecO -1) 

=(_1+ cos 0X-1-1) 

sinO sinO cosO 

=(1+.COSOXI-COSO) 

smO cosO 

l-cos 

= 20 

sinOcosO 

sin 

= 20 

sin 0 cos 0 

sinO 

=- 

cosO 

=tan 0 

The graph of y = tan x sin x + cos x appears to be 

the same as the graph of y = sec x . 

tanOsinO+cosO= sinO (sinO)+cosO 

cosO 

sin 20 cos 20 

=--+-- 

cosO cosO 

sin 2 0 + cos 2 0 

= 

cosO 

1 

=- 

cosO 

=secO 

2+5coss 2 5 identi 

75• Is . = cscs+ cots ani entity 

sms 

Graph each side of the equation. 

y =2+Scos.>: 

1 sin r 

Y2 = 2cscx + 5cotx 

4 

-4 

cosx+ 1 

The graph of y = . appears to be the 

smx+tanx 

same as the graph of y =cot x. 

cosO+ 1 1+ cosO 

= .mul. 

sinO+tanO sinO+ cosO 

l+cosO 

- sin 0(1 + JsO) 

_ 

(l+cosO)cosO 

- sin 0(1 + cobi)cosO 

_ (I + cos 0) cos 0 

- sin O(cosO+ I) 

cosO 

=- 

sinO 

= cot 6 

-4 

2+5cosx 

The graphs of y = . and 

smx 

y = 2 esc x +5 cot x appear to be the same . The 

given equation may be an identity. Verify it. 

2+5coss 2 5coss 

----=--+- 

sins sins sins 

= 2 esc s + 5 cot s 

The given statement is an identity. 

159



2 

sec s 

76. Is I + cot 2 s = 2 an identity 

sec s-I 

Gra h each side of the equation. 

I 

y = 1 + 1 

Y I = 1 + cotZx I 1 + sin x "':"l-----"s.,...in-x- II 

Y _ scc Z x 

z-sccZx-l 

sec 2 x 

The graphs of y = I + cot 2 x and y = --:::2"""'- 

sec x-I 

appear to be the same. The given equation may be 

an identity . Verify it. 


sec 2 s sec 2 s 

sec 2 s -I =tan 2 s 

1 cos 2 s 

= cos 2 s . sin 2 s 

I 2 

=--=csc 

sin 2 s 

s 

= l+cot 2 s 


I I 

The graphs of y = . + . and 

I+slnx I-smx 

y =sec 2 x are not the same. The given statement 

is not an identity. 

79. Is l-tan2 s 

2:= cos 2 s 

1+ tan s 

.. . 

Sin S an identity 

Graph each side of the equation. 

1 tanZ x 

YI := -=-l-+---"tan==-Z..:.:.x 

77. 

l-tan 2 x 

The graphs of y = 2 

I+tan x and 

y = cos 2 x sin x are not the same. The given 

statement is not an identity. 

-4 

tan x cot x 

. 2 

The graphs of y = and y = 2sm x 

tanx+cotx 

are not the same. The given statement is not an 

identity. 

80. Is sin3 s-cos 3 s . 2 . 2 

= sm s + 2 Sin S cos S + cos s 

sins-coss 

an identity Graph each side of the equation. 

YZ = sinZ x + 2sin x cos x + cosZ x 

-4 

sin 3 x cos 3 x 

The graphs of y = . 

and 

smx-cosx 

y=sin 2 x+2sinxcosx+cos 2 x are not the 

same. The given statement is not an identity. 

160



81. cot 0 84• IS = sec 0 - tan O'd an I entity. ' 

cscO+ 1 

Evaluate each side of the equation using a table of 

cot 0 

values. Let Y 1 = and Y 2 =secO-tanO. 

cscO+ 1 

82. 

The table values are not the same. The given 


The table values are the same for all values for 

which Y 1 and Y2 are both defined. If your table 

contained only points for which they are both 

defined, the equation would appear to be an 

identity. Since there are values for which one is 

defined and the other is not defined, the given 


85. Show that sin (esc s) = 1 is not an identity. 

Lets= 2. 

sin (esc 2) =:: .89 10940 1 

Thus, sin(esc s) ¢. 1 for all teal numbers s, so this 

statement is not an identity . 

83. 

The table values are not the same. The given 

statement is not an identity. . 

The table values appear to be the same. The given 

equation may be an identity. Verify it. 

2 . 2 sin 2 x sin 2 xcos 2 x 

tan x-sm x=-----....,.....- 

cos 2 x cos 2 x 

_ sin 2 x(l- cos 2 x) 

- cos 2 x 

sin 2 x ( 2) 

=--2- l e-cos x 

cos x 

= tan 2 xsin 2 x 

= (tanxsinx)2 


86. Show that ~cos2 s =coss is not an identity. 

Let s = 1. 

~cos21 =.54030231 

cos 1 =.54030231 

2n 

But let s=-. 

3 

1 

coss=- 2 

~cos2 s =~(-~r =JT =± 

Thus, ~cos2 s ¢. coss for all real numbers s, so 

this statement is not an identity. 

87. Show that csct =~1 + cot 2 t is not an identity. 

Lett =1. 

esc =:: 1.1883951 

~r-1+-c-o--=t2 1 

=:: 1.1883951 

But let t= 4. 

esc 4 =:: -1.3213487 

~1 + cot 21 =:: 1.3213487 

Recall that .Ja denotes the positive square root of 

a. 

Thus, esc t ¢. ~1 + cos 2 t for all real numbers t, so 

this statement is not an identity . 

161



88. Show that cos t = ~ I - sin 2 t is not an identity. 

6. cos(-105°) 

= co~-45°+(-600)] 

31r 

Let t=-. 

2 

= cos(-45°)cos(-60°) - sin(-45°)sin(-600) 

Then cos t = O. = cos 45° cos 60° - (-sin 45°X- sin 60°) 

~1-sin2t =~I-(-I) =.Ji 

Thus, cos t *-~1-sin 2 t for all real numbers t, so 

this statement is not an identity. =-- 

4 4 

89. Let tan(J = t and show that sin (Jcos(J =-f-. = 

Section 5.3 

Exercises 


t tan(J 

=.J2 ..!.- (_ .J2x- .J3] 

2 2 2 2 

.Ji.J6 

.J2-.J6 

t +1 4 

7. co{~~)=cos(~ + :) 

t 2 + I = tan 2 (J+ I 

tan(J 1r 1r . 1r . 1r 

= cos - cos - - sm-sm 

= sec2 (J 3 4 3 4 

=tan (Jcos 2 (J I .Ji .J3.Ji 

=-._-_. 

sin (J 2 (J 

=--cos 

2 2 2 2 

cos(J 

.Ji.J6 

= sin (Jcos (J 

=-- 

4 4 

.Ji-.J6 

= 

4 

3. cos 75° 

= cos(30° + 45°) 

1r x .1r .1r 

= cos 30°cos 45° - sin 30° sin 45° = cos -cos-+sm-sm 

6 4 6 4 

4 

=_._--.- .J3.JiI.Ji 

=_._+_. .J3..fiI.J2 

2 2 2 2 

2 2 2 2 

= 

.J6-.Ji 

.J6.Ji 

=-+ 

4 4 

4. cos(-15°) .J6+.Ji 

= 

= cos(30° - 45°) 4 

=cos 30° cos 45° + sin 30° sin 45° 

.J3.JiI.Ji 

9. cos 40° cos 50° - sin 40° sin 50°= cos(40° + 50°) . 

=_._+_.- = cos 90° 

2 2 2 2 =0 

.J6+.Ji 

= 10. cos(_10°) cos 35° + sin(_10°) sin 35° 

4 

= cos(-IO° - 35°) 

5. cos(105°) 

=cos(-45°) 

=cos(60° + 45°) 

= cos(600)cos(45°) - sin(60°) sin(45°) .J2 

=- 2 

=!. .J2 -(.J3X.Ji) 

2 2 2 2 21r 1r . 21r . 1r J21r 1r) 

11. cosScos 10 -smSsm io = CO'S+ 10 

=-- 

.J2.J6 

4 4 51r 

=cos 

.J2-.J6 10 

= 1r 

4 = cos 

2 

=0 

162



7rr 2rr . 7rr . 2rr J7rr 2rr) 

12. =co'\9+9 

cos 

9cos9-sJn 9sJn9 

= cosrr 

=-1 

13. The answer to Exercise 9 is O. Using a calculator 

to evaluate cos 40° cos 50° - sin 40° sin 50° also 

gives the value of O. 

14. The answer to Exercise 10 is ..fi. Using a 

2 

calculator to approximate ..fi gives the value 

2 

.70710678. Using a calculator to evaluate 

cos(-10°) cos 35° + sin(-IO O )sin 35° also gives 

the value .70710678. 

17. cos ~ = sin( ~ - ~) = sin ~; 

19. csc(-14°24') = sec[900-(-14°24')] 

= sec 104°24' 

22. cot 9rr = tan( rr _ 9rr) = tan(- 2rr) 

10 2 10 5 

25. cot 176.9814° = tan(900-176.9814°) 

= tan(-86.9814°) 

26. sin98.0142° = cos(9O° - 98.0142°) 

= cos(-8.0142°) 

28. sin _2;_ = ( _ :) 

. rr rr 2rr 

Since --=-- 

6 2 3' 

sin 2; = cos(; _ 2;) ::;; cos(- : ) 

Answer: cos 

29. 33° ::;; sin 57° 

sin 57° = cos(9O°- 57°) 

= cos 33° 

Answer: cos 

30. 72° ::;; cot 18° 

Since 90° - 18°::;; 72°, 

cot 18° = tan (90° - 18°)::;; tan 72°. 

Answer: tan 

1 

31. cos 70° = -- 

20° 

cos 70° = sin(9O°- 700) 

= sin 20° 

1 

=-- 

csc20° 

Answer: esc 

1 

32. tan 24° =--- 

66° 

Since 90° - 24° ::;; 66°, 

tan 24° = cot(90° - 24°) 

= cot 66° 

1 

= tan 66° . 

Answer: tan 

33. tan 8 = cot(45° + 28) 

Since tan 8 = cot(90° - 8), 

90° - 8 = 45°+28 

38 =45° 

8 = 15°. 

34. sin 8 =cos(28 _10°) 

Since sin 8 =cos(90° - 8), 

90° - 8 =28 - 10° 

100° = 38 

100° 

8=-. 

3 

rr rr 

27. cot-= 

3 --6 

. rrrrrr rr rr 

SInce -::;; - - - cot - = tan - . 

6 2 3' 3 6 

Answer: tan 

163



35. sec 8 = csc(~ + 20°) 

Since sec 8 = csc(90° - 8). 

8 

90°-8 = -+20°. 

2 

~+200=900-8 

2 

8+~+200=90° 

2 

28 + 8 + 40° = 180° 

30= 140° 

140° 

8= 

3 

36. cos 8 = sin(30+ 10°) 

Since cos8 = sin(90° - 8). 

90° - 8 = 30 + 10° 

80°=40 

8 = 20°. 

37. sin(30-15°)=(eosO+25°) 

Since sin 0 = eos(90° - 0), 

sin(30 -15°) = eos(90° - (30 -15°)] . 

90° - (30 -15°) = 8 + 25° 

105° - 30 = 0 + 25° 

80°=40 

20°=0 

38. eot(0 -10°) = tan(20 + 20°) 

Since cot 8 = tan(90° - 0) , 

oot(O _10°) = tan[90° - (0-10 0 )l 

90°_ (0-10°) = 20+20° 

100°-0 =20+20° 

80°=30 

80° 

-=0 

3 

39. cos(O°- 0) = cos 0° cos 0 + sinOsin 0° 

= 1eosO + Osin 8 

=eosO 

40. cos(90° - 0) = cos 90° cosO + sin900sin8 

... = OcosO+ lsinO 

=sinO 

43. cos(O°+8) = cosO°cos8 -sinOosin8 

= 1cos8-0sin8 

= cos8 

44. cos(90° + 8) = cos 90° cos 8 - sin 90° sin 8 

= 0 cos 8 - 1sin 8 

= -sin8 

45. cos(180° + 8) = cos 180° cos 8 - sin 180° sin 8 

= (-1)cos8 - Osin 8 

=-cos8 

46. cos(270° + 0) = cos 270° cosO - sin 270°sin 0 

= OcosO - (-1)sin 0 

=sinO 

1 d si 3 

47. coss=-- an sml=-. 

5 5 

sand 1 are in quadrant II. 

Solve sin 2 s + cos 2 s = 1 for sin s. Since s is in 

quadrant II. sin s > O. 

sins=~l-(-~r = ~ 

Solve sin 2 1 + cos 2 1 = 1 for cos I. Since 1 is in 

quadrant II, cos 1 < O. 

cOSI=-~l-(~r =-~ 

cos(s + I) = cos s cosr - sm ssinr 

=(-~X -~)-( ~X~) 

4 'Wi4 

=--- 

25 25 

4-&./6 

= 

25 

cos(s -I) = eosseOSI + sin ssinr 

=(-~X -~)+( ~X~) 

4 3..fi4 

=-+- 

25 25 

4+&./6 

= 

25 

41. cos(180° - 0) = cos 180° cos 0 + sin 180° sin 0 

= (-1)cos 0 + Osin 0 

=-cosO 

42. cos(270° - 0) = cos 270° cos 0 + sin 270° sin 0 

= OeosO+ (-l)sin 0 

=-sinO 

164



48• sm · s =-2 an d sm si 1 =- -. 1 

3 3 

s is in quadrant II and 1 is in quadrant IV. 

First, find cos s and cos I . 


cos 2 s = 1-sin 2 s 

=1-(%r 

5 

= 

9 

Since s is in quadrant Il, cos s < 0, so 

coss=-~ =- ~. 

cos 2 1 = 1-sin 2 1 

=1-(-~r 

8 

= 

9 

Since 1 is in quadrant IV, cos I> 0, so 

cos 1 = f! = 2.,fi . 

~9 3 

cos(s + I) = cos s cosr -sinssinl 

=(- ~X2~)_(%X-~) 

-2.,JW +2 

= 

9 

cos(s -I) = cos s cosr +sinssinl 

=(- ~X2~)+(~X-~) 

-2.,JW -2 

= 

9 

' 3 d si 12 

49• SIOS=- an SIOI=--. 

5 13 

s is in quadrant I and 1 is in quadrant III 

cos s > 0 and cos 1 < O. 

COSS=~I-(%r =~ 

53 

cos 1 = -~1-(- ~~r = -1 

cos(s + 1)= cos s cos 1- sin s sinr 

=(~X-1~)-(~X-~~) 

16 

= 

65 

cos (s - 1)= cosscosl + sin ssinr 

5 

=(~X-13) + ( %X - ~~) 

56 

=- 

65 

8 3 

so. coss =-- and COsI = --. 

17 5 

sand 1are in quadrant ill. 

First, find sin s and sin I. 

2 

. 2 2 8 225 

sm s =1-cos s = 1- ( - 17 ) = 289 

Since s is in quadrant ill, sin s < 0, so 

sins = -~~~~ = - :~ '. 

2 

. 2 2 3 16 

sm 1=I-cos 1=1- ( -'5 ) = 25 

Since 1 is in quadrant ill, sin 1 < 0, so 

sinl=-~ =-~. 

COS(s+l) =cos s cosr - sins sinr 

=(-1~ X-~)-(-:~X-~) 

24 60 

=-- 

85 85 

36 

=- 

85 

cos(s -I) = cosscos 1+sinssinl 

24 60 

=-+ 

85 85 

84 

= 

85 

165



51. 

. .J5 d. ../6 

S1OS=- an S1OI=-. 

7 8 

sand 1 are in quadrant I. 

cos s > 0 and cos 1 > o. 

COS'=~I-[ ~)' ~~ 

~ (../6)Z. $8 

COSI= 1-""""8" =-8 

cos(s + 1)= cosSCOSI -sinssinl 

cos(s 

=(~X~)-(~)(~) 

2.J638 -.J30 

= 

53. 

Since 1 is in quadrant IV, cos I> 0, so 

.s: 

cosl=--. 

6 

cos(s + 1)= cos s cosr sin ssinr 

=(V;)('T)-(-~Xv;) 

...[6i-~ 

= 

24 

cos(s I) = coss cosr + sin s sinr 

...[6i+~ 

= 

24 

cos 42° = cos(30° + 12°) 

Since 42° = 30° + 12°, the statement is true. 

56 

I) =cos scosr + sins sinr 54. cos (_24°) = cos 16° - cos 40° 

cos (A B) = cosAcosB+ sinAsinB 

=(~X~)+(~X~) 

cos (-24°)= cos(16° - 40°) 

= cos 16° cos 40° + sin 16°sin 40° 

2.J638 +.J30 

¢. cos 16° - cos 40° 

= 

The statement is false. 

56 

52. 

-J2 . .J5 

coss=- and sml=--. 

4 6 

sand 1 in quadrant IV. 

First. find sin s and cos I. 

sin z s = l-cos Z s 

55. cos 74° = cos 60° cos 14° + sin 60° sin 14° 

cos(A + B) = cos A cos B-sin Asin B 

cos 74° = cos(6O°+ 14°) 

= cos 60° cos 14° - sin 60° sin 14° 


56. cos 140° = cos 60° cos 80° - sin 60° sin 80° 

=1-(V;)' 

cos (A + B) = cos A cos B sin A sin B 

1 

=1- cos 140° = cos(600 + 80°) 

8 

7 

= 8 

Since s is in quadrant IV, sin s < 0, so 

sins=-~ =-i=-21 =~. 

57. 

= cos 60° cos 80° - sin 60° sin 80° 

The statement is true . 

x 1r 1r .1r.1r 

cos = cos-cos- SIO -S1O 

3 12 4 12 4 

cos(A + B) = cosAcos B-sinAsin B 

cos"3=co{41r) 

co,, t=l-,in' t= 1-[- ~)' = :~ 

=co{.!!-+ 31r) 

12 12 

=co{~ + ~) 

12 

1r 1r .1r.1r 

= cos -cos - SIO -S1O 

12 4 12 4 

The statement is true. 

166



2n lIn n . lIn . n 

58. cos- =cos--cos-+sm--sm 

3 12 4 12 4 

cos (A - B) = cos A cos B + sin A sin B 

cos ""3 2n = cos (8n) 12 

= cos( lIn _ 3n) 

12 12 

= cos( lIn _ n) 

12 4 

lIn n . l br . n 

= cos--cos- + sm --sm 

12 4 12 4 


59. cos 70° cos 20° - sin 70° sin 20° = 0 

cos(A + B) = cos A cos B - sin A sin B 

cos 70° cos 20° - sin 70° sin 20° = cos(70° + 20°) 

= cos 90° 

=0 


60. cos 85° cos 40° + sin 85° sin 40° = ..fi 

2 

cos A cos B + sin A sin B = cos (A - B) 

cos 85° cos 40° + sin 85° sin 40° = cos(85° - 40°) 

= cos 45° 

..fi 

= 

2 


61. ta{O- ~)=cotO 

Simplify the left side. 

tan(o _ n) = sin( 0 - t) 

2 cos(O-f) 

-sin(f-O) 

- co~f-O) 

-cosO 

=-- 

sinO 

=-cotO 


62. sin(0 - ~) = cos 0 

sin(o- ~)=-sin(~-0)=-COSO 


63. Verify cos( ~ + x) = -sinx. 

n ) n . n . 

co -+x =cos-cosx-sm-smx 

{ 2 2 2 

= 0 cos x-Isin x 

=-sinx 

64. Verify sec (n-x) =-secx. 

1 

sec(n - x) =--- 

cosor - x) 

1 

=------- 

cos n cos x + sin n sin x 

1 

=------ 

(-1) cos x + (0) sin x 

1 

=- 

cosx 

=-secx 

2 

65. Verify cos2x = cos x - sin 2 x. 

cos2x =cos(x+ x) 

= cos x cos x - sin x sin x 

= cos 2 x -sin 2 x 

66. Verify I+cos2x-cos 2 x=cos x. 

2 

From Exercise 65, cos2x = cos x -sin 2 x . 

I+cos2x-cos 2 x=I+cos 2 x-sin 2 x-cos 2 x 

= I-sin 2 x 

2 

=cos x 

67. Verify cos(n+ s-t) = -sinssint -cosscost. 

cos(n+ s - t) = co~n+ (s-t)] 

= cosncos(s--t)-sin nsin(s - t) 

= (-I)cos(s - t) -Osin(s - t) 

= (-IXcosscost + sin ssin t) 

=-cosscost -sinssint 

68. Verify co{~ +S-t)=sin(t-s). 

co{~ +s-t) =co{~ +(s-t)] 

2 

= cos n cos(s_ t)- sin n sin(s - t) 

2 2 

= 0 -Isin(s -t) 

= sin(t -s) 

167



69. Verify cos(a + /3)cos(a - /3)= l-sin 2 a - sin 2 fJ. 


cos(a + fJ)cos(a"- fJ) = (cos a cos fJ -sin asinfJXcosacos fJ + sin asinfJ) 

= cos 2 acos 2 fJ-sin 2 asin 2 fJ +cosacosfJsinasinfJ -sinasinfJcosacosfJ 

= (l-sin 2 fJ)cos 2 a -(I-oos 2 a)sin 2 fJ 

= cos 2 a - sin 2 fJ cos 2 a - sin 2 fJ + cos 2 asin 2 fJ 

= cos 2 a-sin 2 fJ 

= (l-sin 2 a)-sin 2 fJ 

70. Verify cos 4x cos Tx - sin 4x sin Tx = cos l Ix, 76. (a) cos 255° =cos(180°+ 75°) 

cos 4x cos 7x- sin 4x sin 7x =cos (4x+ 7x) == cos 180°cos 75° - sin 180°sin 75° 

= cos l l,r 

= (-1)cos 75° -(O)sin75° 

72. cos 20° =sin (90° - 20°) = sin 70° =-cos75° 

Some other values that would make 

= - cos(45° + 30°) 

cos20° =sin 9 true would be any angle =-(cos45° cos30° - sin 45°sin 30°) 

cotenninal with 70°, such as 430°, 790°, 1150 0 , 

-290 0 , -650 0 • =J~.~_~.!) 

Furthermore, since sin 70 0 = sin 110°, 9 can be 1.2 2 22 

110°or any angle coterminal with 110° (for 

example, 470°, 830°, 1190°, -250°, -610°). = 

.J6-~ 

4 

73. cos 195 0 = cos(180°+ 15°) ~-.J6 

= 

= cos 180°cosl5° - sin180°sin15° 4 

= (-1)cos15° - (0)sin 15° 

=-cosI5° 

(b) cos \1; = cofr - ~) 

74. -



78. Verify tan A =cot (90° - A). 

sinA 

tan A =- 

cos A 

_ cos(90° - A) 

- sin(90° - A) 

= cot(90° - A) 

79. Verify 

- .::....;f(:.-x+_h~)-......:f~(x--"-) =cosx (COSh -1) . (Sinh) 

-SIDX - . 

h h h 

Letf (x) = cos x. 

f(x + h)=cos(x+ h) 

=cosxcosh -sinxsinh 

f(x+h)- f(x) =cosxcosh-sinxsinh-cosx 

h 

h 

cosxcosh - cosx - sinxsinh 

= 

h 

cos x cos h - cosx sin xsinh 

= 

h h 

=cosx 

COSh-1) 

h 

. (Sin h) 

( -SIDX -h 

80. Verify cos 140° + cos 100° + cos 20° = O. 

cos 140° + cos 100° + cos 20° = cos(I20° + 20°) + cos(I20° - 20°) + cos 20° 

= cos 120°cos20° - sin 120°sin 20° + cos 120°cos 20° + sin 120°sin 20° +cos 20° 

= -.!.cos20° - .J3 sin 20° + ( _.!.) cos 20° + .J3 sin 20° + cos 20° 

2 2 2 2 

= - cos 20"+cos 20° 

=0 

81. Since there are 60 cycles per sec, the number of 83. (a) Graph 

cycles in .05 sec is given by 

(.05 sec)(60 cycles per sec) = 3 cycles . p=;cos[2: -C1J 

82. Since V = 163sinca and the maximum value of = ~ co{2H(IO) -10261) 

sin ca is 1, the maximum voltage is 163. 10 4.9 

Since V = 163sinca and the minimum value of 

sin ca is -1, the minimum voltage is -163. 

Therefore, the voltage is not always equal to 115. 

= .04cos(20H -10261). 

4.9 

For x = r, 

4 [201r ] 

P(t) = to cos 4:9 - 1026t 

.057/ 

£-,: r':\('l#,,:;t:;ri; 3~ 

; \: .r. ';i .~ ,~' ',~ I 

o .05 

. 

.i , ~~ ! t ~, - :< .~ , i> ~~ :: 

- .05 

The pressure P is oscillating. 

169



(b) Graph 

p=;co{2~ -ctJ 

= ~cos[21r(r) -1026(10)] 

r 4.9 

=~ coJ 2m- -10.260). 

r \'4.9 

For x = r, 

3 [2Trr 1 

per) = r cos 4:9 - 10,260 

Section 5.4 

Exercises 

3. sin 15° = sin(45° - 30°) 

=sin 45° cos 30° - cos 45° sin 30° 

..fi ~ 

..fi I 

=-._-_. 

2 2 2 2 

.J6..fi 

=-- 

4 4 

.J6-..fi 

= 

4 

The answer is C. 

(c) 

-2 

The pressure oscillates. and amplitude 

decreases as r increases. 

p=;co{2~ -ctJ 

Let r=nll. . 

=.!!....co{2mzll. -ctJ 

nil. Il. 

a 

=- cos(2mz - ct] 

nil. 

= n~ [cos(2mz)cos(ct) + sin(2mz)sin(ct)] 

= n~ [(I)cos(ct) + (O)sin(ct)] 

a 

=-cos(ct) 

nil. 

4. sin 105° =sin(60° + 45°) 

= sin 60° cos 45° + cos 60° sin 45° 

~..fil..fi =_._+_. 

2 2 2 2 

.J6+..fi 

= 

4 

The answer is A. 

5. tan 15° =tan(6QO -45°) 

tan 60° - tan 45° 

= I + tan 60° tan 45° 

~-I ~-I I-~ 

= I+~(I) = I+~ ·I-~ 

~ -3-1+ ·~ -4+2~ 

= =-_..:..:... 

1-3 -2 

-2(2-~) 

= =2-~ 

-2 

The answer is E. 

84. y' =r cos(0 + R) 

= r[cosOcos R-sinOsin R] 

= (rcosO)cosR-(rsinO)sin R 

= ycosR- zsin R 

z' =rsin(O+ R) 

=rco{~ -(O+R») 

=rco{(~ -O)-R) 

= r[cos(~ -0)CosR+sin(~-0)sin R] 

=r[sinOcos R+ cosOsin R] 

= (rsinO)cos R+(rcosO)sin R 

= zcos R+ ysin R 

6. tan 105° =tan(60° + 45°) 

tan 60° + tan 45° 

= 1-tan 60° tan 45° 

.J3+1 .J3+1 

= 1-(.J3XI) =1-.J3 

.J3+ I 1+.J3 4 + 2.J3 

=1-.J3 ·I+~ = -2 

=-2-.J3 

The answer is F. 

7. sin(-105°)=sin(45° - 150°) 

=sin 45° cos 150° - cos 45° sin 150° 

=( ~X-~)-( ~X~) 

= 

The answer is B. 

-J6 -..fi 

4 

170



8. tan(-105°) = -tan105° 

= - tan(60° + 45°) 

tan 60° + tan 45° 

= 1- tan60° tan 45° 

../3 +1 1+../3 

=~·1+../3 

1+2../3+3 

= 

-2 

4+2../3 

= 

-2 

= 2+../3 

The answer is D. 

9. sin ~~ =Sin(: + :) 

.re re re.re 

= sm -cos- + cos-s1o 

4 6 ·4 6 

=_._+_. 

-J2../3 -J2 1 

2 2 2 2 

..f6-J2 

=-+ 

4 4 

..f6+-J2 

= 

4 

10. tan~~ =tan(: +:) 

_ tant+tant 

-1-tan JI..tan JI.. 

6 4 

*+1 1;[ 

=1-*=1i 1 

1+.J3 ../3+1 

= ../3-1 ' .J3+1 

1+2.J3+3 

= 

2 

4+2.J3 

= 

2 

=2+../3 

11. tan 1~ = tan(: - :) 

tant-tant 

. l+tanftanf 

_ 1-4 _1-4 

- 1+(1)4- - 1+4 

3-.J3 

=3+.J3 

3-.J3 3-.J3 

= 3+../3' 3-.J3 

= 

12-6.J3 

6 

=2-../3 

12. sin ~ =sin(: - :) 

.re re re.re 

=S1O-COS--cos-s1o 

4 6 4 6 

=_._-_.- -J2.J3 ..fi 1 

2 2 2 2 

../6 ..fi 

=-- 

4 4 

../6-..fi 

= 

4 

13. sin(- ~~) = Sin(- : - ~ ) 

. (re) re {re). re 

=S1O -"4 cos 3--eo -"4 S10 3 

.re re re.re 

= -S1O- cos - - cos -SID 

4 3 4 3 

..fi 1 -J2.J3 ..fi..f6 

=-_.--_._=--- 

2 2 2 .2 4 4 

-J2-../6 

= 

4 

171



14. tan(- ;;) = tan[- =+ ( - ~)] 22. tan "* + tan t (5rr rr) 

I-tan1ftan t = tan 12+"4 

15. 

tan(-t) + tan(-i) 

= 1-tan(-t)tan(-t) 

_ -1+(-J3) 

- l-(-lX-J3) 

_-I-..[j 

- 1-..J3 

-1-..J3 1+../3 

= 1-13 '1+..J3 

-1-..J3 -..J3-3 

= 

-2 

-4-2..J3 

= 

-2 

=2+..J3 

8rr 

= tan 

12 

2rr 

= tan 3 

=-..J3 

23. cos(300 + 9) = cos 30° cos 9 - sin30°sin 9 

= ..J3cos9-.!.sin9 

2 2 

=: ±(..J3 cos9-sin9) 

..J3cos9 - sin 9 

= 

2 

24. cos(45° - 9) = cos 45°cos 9 + sin45° sin 9 

-Ii 

-Ii ' 

= -cos9+-sm9 

2 2 

sin 76° cos 31° - cos 76° sin 31° 

= sin(76° - 31°) 

..fi(cos9 + sin 9) 

= sin 45° 

2 

..fi 

= 2 

25. cos(600+9) = cos60°cos9-sin60osin9 

= .!.cos9 - ..J3sin9 

16. sin 40° cos 50° + cos 40° sin 50° 

2 2 

= sin(40° + 50°) 

= ±(cos9-..J3sin 9) 

sin 90° 

= 1 cos9-..J3sin9 

= 

2 

17. tan 80° + tan 55° = tan(800+ 550) 

1- tan 80° tan 55° 26. cos(9 - 30°) = cos 9 cos 30° +sin9sin 30° 

= tan 135° 

=-1 

..J3 9 

1'9 

=:-cos +-sm 

2 2 

18. tan 80° - tan(-55°) = tan[800-(-550)] 

..[j cos9+sin9 

= 

1+ tan 80° tan(-55°) 

2 

= tan 135° 

=-1 

31r ) 3rr , 3rr , 

27. co 

{--x =cos-cosx+sm-smx 

4 4 4 

19. tan 100°+ tan 80° = tan(I000+ 800) 

-Ii ..fi . 

=--cosx+-smx 

1- tan 100°tan 800 2 2 

= tan 180° 

=0 

=:- ..fi(-cosx+smx 

inx) 

2 

20. sin 100° cos 10° - cos 100° sin 10° 

=: sin(l000-100) 

=: sin 90° 

= 1 

21. sin rr cos 3rr + cos rr sin 3rr = sin(!E.+ 3rr) 

510 510 510 

, rr 

=sm 2 

=1 

=: ..fi(sinx-cosx) 

2 

172



28. sin(45° + 0) = sin 45° cos 0 + sin 0 cos 45° 

= .J2 sinO+ .J2 cosO 

2 2 

= .J2 (sin 0 + cos 0) 

2 

.J2(sin 0 + cosO) 

= 

2 

tan 0 + tan 30° 

29. tan(0 + 30°) = 1 Ll 300 

-tan u tan 

_ tanO+i 

- l-itan O 

.J3 tanO+ 1 

= .J3 -tanO 

tan -+x 1r ) =_......:t. tan "i + tan x 

30. _ 

( 4 1-tan "itan x 

l+tanx 

= l-tanx 

.{ 1r ) . 1r n . 

31. Sl -+x =sm-cosx+cos-smx 

4 4 4 

.J2 .J2 . 

=-cosx+-smx 

2 2 

..J2(cos x + sinx) 

= 

2 

32. sin(1800-0) = sin 180°cosO-cosI800sinO 

= O(cosO)-(-I)(sinO) 

=sinO 

33. sin(2700 - 0) = sin 270° cos 0 - cos 270° sin 0 

34. 

35. 

= -1(cosO)-O(sinO) 

=-cosO 

tan(1800+ 0) = tan 180° + tan 0 

1- tan 180° tan 0 

O+tanO 

= 1-0 

= tan 0 

tan(3600_ 0) = tan 360° - tan 0 

1+ tan 360° tan 0 

O-tanO 

= 1+ O(tanO) 

=-tanO 

36. sin(n + 0) = sin ncosO+ cos xsin d 

37. 

= O·cosO+(-l)sinO 

= -sinO 

o tan n - tan 0 0 - tan 0 

tan(n- ) = l+tanntanO = 1+ O(tan0) 

=-tanO 

40. If A, B, and C are angles of a triangle, then 

A + B + C = 180° . Therefore, 

sin(A + B+C)= sin 180° = O. 

3. 5 d d . 

41. cosS=-, smt=-, an san tare m 

5 13 

quadrant I. First, find sin s, tan s, cos t, and tan 

t. Since sand t are in quadrant I, all are 

positive. 

sins=~I-(~r =~ 

. 4 4 

sms 5' 

tans=--=-=coss 

! 3 

cos t = ~1_ ( 5 )2 = g 

13 13 

2.. 5 

tant = 11. = 

11. 12 

13 

. 4 12 5 3 63 

sm(s + t) = _. - +_.- = 

513 13565 

. 4 12 5 3 33 

sm(s - t) = - .- - - .- = 

513 13565 

tan s + tan t . 

tan(s+t)=--- 

1-tan stant 

_ l' 

4+2 

IT 

- 1-(tXi!) 

63 

= 

16 

A_...ltan(s 

- t) = 3(AX I2S ) 

1+ 3 IT 

33 

= 

56 

To find the quadrant of s + t, notice that 

sin(s + t) > 0, which implies s + t is in 

quadrant I or II. tan(s + t) > 0, which implies 

s + t is in quadrant I or III. Therefore, s + t is in 

quadrant I. Then, sin(s - t) > 0 and 

tan (s - t) > O. From the preceding, s - t must 

also be in quadrant I. 

173



42. coss = _.!., sint =~, sand t are in quadrant II. 

5 5 

First, find sin s, tan s, cos t, and tan r. Since s 

is in quadrant II, sin s > O. 

sins=~I-cos2 s =~I-(-~r =~ = 2: 

. 2,/6 

tan s = _sl_n_s = _-r_ = -2../6 

coss 

-! 

Since t is in quadrant II, cos t < O. 

cost=-~I-sin2t=-~I-(~r =-~=-~ 

sint ! 3 

tant=--=-=- 

cost --! 4 

sin(s + t) = sin s cost + coss sin t 

=(2:X-~)+(-~X~) 

-8../6 -3 

= 

25 

sin(s - t) = sinscost - cosssint 

=(2:X-~)_(-~X~) 

-8../6 +3 

= 

25 

tans+ tant 

tan(s+t)=--- 

I-tanstant 

_ -2../6+(-t) 

- 1-(-2../6X-t) 

-~ 

- 4-6../6 

4 

_ -8../6-3 

- 4-6../6 . 

tan s - tant 

tan(s-t)=--- 

1+ tanstant 

_ -2../6-(-t) 

- 1+ (-2../6X-t) 

_ -8../6 +3 

- 4+6../6 

To find the quadrant of s + t, notice from the 

preceding that sin(s + t) = 

-s.J6 -3 

25 

< 0 and 

-8../6-3 0 Th .. . 

tan(s+t) = 

L 17 >. e sme IS negative 

4-0"'/6 

in quadrants III and IV, the tangent is positive 

in quadrants I and III. Therefore, s + t is in 

quadrant III. Then, sin(s - t) < 0 and 

tan(s - t) < O. The tangent is negative in 

quadrants II and IV. From the preceding, s - t 

must be in quadrant IV. 

. 2. I. . d II d 

43. SIOS=-, smt=--,s IS 10 qua rant , an t 

3 3 

is in quadrant IV. Since s is in quadrant II, 

cos s < O. 

coss=-~I-(~r =-~ 

Since t is in quadrant IV, cos t > O. 

cost=~I-( -~r = ~ = 2~ 

i 2 2../5 

tans=--=--=-- 

-4 ../5 ·5 ' 

--! 1 -Ii 

tant=-- =---=- 

~ 2.fi 4 

Sin(s+t)=~(-2~-2)~(- ~X-~) 

4.fi ..f5 

=--+ 

9 9 

4.fi +../5 

= 

9 

Sin(s-t)=~e~)-(- '7X-~) 

=--- 

4:J2 ../5 

9 9 

4.fi-../5 

= 

9 

Different forms of tan (s + t) and tan (s - t) 

will be obtained depending on whether tan s 

and tan t are written with rationalized 

denominators. 

174



-~+(-4) 

tan(s+l) =1-(-~)-4) 

or 

-8,/5 -5../2 

= 20-2.J!O 

-i+(-~) 

tan(s+l) = ( ,f ) 

1- -iJ\-~ 

_-4../2-,/5 

- 2.J!O-2 

4../2 +,/5 

= 2-2..[W 

-¥-(-~) 

tan(s-I) = 1+(_2f)-4) 

or 

-g,f5 +5../2 

= 20 +2..[W 

-i-(-~) 

tan(s - I)= ( )( ) 

1+ -i --rlr 

-4../2+,/5 

= 2&+2 

To find the quadrant of s + I, notice that 

sin(s + I) > 0, which implies s + I is in 

quadrant I or II. tan(s + I) < 0, which implies 

s + I is in quadrant II or IV. Therefore, s + I is 

in quadrant II. Also notice that sin(s - I) > 0, 

which implies s - I is in quadrant I or Il, and 

tan(s - I) < 0, which implies s - I is in 

quadrant II or IV. Therefore, s - I is in quadrant 

II. 

' 3. 12. . dId 

44. sms=-, sml=--,slsmqua rant , an I 

5 13 

is in quadrant III. Since s is in quadrant I, 

cos s > o. 

COSS=~l-(~r =~ 

3 

tans= 

4 

Since I is in quadrant III, cos I < O. 

COSI=_~1_(_12)2 

12 

tanl= 

5 

=_2 

13 13 

sin(s + I) =(~)(- 2-) + (iX-g) =_63 

5 13 5 13 65 

sin(s - I) = (~)(_2-) - (iX-g) = 33 

5 13 5 13 65 

1. 12. .il 63 

4 + 5 20 

tan(s + I) = (")('''1) = --r6 =- 

1- t ~ -"2lj 16 

t_ U 33 

tan(s-I)= (1.)(12.)=- 

1+ 4 5 56 

To find the quadrant of s + I, notice from the 

preceding that sin(s + I)= - 63 < 0 and 

65 

tan(s+1)=--O and tan(s-I)=--



To find the quadrant of s + t, notice that 

sin(s + t) > 0, which implies s + I is in 

5 13 

quadrant I or II, and tan(s + I) < 0, which 

I is in quadrant IV. 

implies s + I is in quadrant II or IV. Therefore, Since s is in quadrant Ill, cos s < o. 

s + I is in quadrant II. Also notice that 

sin(s - I) > 0, which implies s - lis in quadrant 

I or II, and tan(s - I) > 0, which implies s - t is 

coss =-~I-(-~r =-% 

in quadrant I or III. Therefore, s - I is in 

quadrant I. 

15 4 

46. cos s = - 17' sin I = '5' s is in quadrant II, and I 

is in quadrant I. 

sin s = ~1-( - : ~ I~ r= 

8 

tans=- 

15 

COSI=~I-(~r =% 

4 

tanl= 

3 

(.!.X~) + (-~X~) = 

sin(s + I) = · 36 

17 5 17 5 85 

sin(s - I) = (.!.X~)-(-~X~) = 84 

17 5 17 5 85 

(-fs)+t 36 

tan(s + I) = 1-(-i\Xt) = n 

(-fs)-t 84 

tan(s - t) = 1+ (--AXt) =-13 

47. sin s = - ~, cos I = Q, s is in quadrant III, and 

Since t is in quadrant IV, sin I < O. 

sinl=-~I-C~r =-1 53 

-t 4 

tans = _j ='3 

-f.,- 

* 

5 

tant=--=- 

12 

sin(s+l) = (_~XI2)+(_2.X-~) = _ 33 

5 13 13 5 65 

sin(s-I)=(_iXI2)_(_2.X-~)=- 63 

513 13 565 

t--& 48-15 33 

tan(s+l) = 1-(tX-n) = 36+20 = 56 

t+-& 48+15 63 

tan(s-I) = I + (tX-n) = 36-20 =16 

To find the quadrants of s + I and s - I, notice 

that sin(s + I) < 0 and sin(s - I) < 0, which 

implies that s + I and s - I are in quadrant III or 

IV. Also notice that tan(s + I) > 0 and 

tan(s - I) > 0, which implies that both s + I and 

s - I are in quadrant I or III. Therefore, both 

s + I and s - I are in quadrant III. 

To find the quadrant of s + I, notice from the 

preceding that sin(s + I) = - 36 < 0 and 

85 

tan(s + I ) = - 36 > O. The e si sme IS negative .. m 

n 

quadrants ill and IV, the tangent is positive in 

quadrants I and III. Therefore, s + I is in 

quadrant ill. Also, sin(s - I) = :: > 0 and 

84 .. ... 

tan ()=-- s -I


n+(-t) 

16 

tan(s + t)= 1-(n-X-t) =- 63 

fi-(-t) 56 

tan(s - t) = ( 5)( 3) = 

1+ IT -'4 33 

Notice from the preceding that 

. 16 16 

S1O(S+t) =--< 0 and tan(s+t);;:;; --< O. 

65 63 

The sine is negative in quadrants III and IV. 

The tangent is negative in quadrants II and IV. 

Therefore, s + t is in quadrant IV. Also, 

. 56 56 

S1O(S-t)=->O and tan(s-t)=->O. The 

65 33 

sine is positive in quadrants I and II, the 

tangent is positive in quadrants I and III. 

Therefore, s - t is in quadrant I. 

.J7. -J3 d d . 

49• cos s = - -, S10 t =-, an s an tare 10 

4 5 

quadrant II. 

Since s is in quadrant II, sin s > O. 

gnS=~l-(- -:)' =~ 

Since t is in quadrant II, cos t < O. 

cost=-~l-( ~)' =-~ 

t 3 3..fi 

tans=--=--=- 

-d}- .J7 7 

:Ii -J3 ~ 

tant= -42 =- ..[fi=---:n: 

Sin(s+t)=(~X-~)+(~X-'7) 

=-3..ffi -.J2f =-(3..ffi +.J2f) 

20 20 

sin(s-t)=(~X- ~)-( ~X-~) 

-3..ffi +.J2f 

= 

20 


(_1:fi)+ (_::&i) 

tan(s+t) = 7 22 

1-(-~)(-~) 

_ -(66.J7 + 7$6) 

- 154- 3.J462 

(_1:fi)_(-1i-) 

tan(s-t) = 7 2 

1+(-¥X-~) 

_-66.J1 + 7$6 

- 154+ 3.J462 

To find the quadrants of s + t and s - t, notice 

that sin(s + t) < 0 and sin(s - t) < 0, which 

mean s + t and s - t are in quadrant III or IV. 

Also notice that tan(s + t) < 0 and 

tan(s - t) < 0, which imply that s + t and s - t 

are in quadrant II or IV. Therefore, both 

s + t and s - t are in quadrant IV. 

.JIT.fi . 

50. coss = -, cos t ;;:;;-, and sand tare 10 

5 6 

quadrant IV. 

gnS=-~l-( ~)' =-~ 

.J14 ../154 

tans=---=-- 

.JIT 11 

sint=-~[-[ "1)' =-.;;: 

_:iiI .J34 

tant = 4- =- .fi =-ffi. 

sin(s+t) =(- ~X'7)+( ~X- .;;:) 

-..fi8 --J374 

= 

30 

sin(s-t) =(- ~X'7)-(~X-.;;:) 

-..fi8 + -J374 

= 

30 

177



(-*)+(-~) 

tan(s + t) = ""::""'"7"""---:::==-- 

1-(- 4r)(-~) 

-../154 -11~ 

=---'----;===- 

11-../2618 

(-4F)-(-ffi) 

tan(S - t) = ....::.........,.----==="- 

1+(-4F)-ffi) 

-/154+11~ 

=---'-----;===- 

11+../2618 

Notice from the preceding that 

. -fiB-../374 

Sln(s+t) = 

< 0 and 

30 

-../154 -11~ .. 

tan(s+t) = ~ >0. The sme IS 

11-"\12618 

negative in quadrants III and IV, the tangent is 

positive in quadrants I and III. Therefore s + t is 

in quadrant III. Also, 

. -.J28 + ../374 

sm(s-t) = 

30 

>0 and 

tan(s-t) = -/154+11~ >0. 

11+../2618 

The sine is positive in quadrants I and II, and 

the tangent is positive in quadrants I and III. 

Therefore, s - t is in quadrant I. 

52. 

53. 

The graph of y =Si{3; + x) appears to be the 

same as the graph of y = - cos x. Verify 

_(31C ) Sin T+x = -cosx. 

. (31C ) _ 31C 31C . 

sm -+x =sm-cosx+COS-SlnX 

2 2 2 

= (-I)cosx+Osinx 

= -cosx 

51. 

The graph of y = sin( ~ + x) appears to be the 

same as the graph of y =cos x. Verify 

sin( ~ + x) = cos x. 

. (1C ) . 1C . 1C 

Sin -+x =sm-cosx+smxcos 

222 

= 1-cosx +sinx ·O 

=cosx 

The graph of y = tan( ~ + x) appears to be the 

same as the graph of y = -cot x . Verify 

tan(~ +x) = -cotx. 

tan(1C + x) = sin(f+ x) 

2 co~f+x) 

_ sinfcosx+cosfsinx 

- cosfcosx-sinfsinx 

_ (lXcosx)+(OXsinx) 

- (OXcosx)-(IXsinx) 

cos x 

=- 

-sinx 

=-cotx 

178



54. 1 + tan x 

Yt=l-tanx 

I+tanx (tr ) 

The graph of y = --- appears to be the same as the graph of y = tan -4 + x . Verify 

1-tanx 

1+ tan x= tan( tr + x). 

I-tanx 4 


tr 

tr ) tan-+tanx 

tan - + x = _---'4'-__ 

( 4 1_ tan tr tan x 

4 

1+ tan x 

= I-tanx 

55. Verify sin2x = 2sinxcosx. 

sin2x = sin(x+ x) 

= sin xcosx + sinxcosx 

=2sinxcosx 

56. Verify sin (x + y) + sin (x - y) = 2 sin x cos y. 

sin (x + y) + sin (x - y) =(sin x cos y + cos x sin y) + (sin x cos y - cos x sin y) 

= 2 sinx cos y 

57. Verify tan(x _ y) _ tan(y _ x) = 2(tan x - tan y) . 

I+tanxtany 

tan(x - y) - tan(y - x) 

tan x- tany tany -tanx 

=------'' 

1+ tan x tan y 1+ tan y tan x 

tan x - tan y - tan y + tan x 

=------''---....;...-- 

I + tan x tan y 

=2(tanx-tany) 

I+tanxtany 

58. Verify sin(210° + x)- cos(120° + x) =O. 

sin(210° +x)-cos(l20° + x) = (sin2100cosx + cos2100sin x) - (cos120°cosx -sin 120°sin x) 

·1 ..[3. (1 ..[3.) 

=-"2cosx-Tsmx- -"2cosx-Tsmx 

=0 

179



cos(a - 13) 

59. Verify . = tana + cot 13. 

cosasmf3 

cos(a - f3) _ cosa cos 13 + sin a sin 13 

cosasinf3 - cosasinf3 

cosf3 sina 

=--+- 

sinf3 cosa 

= cot 13 + tan a 

sin(s+t) 

60. Verify = tans + tant. 

cosscost 

sin(s+t) =sinscost+cosssint 

coss cost cosscost 

sins sint 

=--+- 

coss cost 

= tans+ tant 

61. 

62. 

Verify sin(x- y) = tanx-tany 

sin(x+y) tanx+tany 

tanx-tany = cosx 

tan x + tany 

~_ siny 

cosy 

.sin..I.. + sin y 

cos x cosy 

= sinxcosy -cosxsiny 

sinxcos y + cosxsin y 

_ sin(x- y) 

- sin(x+ y) 

Verify sin(x+ y) cotx+coty 

cos(x - y) 1+ cotx cot y 


~+ cosy 

cot x + cot y = SID X SiiiY 

l+cotxcoty 1+ cosxcosy 

sin X SID Y 

sinycosx+sinxcosy 

sinxsID y 

- sinxsiny+cosxcosy 

smxsmy 

_ sin(x + y) 

- cos(x - y) 

63. 

sint 

sins cos 2 t - sintcost coss 

= 

sint cost 

sintcostcoss+sin 2 tsins 

+-----:---_.:....:..:=..:... 

sintcost 

sinscos 2 t+sinssin 2 t 

= 

sintcost 

= sins(cos 2 t+sin 2 t) 

sintcost 

sins 

= sintcost 

64. Verify tan(a + 13) - tan 13 

1+ tan(a+ f3)lanf3 lana. 

lan(a + 13) - tan 13 

1+ tan(a + 13) tan13 = tan(a+ 13)- 13] = lana 

65. sin 165°= sin(180° -15°) 

= sin180°cosI5° - cos180°sin 15° 

=sinI5° 

=sin(45° - 30°) 

=sin45°cos30°- cos 45° sin30° 

.J2 ~ .J2 1 =_._-_. 

2 2 2 2 

.J6-.J2 

= 

4 

180



66. tan 165° = tan(180° -15°) 

tan 180° - tan 15° 

= 1+ tan 180°tan 15° 

0-tanI5° 

= 1+(0)tanI5° 

=-tanI5° 

= - tan(45° - 30°) 

tan 45° - tan 30° 

= 1+ tan 45° tan 30° 

I-if 

=-1+(1(4) 

1-:f 

=-1+:f 

3-../3 

=- 3+../3 

../3 -3 

= 3+../3 

. ../3-3 3-../3 

= 3+Jj ' 3-:& 

6-J3 -12 

= 

6 

=../3 -2 

67. tan 255° = tan(1800+ 75°) 

tan 1800+ tan 75° 

= 1-tan 180°tan 75° 

= tan 75° 

= tan(45° + 30°) 

tan 45° + tan 30° 

= I - tan 45° tan 300 

I+if 

= 1-4 

3+..[3 3+..[3 

= 3-..[3 ' 3+..[3 

12+6..[3 

= 

6 

=2+..[3 

68. sin 255° = sin(270° - 15°) 

= sin 270° cos 15° - cos 270° sin 15° 

= -cos 15° 

= - cos(45° - 30°) 

= -(cos 45° cos 30° + sin 45° sin 30°) 

=J..fi ...[3 + ..fi .~) 

1.2 2 22 

={~+ '7) 

-J6 -..fi 

= 

4 

69. sin 285° = sin(360° - 75°) 

= sin 360° cos 75° - cos 360° sin 75° 

=-sin75° 

= - sin(30° + 45°) 

= -(sin 30° cos 45° + cos 30° sin 45°) 

=JL..fi + ..[3 . ..fi) 

1.2 2 2 2 

-..fi-~ 

= 

4 

70. tan 285° = tan(36O° - 75°) 

tan 360° - tan 75° 

= I + tan 360° tan 75° 

=-tan75° 

= -tan(300+45°) 

tan 30° + tan 45° 

= 

1- tan 30° tan 45° 

../3+3 ../3+3 

=- 3-..[3' 3+..[3 

12+6-J3 

= 

6 

=-2-../3 

181



_ l br . ( if) 

71. smU=sm if- 12 

73. tan(- 1:;) = - tan(if+ 1~) 

· if . if if 

= smifcos--cosifsm- tan x-e tan-i- 

12 12 12 

if 

· if 

=sm- 

1-taniftan 

12 12 

if 

· (if if) =-tan 

=sm 4"-'6 12 

. if if if. if 

=sm-cos--cos-sm- =-tan(: - :) 

4 6 4 6 

~../3 ~ 1 = tanf-tant 

=_._-_. 

2 2 2 2 l+tanftanf 

~-~ 

= 

4 

= 1-:1i 

3_ 

1+4 

72. tan Ill; = tan(if - 1~) 3-../3 3-../3 

=- 3+..J3 · 3-



76. 

tan(A _ B) = sin(A - B) 

cos(A-B) 

sin AcosB- cosAsinB 

= cos A cos B+ sin A sin B 

sjnAcosB-cosAsin B 

_ cosAcosB 

- cos Acos B±sioAsin B 

cosAcoslf 

sjn Acos B cosAsjn B 

_ cos Acos B cos Acos B 

- cos AcosB + sinAsinff 

cos Acos B cos Acos11 

. A MK 

_ ~-COs1f 

- .IDL1...sin..l!. 

1+ cosA ' cosB 

tanA-tan B 

= l+tanAtanB 

77. Letf(x) = sin x, Show that 

.::....:..-.....:.........:...."'--"- f(x+h)- f(x) = SIn . {COSh-I) + cos {Sinh) --. 

h . h h 

f(x+h)- f(x) sin(x+h)-sinx 

h 

h 

sinxcosh + sinh cosx - sin x 

h 

= SIn . {COSh-I) +cosx (Sinh) - 

h 

h 

78. Since angle /3 and angle ABC are 

supplementary, the measure of angle ABC is 

180 0 - /3. 

79. a +(180 0 - /3)+ 9 = 180 0 

a-/3+9=0 

9=/3-a 

80. tan 9 = tan(/3 - a) 

_ tan /3- tan a 

- I + tan /3tan a 

82. x + y = 9, 2x + Y = -I 

Change the equations to slope-intercept form to 

find their slopes. 

x+y=9 

y=-x+9 

m2 =-1 

2x+ y =-1 

y=-2x-1 

ml =-2 

tan9= m2 -ml 

l+m2ml 

_ -1-(-2) 

- 1+ (-IX-2) 

I 

=- 

1+2 

I 

= 

3 

9=18.4 0 

83. 5x - 2y + 4 = O. 3x + 5y = 6 

Change the equations to slope-intercept form to 

find their slopes. 

5x-2y+4=0 

-2y=-5x-4 · 

5 

y=-x+2 

2 

5 

m\= 2 

3x+5y =6 

5y=-3x+6 

3 6 

y=--x+ 

5 5 

3 

m2=- 5 

. tan /3- tan a 

81. From Exercise 80. tan 9 =---'--- 

1+ tan /3tan a 

Let tan a = m\ and tan /3 = m2 . Then 

tan 9 = m2 - ml . 

l+mlm2 

183



84. e=20sin(:t - ~) 

=2{sin(:t)cos( ~ ) - cos(:t)si{~)] 

= 2{sin( :t }O) _ cos( :t)(1)] 

(b) 

F =.6Wsin(0+90)0 

sin 12° 

= .6W(sin 0°cos 90° + sin 90° cos 0°) 

sin 12° 

=--Wcos .6 0 

sin 12° 

=2.9WcosO 

=-20CO{:t) 

85. (a) Graph y = 30 sin 120m + 40 cos 120m 

over the interval 0 :5 t :5 .05. 

For x = t, 

V = V 1 + V 2 = 30 sin 1201r1 + 40 cos 1201T1 

(c) F will be maximum when cosO = lor 

o= 0°. (0 = 0° corresponds to the back 

being horizontal which gives a maximum 

force on the back muscles. This agrees 

with intuition since stress on the back 

increases as one bends farther until the 

back is parallel with the ground.) 

Section 5.5 


-60 

(b) Using the TRACE feature on your 

calculator, the amplitude is 50 so let 

a = 50. Estimate the phase shift by 

approximating the first r-intercept where 

the graph of V is increasing. This is located 

at t = .0142. 

sin(120m + q,) = 0 

sin[1201r(.0142) + q,] = 0 

[1201r(.0142) + q,] = 0 

q, = -1201r(.0142) 

= -5.353 

Thus V = 50 sin (120m - 5.353). 

(c) 50 sin (l201r t - 5.353) 

= 50[sin 1201r t cos 5.353 - cos 1201rtsin5.353] 

= 50[sin 1201r t(.5977) - cos 1201rt(-.8017)] 

= 29.89sin 1201r t + 40.09 cos 1201rt 

= 30 sin 1201r t + 40 cos 1201r t 

I. Let B =A in the identity for sin A cos B. 

sin A cos A = .!.[sin(A + A)+ sin(A - A)] 

2 

sin Acos A = .!.[sin2A +sinO] 

2 

2sinAcosA = sin2A 

2. Let B =A in the identity for cos A cos B. 

1 

cos A cos A =-[cos(A + A) + cos(A - A)] 

2 _ 

Exercises 

cos 2 A =.!.[cos2A + cos 0] 

2 

2cos 2 A = cos2A + 1 

2cos 2 A-I =cos2A 

F= .6W sin(0 + 90)0 

86. (a) 

sin 12° 

= .6(170)sin(30+90)0 

sin 12° 

=425 Ib 

(This is a good reason why people 

frequently have back problems.) 

184



3 .. 

. 2tan-t 21C 8. cos 2a = 4"' a IS m quadrant III. 

6. D, 2 = tan- =-..J3 

I-tan -t 3 

cos2a = 2cos 2 a-I 

3 2 

-=2cos a-I 

7. cos 2()= ~ , () is in quadrant I. 4 

cos 2()= 2cos 2 ()-I 2cos 2 a .i 

4 

~ =2cos 2 ()-I 2 7 

cos a= 

5 8 

2cos 2 () = ~ + 1= ! Since a is in quadrant III, cos a < 0 

5 5 

cos 2 () = .!- = i 

10 5 

Since () is in quadrant I, cos () > o. 

2 2../5 

cose= ../5 = -5- 

Since () is in quadrant I, sin () > o. 

cosa=-~ =- 2~ =-~ 

Since a is in quadrant III, sin a < 0 . 

sina=-~I-cos2a 

=-~I-i 

sin() = ~1-cos2 e 

=-~ 

1 

=~ =-2Ji 

=H 

1 

../2 

=- 

4 

=.J5 

tana=-=! 

.J5 

=- 

5 

../2 

=-:Ji4 

tan() _ sin() _ :fl _1 

-4 

- cos() - 2# -'2 =7] 

1 1 ..f7 

cot()=--= - =2 = 

tan() t 7 

1 1 

1 1 5 ../5 cot a =--= r;:; = .fi 

sec()=--=-=--=- tana ",7 

cos() ¥ 2../5 2 

1 2../2 2$4 

1 1 sec a = cosa = ---::J7'"= --7 

csc()=-= rc =../5 

sin () :t1. 

5 1 2../2 

csc a =--=---= -2../2 

sina 1 

1 

185



5 1C 2 

9. cos2x=-- -


 

1 3 

cos2x =--=- 

sec2x 5 

tan 2a = sin 2a = - ~ = _ 4../55 

1 3 

cos2a 19 39 

cot2x=--=- 

tan2x 4 

1 39../55 

cot2a=--=-- 

sin2x = tan2xcos2x = (-~X-~) = ~ 

tan2a 220 

1 49 

1 5 

sec2a=--=cos2a 

39 

csc2x=--=sin2x 

4 1 49../55 

csc2a=--=-- 

sin2a 220 


12. cos /3= _E.., sin /3 > 0 

14. tan x = -, 

5. 0 

SIOX < 

13 

3 

sin /3 = ~1- cos 2 2 

/3 

2 2 5 34 

sec x = 1+ tan x = 1+ ( "3 ) = "9 

Since sin x < 0, and tan x > 0, x is in quadrant 

=~I-(-~~r 

ITI, so cos x and sec x are negative. 

5 

= 

~ 

secx=-- 

13 

3 

sin 2/3= 2 sin /3 cos /3 

3 ~ 

12) 

cosx =---=-- 

=2(2.X_ 

~ 34 

13 13 

120 

sinx=cosxtanx=(- ~X~)=-~ 

=- 

34 3 34 

169 

cos2/3 = 2cos 2 /3-1 

sI02x=2smxcosx=2 · . (5~X3~) -- - =- 15 

34 34 17 

=2(- ~~r -1 

cos 2x = 2cos 

2 

x -1 = 2 (9) 34 -1 = - 17 8 

119 

= 

169 

15 

tan2x =- 

tan 2/3= sin2/3 = _ 120 

8 

cos2{3 119 

cot2x=- 

119 

15 8 

cot2{3=- 

17 

120 

sec2x=- 

169 

8 

sec2/3= 

17 

119 

csc2x= 

169 

csc2/3=- 

120 

15. slOa=-7' · J5 cosa> 0 . 

13. tan x = 2, cos x > 0 

2tanx 

tan2x = 2 

2 . 2 

1- tan x 

cos a=l-sm a=l-(J5)2 -7 = 44 49 

2(2) 4 

=1-(2)2 =-3" 

Since cos a > 0, 

Since both tan x and cos x are positive, x must 

cos a = (44 = .J44 = 2..JIT . 

be in quadrant I. Thus, 2x must be in 

""49 7 7 

quadrant ll. 

2 2 16 25 

sec 2x = 1+ tan 2x = 1+ - = 

9 9 

5 

sec2x=- 

3 

cos2a = 1-2sin 2 a = 1-2(J5)2 = 39 

7 49 

sI02a=2smacosa=2 · . (J5X2..JIT)=--- 

- 4../55 

7 7 49 

187



Jj. 0 

16. cosa=-, smrz > 

5 

s;na=~1-cos2a =~1-( "1)' = ~ 

sin2a = 2sinacosa = 2( ~)( ~) = 2: 

cos2a = 2cos 2 a-I = 2(Jj)2 -1 =_.!2. 

5 25 

25• 2 cos 267Io_I=cos267{_s\'n267'!'o 

 

2 2 2 

26. 

=COS2(67{) 

= cos 135° 

..fi 

=- 2 

tan 2a = sin 2a = _ 2../66 

cos2a 19 

19 19../66 

cot2a=---=-- 

2../66 132 

sec2a = _ 25 

19 

25 25../66 

csc2a=--=- 

2../66 132 

17. sin2X=2sinXcosX 

sin Xcos X = .!.sin 2X 

2 

If sin 2X = .4. then sin XcosX = .!.(.4) =.2. 

2 

Therefore, the screen will display .2 for the 

final expression. 

18. cos(2X) = 1-2 sin 2 X 

2sin 2 X=I-cos2X 

If cos (2X) = .3, then 2sin 2 X = 1-.3 = .7. 

Therefore. the screen will display .7 for the 

final expression. 

tan51° 

27. 

I-tan 251° 

. 2tanA 

Since 2 = tan 2A. then 

I-tan A 

28. 

.!.( 2 tan A ) =.!. tan 2A 

2 I-tan 2 A 2 

tan A 1 

--~2- = - tan 2A. 

I-tan A 2 

tan51° =~tan2(51°) 

I-tan 251° 2 

1 

= - tan(102°) 

2 

22. 

2 tan 15° tan 15°+ tan15° 

1-tan 2 15° - 1-tan 2 15° 

= tan(15°+15°) 

= tan 30° 

1 

=Jj 

Jj 

= 3 

29. .!.-.!.sin 2 47.1° = .!.(I-2sin 2 47.1°) 

424 

Since cos2A = 1-2sin 2 A, 

~(I-2sin 2 47.1°) = ~cos 2(47.1°) 

1 

= -cos94.2°. 

4 

. 2 1° (1°) ..fi 

24. 1-2sm 22- = cos2 22- = cos 45° = 

222 

188


29.5° cos 29.5° 

2 21r 

5-cos 5= 

cos 2 A - sin 

2; _ sin2 

2a-sin 2 2a 

cos 5x 

sin A cos 

cos 5x = 

-1 = cos 

= sin 90° 

= 2 sin 

=2('7. 

=1 

= cos 90° 

= 2 cos 

={'7r 

=0 

cos2(600)=cos1200=- 

= cos 2 

=(~r 

1 

=- 

2 

5.5 Double-Angle Identities 

= tan 120° =-.J3 

= 2 tan 60° 

1- tan 2 60° 

2.J3 

= 1_(.J3)2 

2.J3 

=- 

-2 

=-./3 

= cos 1~1r 

41r 

=cos 

3 

1r 

=-cos 

3 

1 

::;;;;:- 

2 

= cos 2 ( 5;)~sin2(5;) 

= cos 2( ~)-(-sin ~r 

=(~r -(- ~r 

1 3 

=-- 

4 4 

2 

=- 

4 

1 

=- 

2 

sin 21r = .J3 

3 2 

= 2' sm-cos 1r 1r 

3 3 

={~X~) 

.J3 

= 

2 

189 

30. .!. sin = J.... (2 sin 29.5° cos 29.5°) 38. tan 2(60°) 

8 16 

tan 2(600) 

= J....sin2(29 .5°) 

16 

= J....sin59° 

16 

. 2 21r -( 2 21r . 2 21r) 

31. sm cos 5-sm 5 

Since 2 A = cos 2A • 

-(cos 2 2;) = -cos(2 . 2;) 

39. cos 2(5;) 

41r 

=-COS-. 

5 

32. cos 2 = cos2(2a)= cos4a 

33. 2 sin 5x 

Since 2 A = sin 2A. 

2 sin 5x sin 2(5x) = sin lOx. 

34. 2 cos 2 6a 2(6a)= cos 12a 

COS2( 5;) 

35. sin 2(45°) = 1 

sin 2(45°) 45° cos 45° 

'7) 

36. cos 2(45°) = 0 

cos 2(45°) 2 45° -1 

-1 

40. sin 2(1r) = 

3 

sm . 2(1r) - 

1 

3 

37. 

2 

cos 2(60°) 60° -sin 2 60° 

-( ~r 

189



41. tan2(-~)=tan(_2;)=~ 45. sin2(- 1~1r) == sin(-ll1r) = 0 

tan2(- ~ ) = - tan2( ~ ) sin2(_ I:) = 2 Sin(_ 1~1r ) co{_ 1~1r ) 

2tan1 

= l-tan 

2 

t 

=0 

= 

2~ 

1_(~)2 

= 2(1)(0) 

46. Sin2( _1~1r)=Sin(-171r)=O 

2~ Sin2(_ 1~1r) = 2Sin(_ 1~1r )co{_ 1~1r) 

=-- 

1-3 = 2(-1)(0) 

=~ =0 

42. cos2(- 9:) = cos( _ 9;) = 0 

47. 

cosz(- 9:) = 2COS 2( _ 9:)_1 

=2('7r -I 

=0 

43. tan2(- ~) = tan(_ 8;) = ~ The graph of y = cos 4 x - sin 4 x appears tobe 

tan2(- ~) = - tan 2( 4;) 

the same as the graph of y = cos 2x. 

cos 4 x-sin 4 x = {cos 2 x+sin 2 xXcos2 x-sin 2 x) 

2 tan-¥ 

= l·cos2x 

= =cos2x 

l-tan 2 -¥ 

2~ 48. 4tan x cos2 x - 2tan x 

= 1 - tao 2 x 

I-(~f 

2~ 

=-- 

1-3 

=~ 

4tanxcos2 x-2tanx 

The grap h 0 f y = 2 appears 

1- tan x 

to be the same as the graph of y = sin 2x. 

2 

4 tan xcos x-2tanx 2tanx (2 2 I) 

-------:;---- = 2 cos x-. 

1-tan 2 x 1-tan x 

= tan 2x cos 2x 

sin2x 

=---cos2x 

cos2x 

=sin2x 

190



49. 

cot 2 x-I 

The graph of y = appears to be the 

2cotx 

same as the graph of y = cot 2x. 

cot 2 x-I ~-1 

= ---.L 

2cotx tanx 

2 

(± - 1)tan x 

_ 

lan x 

- 

(!iih ) tan 2 x 

1- tan 

2 

x 

= 

2tanx 

1 

-l::/X 

1 

=- 

tan2x 

=cot2x 

50. 2tanx 

2 - sec2x 

2tanx 

The graph of y = 2 appears to be the 

2-sec x 

same as the graph of y = tan 2x. 

2tan x 2tanx 

2 - sec 2 x := 1-(sec 2 x -1) 

2tanx 

-1-tan 2x 

= tan2x 

51. Verify (siny+cosyf =sin2y+1. 

(siny+cosy)2 = sin 2 y + cos 2 Y+ 2sinycosy 

=1+sin2y 

2 4 

52. Verify en secx= 2 sec x+sec x 

4 

" 

2+sec 2 x-sec x 


2 4 (-.l..-+ 1)cos 4 x 

sec x+sec x cos 2 x ~ 

2 4 = ( 

2 + sec x - sec x 2 + ---l..- - -.l..-) cos" X 

cos' x 

cos'' x 

cos 2 x+l 

=--.,....----::-- 

2cos 4 x+cos 2 x-I 

cos 2 x+l 

2 

- (2cos 2 x-IXcos x + I) 

1 

=----:::-- 

2cos 2 x-I 

I 

=- 

cos2x 

= sec2x 

53. Verify tan 8k - tan 8k tan 2 4k = 2 tan 4k . 

tan 8k - tan 8k tan 2 4k = tan 8k(l- tan 2 4k) 

54. Ven 

ify "2 2tany 

SID y = 2" 

l+tan y 


2tany 2tany 

1+tan 2y = l+sin~Y 

cosy 

2 sinr 

_ cosr 

- cos 2 r+sin2r 

cos2r 

=2sinycosy 

=sin2y 

2-sec 2 y 

55. Verify cos2y = 2 

sec y 


(2--1-)COS 2 

2 - sec 2 y cos! y Y 

2 

sec Y 

= (_I'_)COS 2 Y 

cos- y 

:=2cos 2 y-l 

=cos2y 

= 2 tan 4k (1_ tan24k) 

l-tan 24k 

=2tan4k 

191



. 2tanO 

56. Venfy -tan20= 2 . 

sec 0-2 


2tanO 2tanO 

sec 2 0-2 - (1+ tan 2 0)-2 

2tanO 

- tan 2 0-1 

=-tan20 

57. Verify sin4a = 4sinacosacos2a. 


4sinacosacos2a = 2(2 sin acosa)cos2a 

= 2sin2acos2a 

= sin2(2a) 

= sin4a 

ify l+cos2x 

58. Ven =cot r , 

sin2x 

2x-I) 

l+cos2x = 1+(2cos 

sin2x 

2sinxcosx 

2cos 2 x 

= 2sinxcosx 

cosx 

=- 

sinx 

=cotx 

59. Verify tan(O - 45°) + tan(O+ 45°) = 2 tan 20 . 

tan(0 - 45°) + tan(0 + 45°) 

tan 0 - tan 45° tan0 + tan45° 

------+----- 

1+ tan0 tan 45° 1- tan0 tan45° 

tan 0 - I tan 0 + I 

= +-- 

I + tan8 I - tan 0 

tan 0 -I tan 0 + I 

= tanO+1 tanO-1 

tan 2 0-2tanO+1-tan 2 0-2 tanO-1 

= 

(tanO+IXtanO-I) 

-4 tan0 

=-..."...- 

tan 

20-1 

4tanO 

= l-tan 2 0 

2(2 tan0) 

=1-tan 20 

= 2tan20 

. l-tan 220 

60. Venfy cot40=--- 

2tan20 

I 

cot40=- 

tan40 

I 

2 ran 26 

l-tan 226 

2cos2a 

61. Verify =cota-tana. 

sin2a 


cosa sina 

cot a - tana =---- 

sina 

cosa 

cos 2 a - sin 2 a 

= 

sinacosa 

= 2(cos 2 a-sin 2 a) 

2sinacosa 

2 cos2a 

= sin2a 

62. Verify sin4r = 4sinrcosr - 8sin 3 rcosr . 

sin4r = 2sin2rcos2r 

= 2(2 sin rcos r)(I- 2sin 2 r) 

= 4sin rcosr - 8sin 3 rcosr 

63. Verify sin 2acos2a = sin 2a - 4 sin 3 a cos a . 

sin2acos2a = (2sinacosa~I-2sin2 a) 

. l-tan 2 x 

64. Venfy cos2x = 2' 

I+tan x 


. 2 

I 2 I-~ 

- tan x cos x 

1+ tan 2 x = 1+"* 

cos x 

cos 2 x-sjn 2 x 

_ cos2x 

- cos 2 x+sin 2 x 

2 

cos x 

=cos 2 x-sin 2 x 

=cos2x 

= 2sinacos~ -4sin 3 acosa 

= sin2a -4sin 3 acosa 

192



65. Verify tan s + cot s = 2 esc 2s . 68. Verify 

sins coss 

coss sins 

sin 2 s + cos 2 S 

tanO+tann 

= 

tan(0+ n) =---- 

cosssins 

l-tanOtann 

tanO+O 

I 

= 

= 

l-tanO ·O 

cosssins 

=tanO 

2 

= sin(n-O) = sin ncosO - cosnsinO 

2cosssins = O· cosO+ I ·sinO 

2 

=- 

=sinO 

sin2s 

= 2csc2s 222 

= 0 .cos 0 + I .sin 0 

cota-tana 2 

66. Verify = cos a. 

cota+tana 

=sinO 

tan s + cot s =--+- cotOtan(O+n)-sin(n - O)co{~ -0) = cos () 

cos(n - 0) = cos n cos0+ sin!!..sin0 

cota - tan a ~- sina cotOtan(o+n)-sin(n-O)co{~-0) 

sma cosa 

cota+tana ~~~g + ~~~ = cot 0 tan 0 - sin OsinO 

cos 2 a-sin 2 a = l-sin 2 0 

= cos 2 a+sin 2 a = cos 2 0 

= cos 2 a -sin 2 a 

=cos2a 

In Exercises 69-72, other forms may be possible. 

67. Verify I + tan x tan 2x = sec2x. 

1+tanxtan2x = 1+ tanx( 2 tan: ) 

I-tan x 

69. cos3x 

=cos(2x+x) 

= cos2xcosx -sin 2xsinx 

2 tan 2 x = (1- 2sin 2 x)cosx - (2sin xcosx)sinx 

I +--....,..... 

- l-tan 2 x =cosx-2sin 2 xcosx-2sin 2 xcosx 

l-tan 2 x+2tan 2 x 

= = cosx-4sin 2 xcosx 

l-tan 2 x 

l+tan 2 x 

=cosx(l-4sin 2 x) 

-1-tan 2x 

=cosx[l-4(I-cos 2 x)] 

005 2 X:,in2 x 

_ cos x =cosx[-3+4cos 2 x] 

- cos 2 x-sin 2 x 

cosFx 

I 

=- 

cos2x 

=sec2x 

=-3cosx+4cos 3 X 

70. sin 4x = 2sin 2x cos 2x 

=2(2sinxcosx)(cos 2 x-sin 2 x) 

=4sinxcos 3 x-4sin 3 xcosx 

71. tan 3x = tan(2x + x) 

tan 2x+ tan x 

= I - tan 2x tan x 

~+tanx 

_ I-tan x 

-I-~tanx 

I-tan x 

2 tan x + tan x - tan 3 x . 

= I - tan 2 x - 2 tan 2 x 

3tanx-tan 3 x 

=----..,.,- 

1-3tan 2 x 

2 

193



72. cos4x = 2 cos 2 2x-l 

=2(2coS 2 X-It -1 

=2(4COS 4 x-4cos 2 x+l)-1 

=8cos 4 x-8cos 2 x+l 

73. ¢ = 47° 12', h = 387.0 ft 

Use a calculator and substitute these values in 

g =978.0524(1 + .005297sin 2 ¢ 

- .0000059 sin 2 2¢) - .()()()()94h. 

g =978.0524[1 +.005297sin 2 47°12' 

- .OOOOO59sin 2(2 .47°12')] 

- .000094(387.0) 

=980.799 em per sec 

2 

74. ¢ = 68° 47' = 68.78333°, h = 1145 ft 

Use a calculator and substitute these values in 

g =978.0542(1 + .005297sin 2 ¢ 

-.OOOOO59sin 2 2¢) - .000094h. 

g = 978.052~1 + .005297 sin 2 68.78333° 

- .OOOOO59sin 2(2 .68.78333 0 ) ] 

- .000094(1145) 

=982 .444 em per sec 2 

75. cos 45° sin 25° 

=.!.[sin(45° + 25°) - sin(45° - 25°)] 

2 

= .!.(sin70° -sin20°) 

2 

76. 2sin74°cosH4° 

= 2·.!.[sin(74° + 114°)+ sin(74° -114°)] 

2 

=sin 188°+ sin(-400) 

= sin 188°+ (-sin 40°) 

= sin 188° - sin 40° 

77. 3 cos 5x cos 3x 

=3· .!.[cos(5x +3x)+ cos(5x - 3x)] 

2 

3 

=-(cos8x+ cos2x) 

2 

78. 2 sin 2x sin 4x 

=2 ..!.[cos(2x - 4x) - cos(2x +4x)] 

2 

=cos(-2x)-cos6x 

=cos 2x - cos 6x 

79. sin(-8)sin(-38) 

=~[CO~-8 - (-38)] - co~-8 + (-38)]) 

=.!.[cos(28) - cos( --48)] 

2 

1 

=-(cos28 - cos 48) 

2 

80. 4 cos 8a sin(--4a ) 

= 4 ·±[sin[8a + (--4a)]-sin[8a - (--4a)]J 

=2(sin 4a - sin 12a) 

= 2sin4a - 2sin12a 

81. -8 cos 4y cos 5y 

=-8· '2 1 [cos(4y + 5y)+ cos(4y-5y)] 

= --4[cos9y+ cos(-y)] 

=--4(cos9y+ cosy) 

82. 2 sin 3k sin 14k 

1 

=2 ·-[cos(3k -14k) -cos(3k + 14k)] 

2 

=cos(-llk)- cos(17k) 

83. (a) 

= cosHk -cos 17k 

= 

32 

2v 2 sin8cos8 

= 

32 

=v 2 sin(28) 

32 

Which is dependent on both the velocity 

and angle at which the object is thrown. 

(b) D = (36)2 sin(2· 30) 

32 

= (1296)(0.866) 

32 

"" 35 ft 

194



84. From Example 6, 

(163sin 1201l' 1)2 

W =-'------'- 

15 

W "" 1771.3(sin 1201l'1)2. 

Thus, 

1771.3(sin 1201l'1)2 

=1771.3sin 1201l' 1.sin 1201l'1 

=(1771.3)(~)[cos(1201l'1-1201l'1) 

-cos(1201l'1 + 1201l'1)] 

=885.6(cos 0 - cos 2401l' 1) 

=885.6(1- cos 2401l' 1) 

=-885.6 cos 2401l' 1+ 885.6. 

If we compare this to W =a cos(lOt) +c, 

then a = -885.6, m =2401l', and c = 885.6. See 

the answer graph in the back of the textbook. 

85. (a) Graph W =VI =[163sin(1201l'1)] 

[1.23sin(1201l'1)] over the interval 

0::; 1 ~ .05. 

For x = t, W = VI = 

(163 sin 120711)(1.23 sin 120171) 

3OOr1/~~70 77:' 

-so 

(b) The minimum wattage is 0 and the 

maximum wattage occurs whenever 

sin(1201l'1)=1. This would be 

163(1.23) = 200.49 watts. 

(c) [I63sin(1201l'1)I 1.23sin(1201l'1)] 

=200.49sin 2(1201l' 1) 

=2oo.49[~(I- cos 2401l' 1)] 

=-100.245 cos 2401l' 1+ 100.245 

Then a =-100.245, m =2401l', and 

c =100.245. 

(d) The graphs of 

W =[I63sin(1201l'1)11.23sin(1201l'1)] and 

W =-100.245 cos 2401l' 1+ 100.245 are the 

same. 

(e) The graph of W is vertically centered 

about the line y = 100.245. An estimate for 

the average wattage consumed is 

100.245 watts. (For sinusoidal current, the 

average wattage consumed by an 

electrical device will be equal to half of 

the peak wattage.) 

86. (a) The period is equal to 

21l' 21l' I 

-=--=b 

21l'm m 

Section 5.6 

Exercises 

1. Since 195° is in quadrant III and since the sine 

is negative in quadrant Ill, use the negative 

square root. 

2. Since 58° is in quadrant I and since the cosine 

is positive in quadrant I, use the positive square 

root. 

3. Since 225° is in quadrant III and since the 

tangent is positive in quadrant III, use the 

positive square root. 

4. Since -10° is in quadrant IV and since the sine 

is negative in quadrant IV, use the negative 

square root. 

s, c; 

. 150 ~1-COS300 

SIO = 

2 

=~1-24 

=~2-~ 

2 ·2 

=-=---=,;.- 

~2-~ 

.J4 

=-=-_.:... ~2-~ 

2 

0 _1_-_c_os_3_0_0 

6. A; tan 15 = sin 30° 

I_di 

=-t 

2-~ 2 

=-_. 

2 I 

=2-~ 

195



7. D; 

Jt+COS" 

8 2 

=~1+24 

x 

cos-= 4 

=~2+.J2".!. 

2 2 

~ 

- -J4. 

=~ 

2 

10. 

B; cos 67.5° = 1+ cos 135° 

2 

=Jl+(~f) 

=~2-.J2 ..!. 

2 2 

-~ 

- -J4 

=~ 

2 

8. E; 

9. F; 

tan(- 1C) _ 1-cos(-~) 

8 _ ( 1C) 

SIn -'4 

_ l-cos(lf) 

- sin(lf) 

_1-4 

r:»: 

2 

_2-.J2 2 

--2---.J2 

_ -(2 - .J2) .J2 

- .J2 "12 

-2.J2 +2 

= 

2 

=1-.J2 

tan 67.5° = I-cos 135° 

sin 135° 

_ 1-(-4) 

-::li 

2 

_2+.J2 2 

--2-'.J2 

2+.J2 .J2 

= .J2 '.J2 

2.J2 +2 

= 

2 

=1+.J2 

11. sin 67.5° =sin(13:0) 

Since 67.5° is in quadrant I, sin 67.5° > O. 

sin 67.5° = 1- cos 135° 

2 

l+cos45° 

2 

=~1+2"f 

=~2+4.J2 

=~ 

2 . 

12. sin 195° =sine~OO) 

Since 195° is in quadrant III. sin 195° < 0, 

sin 195° =_~I-CO;3900 

=_~I-C~S300 

=_~1-2~ 

=_~2~~ 

-s:» 

2 

196



13. 

14. 

15. 

3900) 

cos195° = cos -2 ( 

Since 195° is in quadrant III, cos 195° < O. 

1+cos3900 

cos 195° = -./---- 

2 

1+ cos 30° 

= 

2 

=_~1+2,Jf 

=_~2+4.[3 

-~2+.[3 =----:._- 

2 

3900) 

tan 195° = tan -2 ( 

sin 390° 

= 1+ cos 390° 

sin 30° 

= 1+ cos 30° 

- ! 

-~ 

T 

1 

=2+.[3 

1 2-.[3 

= 2+.[3 ' 2-.[3 

2-.[3 

=-- 

1 

=2-.[3 

3300) 

cos 165° :::; co -2 { 

Since 165° is in quadrant II, cos 165° < O. 

1+cos330° 

cos165° =-./---- 

2 

1+ cos 30° 

= 

16. sin 165° =sine~~O) 

Since 165° is in quadrant II, sin 165° > O. 

sin 1650=~1-CO;3300 

=l-C~S300 

=~1-2,Jf =~2-4..J3 

~2-.[3 

=-=---~ 

2 

17. tan - 

x ) 

=-- 

1-cosx 

( 2 sinx 

If cos x = .9682458366 and sin x = .25. then 

tan(~) =1-.9682458366 

2 .25 

=.1270166536 

Therefore, the screen will display .1270166536 

for the final expression if no round off errors are 

introduced. 

Most trigonometric function values are 

transcendental numbers. that is. they are 

nonrepeating, nonterrninating decimals. When 

we use the value given on the calculator screen 

for these numbers, we are not using their exact 

value; instead. we are.using a rounded form of 

the number. Since the calculator actually stores 

more significant digits than is displayed. round 

off errors can be introduced when we use only 

the numbers shown on the display. 

18. tan(~) =_l_-_c_os_x_ 

2 sin x 

If cos x = -.75 and sin x = .6614378278, then 

x) 1-(-.75) 

( 

tan "2 = .6614378278 

=2.645751311. 

Therefore. the screen will display 2.645751311 

for the final expression if no round off errors are 

introduced. 

Most trigonometric function values are 

transcendental numbers. that is. they are 

nonrepeating, nonterminating decimals. When 

we use the value given on the calculator screen 

for these numbers. we are not using their exact 

value; instead. we are using a rounded form of 

the number. Since the calculator actually stores 

more significant digits than is displayed. round 

off errors can be introduced when we use only 

the numbers shown on the display. 

197



20. Show that .fi -1 =~3 - 2.fi ' Notice that 

.J2 -1>0 and ~3-2.fi >0. 

(.fi-If =2- z-Ii +1=3- 2.J2 

(~3-2-J2Y =3-2-J2 

If a> 0 and b > 0 and a 2 =b 2 • the a =b. Since both 

are positive and their squares are equal, by the above 

theorem, -J2 -1 =~3 - 2.J2 ' 

21. Find cos ~, given cos 0 = .!., with 0 < 0 < 1r , 

242 

0



25. Find sin a , given tan a = 2, with 0 < a O. 

sec 2 a = tan 2 a + 1 

Thus, cos a is positive. 

2 

cos-= a ~I +cosa 

2 2 

=(2)2+ 1 

=5 ~~I-: 

seca={5 

1 

cos a = see a 

1.J5 

= {5 =5 

1f a 1f 

O



28. Find cot fJ if tan fJ = - ..[5, with 

2 2 

90° < fJ < 180°. 

Use identities to find sin fJ and cos fJ . 

sec' p= I+tao' p= 1+(-~r =~ 

Since 90° < fJ < 180°. sec fJ < O. 

. 3 

secfJ=- 2 

2 

cosfJ=- 3 

sin fJ = cos fJ tan fJ = ( - ~X-V;J= ~ 

fJ 

cot =~ 

2 tanT 

= 1+cosfJ 

sinfJ 

1+(-t) 

= ~ 

1 

=..[5 

..[5 

= 5 

29. Find sin8, given cos28=~, 8 is in quadrant I. 

5 

Since 8 is in quadrant I. sin 8 > O. 

sin 8 = sin 1.(28) 

2 

=l-~S28 

=~1~~ 

=..[5 

5 

30. Find cos 8. given cos 28 = .!., 8 is in 

2 

quadrant II. Since 8 is in quadrant II, cos8 is 

negative. 

31• FIn 

cos8 = _~1 + C;S28 

=_~I:t 

=-~ 

.J3 

=- 2 

· d cos x, given . cos 2x =--, 5 -



38. sin 158.2° = tan 158.2° = tan 79.10 

1+cosI58.2° 2 

46. 

1+ cos18x 18x 9 

39. ± =cos-=cos x 

2 2 

1+ cos 20a 2Da 10 

40. ± = cos-- = cos a 

2 2 

l-cosSO 80 

41. ± = tan- = tan40 1-cosx 

l+cos80 2 The graph of y = . appears to be the 

smx 

l-cos5A 5A 

x 

same as the graph of y =tan-. 

42. ± = tan 

1+cos5A 2 

2 

1- cos x _ 1- cos 2(t ) 

l+cos(t) t x sin x - sin 2(t) 

43. ± = cos- = cos 

2 2 8 = 1-[1-2sin2t] 

2sinfcost 

~ 1- cos 1Jl. .3fl. 30 

44. ± 2 5 = sin t = sin 10 2sin 2 f 

=------'= 

2sintcosf 

45. 

sin x 

=--"£. 

cost 

=tant 

47. tan£-cot.:!. 

2 · 2 

cot':!' - tan .:!. 

2 2 

sinx 

The graph of y = appears to be the 

l+cosx 

x 

same as the graph of y = tan-. 

2 

sinx _ sin2(f) 

l+cosx - l+cos2(t) 

2sin(t)cos(t) 

= ~[2 cos 2 (t )-1] 

= 

2sin(f)cos(f) 

2cos 2(f) 

sint 

=- 

cosf 

x 

= tan 

2 

tan..!".+cot.A 

The graph of y = 2 2 appears to be the 

cot'f - tan t 

same as the graph of y =sec x. 

201



48. 

sin.L cos.L 

.:=.l...+~ Verif 2 x (I +cosx)2 

tan1" + cot t _ cost sint . sinfcost 50. en y cot -= 2 

2 sin x 

cotf-tanf - sint _ cost sintcost 

cost sint 

cot 2 ~ =[_I)2 

_ sin 2 A 2 + cos 2 ..\: 2 2 tant 

-( 2 x .2..I.) I 

- cos 2-sm 2 

I 

=C:i:~J2 

- -cos2(t) 

I 

_ (I + cosx)2 

=-- 

cosx - sin 2 x 

=-secx 

of . 2 x tanx-sinx 

51. Ven y sm -= . 

2 2tanx 


° 

sm 2 -= x (±PP-cosx)2 

2 2 

I-cosx 

= 

2 


. .sin.L' 

tanx-smx = cosx -smx 

2 tan x 2 ~~~ 

The graph of y = 1-8sin 2 ~cos2 ~ appears to 

2 2 

sinx -cosxsinx 

= 

be the same as the graph of 

2sinx 

y =cos h. I-cosx 

= 

2 

1-8sin 2 ~cos2 ~ = 1-2(4sin 2 ~cos2~) 

~r 

2 2 2 2 

. sin 2x 2 x . 2 x 

52. Venfy --=cos --sm -. 

= 1-2(2 sin ~ cos 2sinx 2 2 


sin2x 2sinxcosx 

= 1-2[sin2(~)r --= =cosx 

2sinx 2sinx 


= 1-2sin 2 x 

2 x . 2 x I + cos x 1-cos x 

=cos2x cos --sm -=-- 

2 2 2 . 2 

2 

2cosx 

49. Verify sec 2 ~ =-- =-- 

2 I+cosx 

2 

= cos x 

2 x I 

sec -=-- 

2 cos 2 t 2 

53. Verify -- 

I 

I+cosx 

2 

= 

(±~I±CfsX ) 

1+:'. -tan2~~ [+:0" 

(± ::::J 

I 

= I+cfsx 2 I-cosx 

= 

2 

I +cosx I +cosx 

=--- 2-I+cosx l+cosx 

I+cosx = =-- 

I+cosx I+cosx 

=1 

202


54. Verify tanr = esc r 

2 

r sinr 

tan 2" = 1+ cos r 

- cot r ' 57. 

(I-COSr) 

1- cos r 

_ sinr-sinrcosr 

- I-cos 2 r 

_ sinr-sinrcosr 

- sin 2 r 

= cosr 

sinr sin v 

=cscr-cotr 

55. V 

ify 1 2 9 

en -tan -=-- 

2cos9 

2 I+cos9 

I-tan2~=1-( sin9 )2 

2 1+cos9 

_ (1+ cos 9)2 - sin 9 

- (1+ cos 9)2 

1+ 2cos9 + cos 2 9 - sin 2 9 

= 

(1+ cos 9)2 

_ 1+ 2cos9+ cos 2 9 -(I-cos 2 9) 

58. 

A sin A 

tan-=-- 

2 I+cosA 

sin A I-cosA 

= 1+ cos A 1- cos A 

_ sinA(I-cosA) 

- I-cos 2 A 

_ sinA(l- cosA) 

- sin 2 A 

I-cosA 

= 

sin A 

· a 1 

sm-= 

2 m' 


3 

m= 2 

, a 1 2 

sm-=-= 

2 t 3 

· 2 a I-cosa 

sm -= 

2 2 

so 

56. 

- (I+cos9f 1 

1+ 2cos 2 9-1 +2cos9 

= 

(1+ cos 9)2 

_ 

- 

2 cos 9(1+ cos 9) 

(1+ cos 9)2 

59. 

5 

m= 4 

cosa= 9 

a =84°. 

2cos9 

= 

sin a =..!..= ±= .8 

1+ cos 9 2 m 5 

!!. = 53° 

I- tan2 t 

Verify cosx = 2 ' 

1+ tan t 

I-tan2t _I-~ 

2 - .1.=.lm.I. 

1+ tan t 1+ licOsx 

_ (I+cosx)-(I-cosx) 

- (I+cosx)+(I-cosx) 

2cosx 

=--=cosx 

2 

60. 

2 

a « 106° 

· a 1 2 

sm"2=;;;' m = 

, a 1 

sm-= 

2 2 

· 2 a I-cosa 

SID -= 

2 2 

(±r = I-~osa 

2(~ ) =I-cosa 

1 

--I=-cosa 

2 

1 

--=-cosa 

2 

1 

-=cosa 

2 

a=60° 

203



5 

61. m:= 

2 

sin a = ..!.. :=~ :=.4 

2 m 5 

a"" 23.60 

2 

a ""47° 

62. a =30° 

. 30° 1 

sm-= 

2 m 

sin 15°:=..!.. 

m 

1 

sin 15° 

""3.9 

63. a=60° 

m:=- 

. a 1 

sm-:= 

2 m 

..!.. =sin 30° :=.!. 

m 2 

m=2 

64. They are both radii of the circle. 

65. It is the supplement of a 30° angle. 

66. Their sum is 180° - 150° =30°, and they are 

equal. 

67. Since triangle ACB is a 30°-60°-90° right 

triangle with AB =2 and CB:= ../3, 

DC= BD+BC=2+../3 

68. In right triangle DAC, 

(AD =(AC +(DC)2 

=12 +(2+../3)2 

=1+4+4../3+3 

=8+4../3 

=6+4../3+2 

=6+2.JU+2 

=(.J6 + ..fi)2 

AD=.J6+5 

69. In right triangle EAD, 

AD 

cosD= 

DE 

cos 150 =.J6 +..fi . 

4 

70. In right triangle EAD, 

(DE)2 =(AE)2 + (AD)2 

4 2 =(AE)2 +(.J6 +..fi)2 

16 = (AE)2 + 8 + 4../3 

(AE)2 =8 - 4../3 

=6-2.JU,+2 

=(.J6 _..fi)2 

AE =.J6- -Ii. 

Thus, 

. D AE 

sm = 

DE 

sin 150= .J6 -..fi . 

4 

71. Since angle ADC is 15°, 

AC 

tanI5°= 

CD 

1 

=../3 +2 

1 (../3-2) 

= (../3 +2) . (Jj-2) 

../3 -2 

=- 

3-4 

=2-../3. 

(J s-» 

72. (a) cos-=- 

2 R 

(J 1-cos l!. 

(b) tan- = 2 

4 sin! 

(c) 

_ 

1- R-b 

R 

- so 

Ii 

R-(R-b) 

= 

50 

b 

= 

50 

(J 12 

tan-= 

4 50 

Using ~ ~ or ITAN"1 

find that 

!!. "" 13.5° 

4 

(J "" 54° 

on a calculator, we 

204



1 

1. secx=- 

cos x 


1 

2. cscx=- 

sinx 


sinx 

3. tanx=- 

cos x 


cos x 

4. cotx=- 

sinx 

The answer is F. 

5. sin 2 x = l-cos 2 x 

The answer is H. 

6. tan 2 x+l=sec 2 x =_1_ 2 

 

cos x 


7. tan 2 x =_1_ 

2 

cot x 

The answer is D. 

8. sec 2 0_ tan 2 0 = _1_- - sin: 0 

2 

cos 0 cos 0 

I-sin 20 

- cos 20 

20 

= cos = 1 

cos 2 0 

2 

9. cot 0 =~ = cos 0 

sec0 COSlT sinO 

10. tan +cot 17 =-- +-- 

cos 20 sin 20 

20 2 

= sin (sin 0+cos 0) 

2 

cos 2 0 sin 2 0 

20 

sin ( 1 ) 

= cos 2 0 sin 20 

1 

= cos 2 0 

2 ~1 2 D) sin 2 0(1 cos 2 0) 


12. csc 2 0+sec 2 0 =_1_ 

2-+_1_2 

sin 0 cos 0 

cos 2 0 + sin 2 0 

- sin 2 0 cos 2 0 

1 

sin 0 1 1 

13. tan 0 - sec 0 csc 0 =-----._ 

cosO cosO sin 0 

sin 2 0 -1 

= sinOcosO 

20 

-cos cosO 

= =-- 

sin 0 cos 0 sin 0 

14. sin(-x) = -sinx 

tan(-x) = -tan x 

cot(-x) =-cotx 

csc(-x) =-cscx 

The functions sine, tangent, cotangent, and 

cosecant satisfy the condition f(-x) = - f(x) . 

15. cos(-x) = cosx 

sec(-x) = sec x 

The functions cosine and secant satisfy the 

condition f(-x) = - f(x) . 

16. cosx =~. x is in quadrant IV. 

5 

2 

. 2 2 3 16 

510 x = 1-cos x = 1- ( '5 ) = 25 

Since x is in quadrant IV, sin x < O. 

sinx=-~=-~ 

sinx --! 4 

tanx=--=-=- 

cosx ~ 3 

1 1 5 

secx=--=-=cos 

x t 3 

1 1 5 

cscx =--=- =- 

sinx -! 4 

1 1 3 

cotx=--=-=- 

tan x -1 4 

1 cos0 1+ cos 0 

11. cscO+cotO=--+--=-- 

sin 0 sin 0 sin 0 

205


206 Chapter 5 

Trigonometric Identities 

5 

17. tan x = - 

4' 

1r 

- < x < 1r 

2 

( 5)2 

2 41 

sec x=tan 2 x+l= -- +1= 

4 16 

Since x is in quadrant II, sec x < 0, so 

.,!4f 

secx=---. 

4 

1 4 4.,!4f 

cosx =--=---=-- 

sec x .,!4f 41 

sin x = cosx tan x = _ 4.,!4f (_ ~) = s.,I4f 

41 4 41 

1 4 

cotx=--=- 

tan x 5 

1 41 .,!4f 

cscx = sinx = 5J4i = -5 

18. (8) . 1r . (1r 1r) 

sm 12=sm 4"-6" 

.1r 1r 1r .1r 

=sm-cos--cos-sm 

4 6 4 6 

..fi..[j ..fi 1 =_._-_. 

2 2 2 2 

J6..fi 

=-- 

4 4 

J6-..fi 

= 

4 

cos ~ = cos(: - :) 

1r x .1r.1r 

= cos-cos-+ sm-sm 

4 6 4 6 

..fi..[j 

=_._+_ 

2 2 2 2 

J6..fi 

=-+ 

4 4 

J6+.J2 

= 

4 

tan ~ = tan(: - :) 

_ tanf-tanf 

..fi .- 1 

- l+tanftanf 

1-4 

= 1+(I)..[j . 

3 

3-..[j 3-..[j 

= 3+..[j ·~ 

12-6.J3 

= 

6 

= 2-..[j 

(b) In this exercise .!!.... is in quadrant I, where 

12 

all circular functions are positive. 

sin ~ = sin[; ) 

= ~1-COS1t 6 

2 

=~1-2~ 

=~2-4..[j 

=~ 

2 

cos~ =co{;) 

=~I+~ost 

=~1+./3 

.:..:.-..L.. 

2 

=~2+4..[j 

~ 

= 

2 

tan ~ = tan[; ) 

= sint 

l+cost 

_ 1. 2 _ 1 

- 1+./3-UJ3 

2 

1 2-..[j 

= 2+.J3 · 2-..[j 

=2-..[j 

206



19. Use the fact that 165° = 180° - 15°, 

sin 165° =sin 180° cos 15° - cos 180° sin 15° 

=sin 15° 

=sin(45° - 30°) 

=sin 45° cos 30° - cos 45° sin 30° 

~..fj -fi 1 

=2'2-2'2 

-I6-..fi 

= 

4 

cosl65° = cos 180° cos 15° + sin 180° sin 15° 

=-cos 15° 

=-cos(45° - 30°) 

=-(cos 45° cos 30° + sin 45° sin 30°) 

=J..fi ...fj + ..fi.!) 

1.2 2 22 

-J6 --fi 

= 

4 

tan 180° - tan 15° 

tan 165° = ...:.;,;;~-__-":"":':' 

I + tan 180° tan 15° 

=-tanI5° 

tan 45° - tan 30° 

= 1+ tan 45° tan 30° 

I-lj 

=-~ 

5 0 

sec16 

3-..fj 3-..fj 

=-3+$ '3-Jj 

12-6..J3 

= 

6 

=-2+..fj 

I 

cos 1 650 

I 

=-2+$ 

1 -2-..fj 

= -2+..fj -2-..fj 

-2-..fj 

= 4-3 

=-2-..fj 

20. cos 2 10° = cos(1500+600) 

= cos 150° cos 600 - sin 150° sin 60° 


21. sin 35° =cos(90° - 35°) 

= cos55° 


22. tan(-35°) =cot[900 -(-35°)] 

= cot 125° 

The answer is J. 

23. - sin 35° =sin(-35°) 


24. cos35° = cos(-35°) 

The answer is I. 

150° 

25. cos 75° = cos- 

2 

I +cosI50° 

= 

2 


26. sin 75° =sin(15° + 60°) 

= sin 15°cos6O° + cos 15°sin 60° 

The answer is H. 

27. sin 300° = sin 2(150°) 

= 2 sin 150° cos 150° 

The answer is D, 

207



28. cos 300° = cos 2(150°) 

= cos 2 150° -sin 2 150° 

The answer is G. 

29. Since sin x = 0.8, sin 2 x = 0.64. Then 

cos 2 x = 1-sin 2 x = 1-0.64 = 0.36, and 

cos x = ..J0.36 = 0.6 (take the positive root 

. 0 Tr 

since



. 2 I . . d II' 

32. smy = --. cosx = -- X IS m qua rant , y IS 

3 5 

in quadrant III. First, find sin x, cos y, tan x, and 

tan y. Since x is in quadrant II, sin x is positive. 

sinx=~I-cos2 x 

=~ 

2../6 

=- 

5 

Since y is in quadrant III, cos y is negative. 

cosy = -~I-sin2 y 

=-FH) 

=-~ 

~ 

=- 

. 3 ¥ 

tan x = smx =--=-2../6 

cos x -t 

siny -t 2 2~ 

tany = cosy = _{- = ~ =-5 

sin(x + y) =sin xcosy+ cosxsin y 

=(2~X- ~)+(-~X-~) 

-2-J3Q +2 

= 

15 

cos(x - y) =cosxcosy+sinxsiny 

=(-~X-~)+(2~X-~) 

~ -4../6 

= 

15 

_ ta_n_x_+_ta_ n--,Y:.... 

tan(x+y) = l-tanxtany 

-2../6+M 

= 5 

1-(-2../6X2f5) 

-10../6 +2~ 

=----'-----==- 

5+4-J3Q 

To find the quadrant of x + y. notice that 

sin(x + y) < 0, which indicates that x + y is in 

quadrant III or IV. Also, tan(x + y) < 0, which 

indicates that x + y is in quadrant II or IV. 

Therefore, x + y is in quadrant IV. 

33. Find sin (x + y), cos (x - y), and tan (x + y), 

I 

• 

. . 1 4. . d 

given Sin x = -, cos y = -, X IS m qua rant 

10 5 

y is in quadrant IV. 

Since x is in quadrant I, cos x > O. 

cosx=~I-sin2 x 

=~1-1~ 

=~:~ 

3M 

=- 

10 

Since y is in quadrant IV, sin y < O. 

siny = -~I-cos2 x 

=-~I-~~ 

=_~25;16 

3 

=- 

5 

sin(x+ y) = sinxcosy+ cosxsiny 

=(-.!.-X~) + 3.J11 (-~) 

10 5 10 5 

4 9-../li 

=--- 

50 50 

4-9.J!f 

= 

50 

cos(x- y) =cosxcosy+sinxsiny 

=( ~X~)+C~X -~) 

12M 3 

=--- 

50 50 

12M-3 

= 

50 

209



sin(x + y) 

tan(x + y) = _':'-""""--7 

cos(x+ y) 

_ 4-9P 5 

- 120Ji1+3 

50 

4 - 9.Ji1 12.Ji1 - 3 

= 12.Ji1 +3 . 12.Ji1-3 

4S.Ji1 -12 -l1(IOS) + 27.Ji1 

=_..:...-_--:---:~..:--.:.....- 

144(11)-9 

75.Ji1-1200 

= 

1575 

_ 75(.Ji1) -75(16) 

- 75(21) 

.Ji1-16 

= 

21 

To find the quadrant of x + y, notice that 

sin(x + y) < 0, which implies x + y is in 

quadrant III or IV. Also tan(x + y) < 0, which 

implies that x + y is in quadrant II or IV. 

Therefore, x + y is in quadrant IV. 

2. I. . d IV .. 

34. cosx=-, smy=--, IS In qua rant ,y IS In 

9 2 

quadrant III. First, find sin x, cos y, tan x, tan y. 

Since x is in quadrant IV, sin x is negative. 

sinx = -~I-cos2 x 

=-R%) 

=-~ 

..ff7 

=-- 9 

Since y is in quadrant III, cos y is negative. 

cosy = -,It -sin 2 y 

=-FG) 

=-l 

.J3 

=- 2 

-lll 

sinx 

..ff7 

tanx=--=--=-- 

cos x ~ 2 

siny --!- 1 .J3 

tany=--=--=-=cosy 

_li .J3 3 

2 

sin(x+ y) = sinxcosy+cosxsiny 

~231-2 

=....;"....- 

IS 

cos(x- y) = cosxcosy+ sinxsiny 

..ff7 -2.J3 

= 

IS 

_ tan ____ta x + _n -,y~ 

tanx+y ( ) = l-tanxtany 

_.ffi +..[3 

_ 2 3 

- 1-(-~)(~) 

_ 2.J3 - 3..ff7 

- 6+~231 

or 

sin(x+y) ~231-2 

tan(x + y ) = = ----,=---= 

cos(x + y) -..ff7- 2.J3 

To find the quadrant of x + y, notice that 

sin(x + y) > 0, which indicates that x + y is in 

quadrant I or II. Also, tan(x + y) < 0, which 

indicates that x + y is in quadrant II or IV. 

Therefore, x + y is in quadrant II. 

35. Find sin 9 and cos 9, given cos 29 = - ~, 

4 

90 0 < 29 < IS0 0 • 

Since 29 is in quadrant II, 9 is in quadrant I. 

cos 29 = 1-2 sin 2 9 

-~=1-2sin29 

4 

7 2· 2 

--=- sm 9 

4 

2. =sin 2 9 

S 

Since 9 is in quadrant I, sin 9 > 0. 

Jt =sin9 

sin9 = .J14 

4 

210



36. cos 2B = .!., B is in quadrant IV. Since B is in 39. Find cos !!.., given cosO = _.!.., 90 0 < 0 < 180 0 • 

8 

2 2 

quadrant IV, sin B is negative and cos B is 

positive. 

Since 0 is in quadrant Il, !!.. is in quadrant I, so 

2 

sinB=_~I-C;S2B =_~I;t =_~ =_ ~ 

COSB=P+COS2B =~I+i = f9 =~ 

2 2 "16 4 

37. Find sin 2x and cos 2x, given tan x = 3, 

40. Find sin~, cosA=-~,900 0 and sin x < 0, then x is in 

A 

quadrant ill and 2x is in quadrant I or n. 

Since 90° < A < 180°, 45° < - < 90°, 

2 

2tanx 2(3) 6 3 

tan2x= =--=--=- 

l-tan 2 x 1-3 2 . A, , . 

Thus, sm- IS positive. 

8 4 

2 

Since tan 2x < 0, 2x is in quadrant n. Thus, sec 

2x < 0 and sin 2x > O. 

sm-= 

, A ~1-COSA 

2 2 

sec2x = -~1 + tan 2 2x 

=-FH) 

=~ 

6 

=,fI 

=- 

8 

..fi../2 

5 

=- =2../2 4 

'"J2 

1 4 

$4 

cos2x =--=- 

=- 

sec2x . 5 

4 

. r-p4)2 3 

sm2x = v 1 -l-s) ='5 

41. Find tan x, given tan 2x = 2, rr < x < 3rr . 

2 

tan2x = 2 

5, 0 

2tanx =2 

38• secy = -'3' smy> 

l-tan 2 x 

1 3 

cosy=--=- 

2 tan x = 2 - 2 tan 2 x 

secy 5 

tan 2 x+tanx-l=O 

-1 ±.Jl+ 4 -1 ±../5 

SinY=~I-COS2Y=~I-(-~r =~ 

tan x = =-~- 

2 2 

sin2y = 2sinycosy 

Since x is in quadrant III, tan > o. 

-1+../5 

tanx=-- 

=2(~X-~)=-~: 

2 

cos2y = cos 2 y-sin 2 y 

=(-~r -(~r 

9 16 

=-- 

25 25 

7 

=- 

25 

211

 



43. 45. s ~ 

Y I =Y-::~ ~J 

4 

44. 

sin2x+sinx be sinx 

The graph of y = appears to The graph of y = appears to be the 

cos2x - cos x 

1-cosx 

x 

x 

the same as the graph of y = cot -. same as the graph of y = cot - . 

2 2 

sinx 1 

sin2x+sinx 2sinxcosx +sinx 1-cos x - 1-9 

= OSX 

SID X 

cos2x -cosx 2cos 2 x -1-cosx 1 

sin x(2 cosx + 1) 

=- 

= tan! 

(2cosx + 1Xcosx -1) 

x 

sinx 

=cot 

= 2 

cosx-1 

sinx 46. cos X sin 2x 

= 1-cosx 1 +cos2x 

1 

= I-wsx 

Sin X 

1 

=- 

tan! 

x 

=cot 

2 

-4 

cosx sin2x b h 

The grap h 0 f y = appears to e t e 

1+cos2x 

same as the graph of y =sin x. 

cosxsin2x cos x ·2sinxcosx 

-----=-----.,;:-----., 

1+cos2x 1+2cos 2 x-I 

2sinxcos 2 x 

= 2cos 2 x 

=sinx 

1-cos2x 47. 2(sin x - sin 3 x) 

The graph of y = . appears to be the YI = --'---cos-x---'-- I 

sm2x 

same as the graph of y = tan x. 

4 

----=tan 1-cos2x (2X) 

sin2x 2 

= tan x 

-4 

212



48. 

2(sinx-sin 3 x) 

The graph of y = 

appears to be 

cosx 

the same as the graph of y = sin 2x. 

2(sinx-sin 3 x) 2sinx(l-sin 2 x) cosx 

---:._------"- = 

._ 

cosx cos x cosx 

_ 2sinxcosx(l-sin 2 x) 

- cos 2 x 

_ sin2x(l-sin 2 x) 

- l-sin 2 x 

=sin2x 

The graph of y = esc x - cot x appears to be the 

x 

same as the graph of y = tan- . 

2 

1 cosx 

cscx-cotx =---- 

sinx sinx 

l-cosx 

= sinx 

x 

=13n 

2 

2 Y 

2 

49. Verify sin 2 x - sin 2 y = cos - cos x. 

sin 2 x-sin 2 y= (1- cos 2 x) - (l-cos 2 y) 

=1-cos 2 x-l+cos 2 Y 

2 2 

=cos y-cos X 

2x-sin2x 

. 3 cos 

50. Venfy 2cos x-cosx=----- 

secx 


cos 2 x-sin 2 x cos 2 x-sin 2 x 

-----=-----:--- 

secx ~ 

cosx 

= cos3x -sin 2 xcosx 

3 

= cos x -(1-cos 2 x)cosx 

= 2 cos 3 x - cos x 

sin 2 x 2 x 

51. Verify = cos - . 

2-2cosx 2 


sin 2 x 1-cos 2 x 

----=--- 

2-2cosx 2{I-cosx) 

_ (1-cosx){1 + cosx) 

- 2{I-cosx) 

l+cosx 

= 

2 


2 x l+cosx 

cos -=-- 

2 2 

52. Verify si~ 2x = _2_. 

smx secx 

sin 2x 2sin x cosx 

--= 

sinx sinx 

=2cosx 

2 

= I 

cos x 

= 2 

sec x 

53. Verify 2cosA -secA = cosA _ tan A . 

cscA 


tan A ~1 

cosA---=cosA-- 

cscA 

sr.hsin 

2 A 

=cosA--- 

cosA 

2 

= cos A-(I-cos 2 A) 

cos A 

2cos 2 A-I 

= 

cos A 

1 

=2cosA-- 

cos A 

= 2cosA -secA 

ify 2 tan B 2 

54. Ven ---=sec B. 

sin2B 

2 tan B 2 . ..s.in..8. cos B 

---=--=~sin2B 

2sinBcosB 

1 

= cos 2 B 

=sec 2 B 

213


214 

Chapter 5 Trigonometric Identities 

55. 

56. 

57. 

Verify 1+ tan 2 a = 2 tanacsc2a . 

2tana 

2tanacsc2a =-- 

sin2a 

2~~ 

=-----"="-- 

2sinacosa 

2sina 

=------=- 

2sinacos 2 a 

1 2 

=--2- ;: sec a 

cos a 

=1+tan 2a 

Verify tan A + tan B+tanC = (tan AXtanBXtanC) 


tan A + tan B+ tan C;: tan A + tanB+ tan(I80 -(A + B)) 

= tan A + tanB- tan(A + B) 

tan A + tan B 

= tan A + tan B - ---- 

I-tanA tan B 

tan A(I- tan A tan B)+ tan 8(1- tan A tan B) - tan A - tan B 

l-tanA tan B 

tan A -tan 2 A tan B+ tan B-tanA tan 2 B-tanA -tan B 

= 

I-tanA tanB 

-tan 2 AtanB-tanAtan 2 B 

;: 

l-tanA tan B 

tan A tan B(tan A + tan B) 

I-tanA tan B 

;: _ tan A tan B. (tan A + tan B) 

I-tanA tan B 

= -tan A tan Btan(A + B) 

= tan Atan B(-tan(A + B)) 

= (tan A)(tan B)(tanC) 

Verify [tan(45° + A)][tan(45° - A)];: 1. 

Work with left side. 

[tan(45° + A)!tan(45° - A)] 

0+ 0 

;: [sin(45 A)Isin(45 - A)] 

cos(45° + A) 

cos(45°-A) 

_ (sin 45° cos A + cos 45° sin AXsin 45° cos A -cos45°sinA) 

- (cos 45° cos A -sin 45° sin A)(cos45°cos A +sin45°sin A) 

sin 2 45°cos 2 A -sin45°cos45°sinAcosA + sin 45° cos 45° sin AcosA -cos 2 45°sin 2 A 

- cos 2 45°cos 2 A +sin45°cos45°sin AcosA - sin 45° cos 45° sin AcosA -sin 2 45°sin 2 A 

sin 2 45° cos 2 A - cos 2 45°sin 2 A 

- cos 2 45° cos 2 A - sin 2 45°sin 2 A 

_(~t cos 2 A-(~t sin 2 A 

-(~t cos 2 A-(4f sin 2 A 

=1 

214


Cbapter 5 Review Exercises 215 

. 2cotx 2 

58. Venfy --=csc x-2. 

tan2x 

2cotx 

2 

tan 2x = tan x(~) 

I-tan x 

l-tan 2 x 

= tan 2 x 

I-~ 

~ 

= . 2 

=2; 

cos 2 x-sin 2 x 

= 

sin 2 x 

1-2sin 2x 

= sin 

2 

x 

=csc 2 x-2 

59. Verify tanOsin20 = 2 -2cos 2 0 

tan Osin20 = tan 0(2 sin 0 cosO) 

sinO . 

=--(2smOcosO) 

cosO 

= 2sin 20 

= 2(I-cos 2 0) 

=2-2cos 20 

60. Verify esc A sin 2A - sec A = cos 2A sec A . 

esc Asin 2A - sec A = _1_(2sin AcosA) __1_ 

sin A 

cos A 

2cos 2 A-I 

= 

cos A 

cos2A 

=-- 

cos A 

= cos2A sec A 

61. Verify 2 tan x esc 2x - tan 2 x = 1. 

2tanxcsc2x-tan 2 x 

I 2 

=2tanx---tan x 

sin2x 

I 2 

= 2 tan x---tan x 

2sinxcosx 

sinx sin 2 x 

= sinxcos 2 x - cos 2 x 

sin 2 x 

= cos 2 x - cos 2 x 

l-sin 2 x 

= cos 2 x 

2 

cos x 

= cos 2 x 

=1 

. 2 1-tan 2 0 

62. Venfy 2cos 0-1 = 2 ' 

I+tan 0 


2 I-~ 

I - tan 0 

cos 2 (} 

l+tan 20 = I+~ 

cos e 

cos 2 0 - sin 2 0 

= cos 2 0+sin 2 0 

= cos 2 0-sin 2 0 

= cos20 

= 2cos 2 0-1 

2 ll 

2tanOcos 20-13nO 

II 

63. Verify tan u cos u = 2 

I-tan 0 


2 tan Ocos 2 0 - tan 0 tan0(2cos 20-1) 

l-tan 20 l-tan 20 

tan 0(2cos 2 0-1) 

= cos 2 (I-sin 2 (I 

cos 2 (} 

_ cos 2 otan0(2 cos 2 0-1) 

- 2cos 20-1 

20tanO 

=cos 

= tanOcos 2 0 

64. 

Verify 

en -cot-= 

x _s_in_2_x_+_s_in_x_ 

2 cos 2x - cos x 


sin2x + sin x 2sinxcosx +sinx 

cos2x-cosx - (2cos 2 X-I)--cosx 

2 sin x cos x + sin x 

= (2cos 2 x-cosx-I) 

sin x(2 cos x + I) 

=..,....-~-----'-....,... 

(2 cos x + IXcosx -I) 

sinx 

= cosx-I 

sinx 

= I-cosx 

I 

=-- x 

tan 

2 

x 

=-cot 

2 

215



3 COS 2 x-sin 2 x 

65. Verify 2cos x-cosx=---- 

sec x 


cos 2 x-sin 2 x 2cos 2 x-I 

secx 

= -..l. 

cosx 

= (2cos 2 x-l)cosx 

= 2cos 3 x-cosx 

66. Verify sin 36 = sin 6 - cos 2 6sin 6. 


sin6 - cos 2 6sin 6 = sin6 - (1- sin 2 6)sin 6 

= sin 6 - sin 6 + sin 3 6 

= sin 3 6 

67. Verify cos 2 x+sin 2 x(I-2sin 2 y)=cos 2 y+sin 2 y(l-2sin 2 x). 

Show that the left side minus the right side equals equals O. 

2 2 

cos x +sin 2 x(l- 2sin 2 y) - cos y -sin 2 y(1-2sin 2 x) 

=cos 2 x+sin 2 x-2sin 2 x,sin 2 y-cos 2 y-sin 2 y+2sin 2 xsin 2 y 

= (cos 2 x+sin 2 x)-(cos 2 y+sin 2 y) 

=1 -1 

=0 

68. 

. 2 sec2a-l 

69. Venfy sec a-I = . 

sec2a+l 


sec2a-l roba- 1 

sec2a+ 1 coJ2a + 1 

J -1 

_ cos 2 a-sin 2 a 

- J + 1 

cos 2 a-sin 2 a 

l-cos 2 a+sin 2 a 

- l+cos 2 a-sin 2 a 

2sin 2 a 2 

= 2 = tan a 

2cos a 

=sec 2 a-I 

70. Verify sin 31 + sin 21 = tan ~ . 

sin 31 - sin 21 tan t 

Use the identities in Section 5.5 Connections 

. 31 + 21 31 - 21 

with A=-- and B=--. Then 

2 2 

A+B=31 and A-B=21. 

2sin(1!.¥L) cos(1ft) 

sin 31 + sin 21 

sin31-sin21 - 2cose';2')sine';2,) 

sin~cosL 51 I 

= 2 2 = tan-cotcos~sinL 

2 2 

2 2 

tan~ 

=__ 2_ 

tant 

. 2 tan26 

71. Venfy tan 46 = 2' 

2 -sec 26 

tan46 = 2 tan 26 

1-(sec 2 26 -1) 

2 tan 26 

- 1- sec 2 26 + 1 

2tan26 

- 2 -sec 2 26 

216



72. 

73. 

Verify 2cos 2 '::'tanx = tanx+sinx. 

2 


. sinx . 

tanx-e sm r =--+SIOX 

COSX 

= sin X(I + _1_) 

cos x 

= sinx(l+ ~ ) 

2cos "!-I 

. (_2C_os_ 2,*"t_-_ I+_I) 

=SlOX 

2cos 2 f - 1 

2sinxcos 2 t 

= 2cos 2 f - 1 

2 x sinx 

=2cos -'- 

2 cosx 

= 2 tanxcos 2'::' 

2 

Verify tan(1+ :) = sec x + tan x . 


tan('::'+~) = tant+tant 

2 4 1-tanttant 

=tant+ 1 

I-tant 


I sin x 

secx+tanx=--+- 

cos x COSX 

_ cos 2 t+ sin2 t+2sintcost 

- cos 2 t-sin2 t 

_ 

(cost+sint)2 

- (cosf-sintXcost+sint) 

= cost+sint 

cost-sint 

cost + sin! 

_ cost cost 

- cost _ sinf 

cosf cost 

=I+tant 

I-tant 

74. 

75. 

76. a. 

. 1 x 1 x 

Venfy -cot---tan-=cotx. 

2 2 2 2 

'!'cot'::' _.!. tan'::' =.!.( 1+ cosx) _.!. (1-cosx) 

2 2 2 2 2 sin x 2 sin x 

=(l+cosx)-(l-cosx) 

2sinx 

2cosx 

=-- 

2sinx 

=cotx 

From Exercise 56, 

tanA+ tanB+ tanC = (tan AXtanBXtanC). 

Let C be the largest angle of D.ABC. 

Then 90° < C < 180° (since the angles of a 

triangle sum to 180°) and tan C < O. (since 

tan x is negative for 90° < x < 180°) Since 

C > 90°, then A + B < 90°, so 0 < A < 90° and 

o

negative angle). Then tan A > 0 and 

tan B > 0 (since tan x is positive for 

0° < x < 90°), which means that 

(tan AXtan BXtan C) is the product of two 

positive values and one negative value, and 

hence is a negative number (which is < 0). 

Since (tan AXtanBXtanC) < 0, 

tan A + tanB+tanC < O. 

Although the table indicates that the 

functions may be equal, the graph clearly 

demonstrates that cos x = cos 4x is not an 

identity. 

217



b. x · Vi ·:· Y2 Chapter 5 Test Exercises 

-0 

c.: ~ 

0 " 

3.1lt1& 0 , .0 

-, .' 

Un2: 0 . . ~: .~ . 3n 

'.'12:ltl "2["n 0 1. Since - < x < 21f, sinx < 0 and cos x > O. 

1,".5&& 0 · ' "1E"U .. 2 

1!.01 2[~1) 0 

.11.15: -"E·l~ 1E~13 Then 

~.., ' . .,J 

K=i!t 

Although the table indicates that the 

functions may be equal, the graph clearly 

demonstrates that sin (x + n) =.5 sin x is 

sec2x=tan2x+l=(-~r +1 

25 

=-+1 

36 

61 

= 

36 

~ kith . . 

S0, sec x = 6' ta ng e positive square 

6 6~ 

root. Thus, cos x = 177 =--. 

....61 61 

. 5 6~ 5.J6l 

smx = tan x cosx = -_ .--=- 

6 61 61 

2. tan2x-sec2 x=(sinx)2 _(_1_)2 

not an identity. cosx cos x 

c. Y I = 4sin x cos x 

Y 2 = 2sin 2x 

sin 2 x-I 

- cos 2 x 

2 

cos X 

= - cos2 x 

=-1 

3. Since x is in quadrant ill, cos x < O. Then 

sin 2 x + cos 2 x = 1 

cos 2 x =I-sin 2 x 

=1-(-~r 

1 

=1- 

9 

8 

= 

9 

cos x = _ f! = -2.,fi 

The table indicates that the functions may 

'/'9 3 

be equal, the graph confirms that 

4sin x cos x = 2sin 2x is an identity. 

218



Since y is in quadrant II, sin y > o. Then 

sin 2y+cos2y=1 

. 2 I 2 

S10 y= -cos Y 

=1-(-~r 

=1-~ 

4. sin(-22.50) = ±~ 1-cOi-45°) 

=+~1_.J2 __2_ 

- 2 

= ±~2-4..fi 

25 

21 

=+~ 

= 

- 2 

25 Since -22.5° is in quadrant IV, sin(-22.5°) is 

sin y = (2l =..fii 

V"25 5 

sin(x + y)= sinxcosy + cosxsiny 

-~2-.J2 

negative. sin(-22.5°) = ----'-- 

2 

5. Y I = sec x - sin x tan x 

2 2...[42 

=--- 

IS IS 

2-2...[42 

= 

IS 

-4 

cos(x - y)= cos x cosy+ sin xsiny 

. _: .. 

~. ~-, 

-.~ 

The graph of sec x - sin x tan x appears to be the 

same as the graph of cosx. 

secx-s1Oxtanx=---s1OX 

. I. (sinx) -- 

4.J2 .J2! cosx cosx 

=---- 

15 15 1-siri 2 x 

= 

4..fi -..fii 

cosx 

= 

15 cos 2 x 

=- 

cosx 

tan(x + y)= _ta_n_x_+_ta_n.....:y~ 

=cosx 

1- tan x tan y 

_ ~+(-4I) 

- 1-(1)(- ~) 

2.J2 -4.J2! 

=---.==- 

8+...[42 

6. Y I = rol X 

Z 

- calx 

x 

The graph of cot - - cot x appears to be the 

2 

same as the graph of esc x . 

x 1+ cosx cosx 

cot - - cot x =-- 

2 sinx sinx 

I+cosx-cosx 

= 

sinx 

=- 

sinx 

=cscx 

219



ify 2 B 1 

7. Ven sec = 2' 

I-sin B 


1 1 2 

--....."...- =---=sec B 

l-sin 2 B cos 2 B 

8.encos Verify 2A = cot A - tan A , 

esc AsecA 


cos A 

sinA 

cot A - tan A _ Siii"A - CiiS'A 

esc A sec A - (si~A)(C~A) 

cos 2 A-sin 2 A 

sinAcosA 

= J 

sin AcosA 

cos 2 A -sin 2 A sin A cos A 

= sin AcosA 

=cos 2 A -sin 2 A 

22A 

=cos 

9. The cofunction identities in Chapter 2 applied 

only to acute angles . The cofunction identities 

in this chapter apply to any angles and to any 

real numbers. 

10. (a) V =163sin on, Since sin x=co{~ - x). 

V=163CO{~ -Wt). 

(b) If V =163CO{ ~ -1201r t} the maximum 

voltage occurs when cos (~ -1201r t) = 1. 

Thus. the maximum voltagae is 

V =163 volts. 

cos (~ -1201r t) =1 when 

!!.- 1201rt =2k1r. where k is any integer. 

2 

The first maximum occurs when 

1r -1201r t =0 

2 

1r 

- =1201rt 

2 

1 1r 

-_·_=t 

1201r 2 

1 

--=t 

240 

The maximum voltage will first occur at 

1 

--sec. 

240 

220


CHAPTER 6: INVERSE TRIGONOMETRIC 

FUNCTIONS AND 

TRIGONOMETRIC EQUATIONS 

Section 6.1 


j-I(x) = x-5; HAPPINESS IS STAYING OUT OF 

2 

HOSPITALS AND COURT ROOMS 

Exercises 

1. (a) [-I, 1] 

(b) [_1r ~J 

2'2 

(c) increasing 

(d) -2 is not in the domain. 

6. 

7. 

8. 


y = arcco{'7) 

-13 

cos y = -,OS; y S; 1r 

2 

y is in quadrant I. 

1r 

y= 6 

y=tan- Il 1r 1r 

tan y = 1 - - < Y < 

'2 2 

y is in quadrant I. 

n 

y= 4 

y = sin-I 0 

. 1r 1r 

sm y = 0 - - S; y S; 

'2 2 

y=O 

2. (a) [-1, 1] 

(b) [O,1r] 

(c) decreasing 

41r . . th 

(d) - IS not In e range. 

3 

3. (a) (~,oo) 

9. y = cos-I (-1) 

cosy=-l, OS;yS;1r 

y=1r 

10. y = arctan(-1) 

1r 1r 

tany=-l, --


222 Chapter 6 Inverse Trigonometric Functions and Trigonometric Equations 

14. y=arcsm . (-T 

~) 

21. y = coC1(-I) 

coty=-I,O



... 

28. y = arcsec 2 

Jr 

sec y = 2,0:S y:S x, y:;t: 2" 

S· Jr 2 x 

mce sec 3" = , y = 3"' 

29. O=arctan(-I) 

tan 0 = -1, - 90° < 0 < 90° 

ois in quadrant IV. The reference angle is 45°. 

Thus, 0= -45°. 

30. 0 = arccos(- ~) 

1 

cosO =--,Oo:s O:S 180° 

2 

ois in quadrant II. The reference angle is 60°. 

Thus, 0 = 180° - 60° = 120°. 

31. 0 = arcsin(- ~) 

sin 0 = - .J3 ,-90° :s 0 :s 90° 

2 

ois in quadrant IV. The reference angle is 60°. 

0=-60°. 

32. 0 = arcsin(- ~) 

sin 0 = - .fi,- 90° :s 0 :s 90° 

2 

ois in quadrant IV. The reference angle is 45° . 

0=-45° 

l( 

33. 9 =COC - ~) 

cot 0=_.J3 0°



For Exercises 43-48, be sure that your calculator is in 

radian mode. Keystroke sequences may vary based on 


SO. 

43. y = arctan 1.1111111 

Enter 1.1111111 El ~ 

orB ~ 1.1111111 ~ 

Display: .83798122 

arctan 1.1111111 = .83798122 

44. y = arcsin .81926439 

Enter: .81926439 G ~ 

orB ~.81926439 ~ 

Display: .96012698 

arcsin .81926439 = .96012698 

Domain=(~, -~]u [~, DO) 

Range=[-~, 0)U(0, ~] 

45. 

y=coC 1(-.92170128) 

Enter: 

.92170128 B §) B ~ m [nl 

or 

B ~ GJ .92170128 0 m Inl ~ 

Display: 2.3154725 

coC I (-.91270128) = 2.3154725 

46. y=sec-I(-1.2871684) 

Enter: -1.2871684 §) G §) 

orB §J GJ 1.2871684 0 

Display: 2.4605221 

sec-I (-1.2871684) =2.4605221 

47. y =arcsin .92837781 

Enter: .92837781 B 8 

or B ~ .92837781 ~ 

Display: 1.1900238 

arcsin .92837781 = 1.1900238 

48. y =arccos .44624593 

Enter: .44624593 ~ §) 

or B §) .44624593 ~ 

Display: 1.1082303 

arccos .44624593 = 1.1082303 

49. 

51. 

52. 

53. 

Domain: (-DO, - 2] U [2, DO) 

Range=[0, ~) U (~ , 1r] 

flf-I(x)] = ~ x; 4) 

=\X;2)_2 

=x+2-2 

=x 

f-I[f(x)] =f- I[3x - 2] 

(3x-2)+2 

= 

3 

3x 

= 

3 

=x 

In each case the result is x. The graph is a straight 

line bisecting the first and third quadrants 

(y =x). 

Domain: (- DO, DO) 

Range: (0, 2n) 

It is the graph of y = x. 

224



54. 

tan(arccos ~) =tan m=~ 

55. 

It does not agree because the range of the inverse 

f . . ( 1r 1r) 

tangent unction IS -"2'"2 ,not (-.co, 00),as was 

the case in Exercise 53. 

58. sin ( arccos±) 

1 1 

Let 0 = arccos-, so that cosO =-. 

4 4 

Since arccos is defined only in quadrants I and II, 

and ± is positive, 0 is in quadrant I. Sketch 0 and 

label a triangle with the side opposite 0 equal to 

~42 _1 2 =..m. 

y 

---!...;.....L---_ x 

.( 1). 0 ..m 

S10 arccos'4 =S1O =7 

56. 

59. cos(tan-I(-2)) 

Let co =tan- I (-2), so that tan m =-2. Since 

tan -I is defined only in quadrants I and IV, and 

-2 is negative, co is in quadrant IV. Sketch CO and 

label a triangle with the hypotenuse equal to 

~(_2)2 + 1 =$. 

y 

57. ta{arccos ~) 

3 3 

Let m=arccos '4' so that cos m='4' Since 

arccos is defined only in quadrants I and II, and 

~ is positive, cois in quadrant I. Sketch co and 

label a triangle with the side opposite CO equal to 

~42 _3 2 =.,fi. 

y 

__-!'-..I.-_--"L...-_ x 

3 

60. 

---r-r-.---- x 

- 2 

cos(tan-1(_2)) =cos co = $ 

5 

Le 0 . -l( 1) that si 1 

t = sm - S ' so at S10 0 =- S · 

Since arcsin is defined only in quadrants I and IV, 

1 

and -Sis negative, 0 is in quadrant IV. Sketch (J 

and label a triangle with the side adjacent to 0 

equal to ~52 _(_1)2 =..{i4 = 2-/6. 

225



y 

--;-=::-,--...,...-'X 

-I 

sec( Sin 

.-I( -'5 I)J Ll 5 5../6 

=secl7=2J6=12 

Let m= arcsin ( - ~). so that sin m = - ~. Since 

arcsin is defined only in quadrants I and IV, and 

-~ is negative, mis in quadrant IV. Sketch m 

3 

and label a triangle with the side adjacent to m 

equal to ~32 - (_2)2 = $. 

y 

co{arcsin(- %)J= cot m= -;: 

62. cos (arctan j) 

Let 6 = arctan ~, so that tan 6 = ~. Since arctan 

3 3 

is defined only in quadrants I and IV, and ~ is 

3 

positive, 6 is in quadrant I. Sketch 6 and label a 

triangle with the hypotenuse equat to 

~82 +3 2 =m. 

y 

---f-"--..J..---__ x 

8 

co{arctan~) = cos 6 = ~ = 3m 

3 v73 73 

63. sec(sec- 12) 

Since secant and inverse secant are inverse 

functions, sec(sec- 12) = 2. 

64. esc (esc-I .J2) = esc ==.J2 

65. arccos (cos 1r) = arccos .J2 = 1r 

4 2 4 

66. arctan(tan(- =)) = arctan(-I) = - = 

67. arcsin(sin 1r) = arcsin .,fj = 1r 

323 

68. arccos(cos 0) = arccos 1 = 0 

69. sin(2 tan-I 1:) 

Let m= tan-I .!3., so that tan t» = .!3.. 

5 5 

Since tan-I is defined only in quadrants I and IV, 

an d - 

12 . . . .. d I 

IS positive, m IS In qua rant . 

5 

Sketch m and label a right triangle with the 

hypotenuse equal to ~122 + 52 = 13. 

y 

12 

---+~--I---. X 

. 12 

slnm= 

13 

5 

cos m= 13 

sin (2 tan-I 1:) = sin(2m) 

=2sin m cos m 

=2 (.!3.X2.-) 

13 13 

120 

= 

169 

226



70. Cos(2sin 1 ±) 

Let 0 =sin- 1 !-, so that sinO =!-. 

4 4 

Since sin- 1 is defined only in quadrants I and IV, 

and !- is positive, 0 is in quadrant I. Sketch 0 

4 

and label a triangle with the side adjacent to 0 

equal to ~42 _1 2 =.Jl5. 

y 

y 

--I'-.l----L---.x 

3 

cosro = 

5 

. 4 

Slnro = 

5 

4 

co{2 arctan ~) =cos(2ro) 

_--!"oo::;,,=.o-=_-..L__ x 

cO{2sin- 

1±) =cos 20 

71. co{2 arctan ~) 

=1-2sin 2 0 

=1-2(±Y 

7 

= 

8 

Let ro=arctan i, so that tan ro =i. Since 

3 3 

arctan is defined only in quadrants I and IV, and 

i is positive, ro is in quadrant I. Sketch ro and 

3 

label a triangle with the hypotenuse equal to 

~42 +3 2 =5. 

72. tan(2cos 1 ±) 

=cos 2 to - sin 2 ro 

=(~r -(~r 

9 16 

=-- 

25 25 

7 

=- 

25 

1 

Let 0 =cos- 1 !-, so that cos 0 =!-.Since cos 

4 4 

is defined only in quadrants I and II, and !- is 

4 

positive, 0 is in quadrant I. Sketch 0 and label a 

riangle with the side opposite 0 equal to 

~42 _1 2 =.Jl5. 

y 

e 

1 

$5 

tan 0 =-=.Jl5 

1 

227



74. cos (2 arctan (-2» 

tan(2 cos -I i) = tan 29 Let 9 = arctan (-2), so that tan 9 = -2. 

2 tan () 

- l-tan 29 

2.J15 

= 2 

1-(.J15) 

2.Jl5 

=- 

-14 

J15 

=-- 

and -2 is negative, 9 is in quadrant IV. 

Sketch 9 and label a triangle with the hypotenuse 

equal to ~(_2)2 +1 2 =$. 

Since arctan is defined only in quadrants I and IV, 

y 

7 

73. sin(2 cos-I ~) 

Let CO = cos-I .!., so that cos co = .!.. Since 

5 5 

cos-I is defined only in quadrants I and II, and 

t is positive, (0 is in quadrant I. Sketch co and 

label a triangle with the side opposite CO equal to 

~52 _1 2 =.J24 = 2../6. 

y 

1 $ 

cos9="J5=s 

cos (2 arctantzj) =cos 29 

= 2cos 2 9-1 

=2(JsY -I 

=~-1 

5 

3 

=- 

5 

75. tan(2 arcsi{ ~)) _ 

---~---_JC 

. 2../6 

SIn (0=- 

5 

1 

cos co= 

5 

sin(2 cos-I ~) = sin2co 

= 2sin co cos co 

={2~X~) 

4../6 

=- 

25 

Let co =arcsin(- ~). so that sin co = ~ . 

Since 

arcsin is defined only in quadrants I and IV, and 

-~ is negative, Cl> is in quadrant IV. Sketch Cl> 

and label a triangle with the side adjacent to Q) 

equal to ~52 _(_3)2 =4. 

y 

--~r----~"''' 

3 

tan (0 =- 

4 

228



tan(2 arcsin( ~) ) = tan(2m ) 

76. sin(2 arccos %-) 

2tanm 

-1-tan 2m 

2(-t) 

= 2 

I-(-t) 

-i 16 

= 1--& ·16 

-24 

=- 

16-9 

24 

=- 7 

2 2 

Let 0 = arccos , so that cos 0 = . 

9 9 

Since arccos is defined only in quadrants I and II, 

and ~ is positive, 0 is in quadrant I. 

9 

Sketch 0 and label a triangle with the side 

opposite 0 equal to ~92 - 2 2 =.,fi7. 

y 

77. sin(sin-I~+tan-I(-3») 

Le 

. -I 1 1r < e: 1r 

t ml =sm 2' -'2 - ml ~'2. 

. 1 . . dr I 

sm ml = 2; CO) IS 10 qua ant . 

y 

.J3. 

cosml =-, smml 

2 

1 

= 2 

Let m2 = tan I(-3), - ~ < m2



78. cos(tan-I 2.-cot-I i) 

12 3 

-I 5 f3 -1 4 

Let a = tan -, = cot 

0 

12 3 

a and f3 are both in quadrant 1. 

y 

y 

y 

12 

--4--1,,-L--.. x 

12 

o 12 

SIn COl = 

13 

3 5 

__-1";......1;;_...1--_.. X COSC02 = 

4 

13 

3 5) 

0 

cos arCSIn - + arccos 

( 

o 5 12 

SIna=- cosa=- 

5 13 

13' 13 = cos(COl + co2) 

SIn 0f33 =- cos f34 = 

= cos COl cos CO2 - sin COl sin co2 

5' 5 

4 ) =(~XI53)-(~X~~) 

- I 5 -1 

{ 

12 3 20 36 16 

co tan --cot 

=---=- 

=cos(a- {3) 65 65 65 

= COS a cos f3 + sin asin {3 

=C~X~)+C53X~) 

80. 

63 

= 

tan(arccos ~ - arcSin(-~)) 

65 Leta = arccos ~, f3=arcSin(-~} 

79. coJ arcsin ~ + arccos2.) 

'\ 5 13 

 

0 3 TC TC 

Let col =arcSIn-, - - S COl S 

5 2 2 ° 2 

o 3 0 0 dr I 

SInCOI =-, col IS In qua ant 

5 

y 

0 

y 

y 

--k-~-=--.,....-'x 

4 

1 3 

tan a = - tan{3=- 

.[3 ' 4 

tan(arccos ~ - arcSin(-~)) 

4 

COSCOI = 

5 

o 3 

smcol = 

5 

5 

Let co2 =arccos-, 0 S CO S TCo 

13 

5 0 0 

COS co2 =-, co2 IS d I 

In qua rant 0 

13 

230



=tan(a- /3) 

tanatan/3 . 

= I+tana tan/3 

_ i-(-t) 

- I+(JJ)(-t) 

4+J.J3 

= 4.[j 

W-3 

W 

4+3$ 

= 413-3 

4+3.J3 -3-4$ 

=-3+4J3 ' -3-4$ 

-12 - 25$- 36 

= 

9-48 

-48-25$ 

= 

-39 

48+25$ 

= 

39 




81. cos(tan-1.5) 

Enter:5~ ~ ~ 

or§] 8 ~.5~ 

cosuan" .5) =.894427191 

82. sin(cos-1.25) . 

Enter: .25 ~ ~ ~ 

orE! 8 ~ .251§l 

sin(cos- 1 .25) =.9682458366 

83. tan (arcsin .12251014) 

Enter: .122510148 El ~ 

or~ 8 ~ .12251014 

§] 

tan (arcsin .12251014) = .1234399811 

84. cot(arccos.58236841) 

Enter: .58236841 ~ §J ~ \ 

or(~ ~ §] .58236841) 

~ ~ 

cot(arccos .58236841)=.716386406 

85. sin (arccos u) 

Let OJ = arccos u, o.s; OJ .s; n. 

u 

cosOJ=u= 1 

Ifu > 0, then OJ is in quadrant I. 

If u < 0, then OJ is in quadrant II. 

In either quadrant, sin OJ > o. 

y 

u 

y=~ 

Since y > 0, from the Pythagorean theorem, 

y =~I-u2. 

2 

~1- u r:--:;- 2 

sin (0 = =" 1-u 

1 

2 

Then sin (arccos u)= sin OJ= ~1 - u • 

86. tan (arccos u) 

n 

Let s =arccos u, 0 .s; s .s; 1C, s '# -. 

2 

Then coss=u=!!... 

1 

If u > 0, then s is in quadrant I. 

If u < 0, then s is in quadrant II. 

y 

y=~ y =~ 

< ) 

u 

uo 

In either quadrant, y > O. 

2 

By the Pythagorean theorem, y =~1- u . 

~1_u2 

tans =...:....:.--.:.:. 

u 

~1-u2 

tan(arccosu) = tans = ...:....:.--.:.:. 

u 

231



87. cot (arcsin u) 

Let CO = arcsin u, 2E. S; CO S; 2E.. CO ;t: O. 

2 2 

Then sin CO = u =!!... 

1 

Ifu > 0, then co is in quadrant I and 

cot CO > O. 

Ifu < 0, then co is in quadrant IV and 

cot co < O. 

89. sin(sec-I~) 

Let CO = sec-I!!.., 0 S; co S; TC, co;t: TC • 

2 2 

Since co is in quadrant I or II, sin CO > O. 

y 

u 

y=~ 

y ="';';-4 

u 

. ~ 

Sin co = lui 

88. 

~1_U2 

cot CO = -'-- 

u 

~1_u2 

Then cot( arcsin u) = cot CO = 

u 

cos (arcsin u) 

Let s = arcsinu, _!E. S; u S; TC • 

2 2 

Then sin s = u =!!... 

1 

Ifu > 0, then s is in quadrant I. 

If u < 0, then s is in quadrant IV. 

y 

. 

90. 

Th . ( -IU). ~ 

en Sin sec 2' =SID CO = Iu I . 

co{tan-I~) 

-I 3 TC TC 

Let s =tan - - - < S < -. 

u' 2 2 

Since s is in quadrants I or IV, cos s > O. 

y 

U,U>O 

----tE-+---:-.......--x 

---f-;+-~-+-_x 

U,U O. 

coss= = 2 

~u2 +9 u +9 

x =~I_u2 

~1_u2 ~ 

co { tan -13) -= ~~ 2 

coss= ="'II-u 

u u +9 

1 

cos(arcsinu) =coss = ~I- u 2 

232



Thus. 9 = arcsin .fi = 45°. 

91. tan(arcsin ~J 2 

. u 2 +2 The equation of the asymptote is 9 = 45°. 

. U Tr Tr 

Let a> =arcsm I •- - ~ a> ~ - . 

'Vu 2 +2 2 2 1 

Then sin a> = ~. 

2+2 

u 

Ifu > O. then a> is in quadrant I and tan a> > O. 

If u < O. then a> is in quadrant IV and a> < O. 

y 

..[.Tt2 

u 

(b) x=2. 9= tanu 

u 

u.Ji 

tan a> = .fi =-2 

96- 9 tan- (_x) 

x 2 +2 

(a) x= 1. 9= tan 

1 (+-) 

1 +2 

= tan-l(~) 

= 18° 

1 (+-) 

2 +2 

=tan-l(~) 

= 18° 

(c) x = 3. 9 = tan-1(-;-) 

. u J u u.fi 

3 +2 

tan arcsm = tan CO = """"E =- 

( ~u2 +2 ",2 2 1 31) 

= tan C 

= 15° 

92. 

1+ 1 2 

(d) tan(9+a)=-=x 

x 

1 

tana=x 

tan 9 + tan a 

93. All values in the interval [0. Tr] tan(9 + a)=---- 

I-tan9tana 

94. All values in the interval [-1. 1] 2 tan9+f 

;= I-tan9(f) .. 

42 

95. 2 

(a) 9 = arcsin /--."......--- 2 xtan9+1 

2(42 2) + 64(0) -=--- 

x x-tan9 

. 1 2(x - tan 9) =x(x tan 9+ l) 

= arcsm .Ji 

. .fi 

=arcsm 

2 

2x - 2 tan 9 =x 2 tan 9 + x 

2x - x =x 2 tan 9 + 2 tan 9 

x =tan9(x 2 +2) 

x 

tan9=-2 

y2 x +2 

(b) 9 = arcsin ./-...,.-- 

=arcsin IT V~ 

As y approaches 00. 

2y2 +64(6) 9 = tan- I (+-) 

x +2 

v 2 (l.fi 

2v 2 + 384 ="';2 =2"' 

233



(e) 

If we graph Yl =tan- I (+ )using a 

x +2 

graphing calculator, the maximum value of 

the function occurs when x is 1.4142 m. 

y = tan- 1 (x2 : 2) 

o 10 

99. Since the diameter of the earth is 7927 miles at 

the equator, the radius of the earth is 

3963.5 miles . Then 

cos 6 = 3963.5 = 3963.5 and 

20, ()()() + 3963.5 23,963.5 

6 =arcco{ 3963.5 ) "" 80.480 

23,963.5 

The percent of the equator that can be seen by the 

sate II ' . 26 00 2(80.48) 

=. 

44 701. 

ite rs -·1 = 

-10 

360 360 

· 1 

97. a= 2arcsmm 

Dgl:J:ll"~lS1 ' Y:;3390&Jl . 

-.5 

Radian mode 

(a) m= 1.2 

a = 2 arcsin(_I_) =113° 

1.2 

(b) m =1.5 

a = 2arcsin(_I_) =84° 

1.5 

(c) m = 2 

a = 2arcsin(i) = 60° 

(d) m =2.5 

a = 2 arcsin(_I_) =47° 

2.5 

98. Let a be the triangle to the "right" of 6 and let p 

be the angle to the "left" of 6. Then 

p+6+a = tr (since the angles sum to z radians) 

and 6= tr- P- a. 

150 (150) 

tan a =--:;-' so a =arctan --; 

tanp =~, so p=arctan( 75 ) 

l00-x 

l00-x 

As a result, 

6=tr-p-a 

=tr - arctan( 75 ) _ arctan( 150). 

l00-x 

x 

100. The amount of oil in the tank is 20 times the area 

shaded in the cross-sectional representation. The 

shaded area is the area of the sector of the circle 

defined by 26 (let 26 be the measure of the angle 

depicted in the figure) minus the area of the 

triangles. 

Section 6.2 

Exercises 

I. D 

2. C 

3. A 

4. B 

Since tan 6 = ../8 =../8, 6 =arctan../8 

1 

Using the Pythagorean theorem, we see that 

b 2 + 1 2 =3 2 

b 2 +1 =9 

b 2 =9-1 =8 

b =../8, where b is the base ofone triangle. 

The area of both triangles is 

A~ =2 ·ibh=2·i(../8~1)=../8. 

Then the sector area is 

As = .!-(26)r 2 = (arct~m ../8)(3)2 (with 6 in 

2 

radians). Thus, the volume of oil in the tank is 

V oil =20(A s - A~) 

= 20 (9 arctan(../8) -../8) 

= 20(11.08 - 2.83) 

"" 165 cubic ft 

234



5. From Example I . 

I 

cosO = -- 

2 

or cosO = 1 

0= 120°. 240° 0 = 0° 

Thus, all solutions can be represented as 

0= 360°·n. 120°+ 360° · n. 2400+360°·n 

where n is an integer. 

6. From Example 4. 

tan x = 1 or tan x = -2 

rr 5rr 

x="4'""4 x=2.0.5.2 

Thus. all solutions can be represented as 

rr 51r 

x = - + 2nrr. - + 2nrr. 2.0 + 2nrr•5.2 + 2nrr 

4 4 

where n is an integer. 

7. 2 cot x + 1= -1 

2cotx =-2 

cot x =-1 

3rr 7rr 

x=- 4 • 4 

8. sinx+2 = 3 

sinx = 1 

n 

x= 2 

9. 2sinx+3 =4 

2sinx = 1 

. 1 

smx= 

2 

n 5rr 

x=- - 6' 6 

1O.2secx+l=secx+3 

sec x = 2 

rr 5rr 

x=- 

3' 3 

11. tan 2 x+3=0 

2 

tan x =-3 

tanx=±H 

The square root of a negative number is not a real 

number. 

No solution. 

12. sec 2 x+2 =-1 

2 

sec x =-3 

secx=±H 

The square root of a negative number is not a real 

number. 

No solution 

13. (cot x -1)(..J3cot x + I) = 0 

cot x-I = 0 or ..J3cot x + I = 0 

cotx = I ..J3cotx =-1 

1r 51r I 

cot x = - ..J3 

x = '4' ""4 

..J3 

cotx=- 

3 

21r 51r 

x=- 

3 • 3 

1r 21r 51r 51r 

x=- - - 

4' 3 • 4 • 3 

14. (cscx+2)(cscx-.J2)=0 

esc x + 2 = 0 or esc x -.J2 = 0 

cscx =-2 cscx =.J2 

71r 1m 1r 31r 

x=- - x=- 

6' 6 4 ' 4 

1r 31r 71r 111r 

x=--- 

4' 4 • 6 ' 6 

15. cos 2 x+2cosx+l=O 

x= 1r 

(cos x + 1)2 = 0 

cosx+l=O 

cosx =-1 

16. 2cos 2 x -.J3cos x = O· 

cosx(2cosx-..J3) = 0 

cos x = 0 or 2cosx -..J3 = 0 

x=- 

1r 

- 

31r 

2cosx= 

..J3 

3 

2' 2 

..J3 

cosx= 

2 

x 111r 

x=- 

6' 6 

111r 

6 

17. -2sin 2 x = 3sinx + I 

2sin 2 x +3sinx + 1= 0 

(2sin x + I)(sinx + 1)= 0 

2 sin x + I = 0 or sin x + I = 0 

sin x =_.!. sinx=-I 

2 

71r l11r 

3rr 

x=- - x = 

6 ' 6 2 

71r 31r l11r 

x=- 

6' 2' 6 

235



18. tan 3 x = 3tanx 23. 2 sin 0 - I = esc 0 

tan 3 x -3tanx = 0 

tanx(tan 2 x 3) = 0 

tanx = 0 or tan 2 x -3 = 0 

2sinO-1 = _1_ 

sinO 

2sin 20-sinO=1 

x=O, n tan 2 2sin 2 O-sinO-1 = 0 

x=3 

(2sin 0 + I)(sin0 -I) = 0 

tanx = ±..f3 2sin0 + I = 0 or sin 0 -I = 0 

rr 2rr 4rr 5rr 

sinO = _! sinO = I 

x=3'3'3'3 

2 

0= 210°,330° 0 = 90° 

n 2rr 4rr 5rr 

0=90°, 210°, 330° 

x=O, 3' 3' n, 3' 3 

2 24. tan0 + I = ..f3+..f3 cot 0 

19. 2cos 4 x = cos x 

2cos 4 x-cos 2 x=O 

tan 0 + I = ..f3+ ..f3 

cos 2 x(2 cos2 x -I) = 0 or 2cos 2 x-I=O 

tan 0 

tan 2 0+ tan 0 =..f3 tanO+..f3 

cos 2 x =0 

cos 2 x=! 

2 tan 2 O+(I-..f3)tanO-.J3 = 0 

.J2 

cosx=O cosx=± 

2 

(tan0 - .J3}



28. sin 2 OcosO = cosO 

29. 

sin 2 0 cos 0 - cos 0 = 0 

cosO(sin 2 0 -1) = 0 

cosO=O or sin 20-1=0 

0=90°, 270° sin 20 = 1 

2tan 20sinO-tan20=0 

sinO= ±1 

IfsinO = I, 0 = 90°. 

IfsinO=-l, 0=90°, 270° 

tan 2 0(2sinO -1) = 0 

tan 

20=0 

or 2sinO-1=0 

tan 0 = 0 sin 0 = 1 

2 

0=0°, 180° 0 = 30°, 150° 

0=0°,30°,150°,180° 

30. sin 2 Ocos 2 0 = 0 

sin 

20=0 

or cos20=0 

sinO =0 

cosO = 0 

0=00,180° 0 = 90°,270° 

0=00,90°, 180°, 270° 

31. sec 2 otan 0 = 2 tan 0 

sec 2 0 tan 0 - 2 tan 0 = 0 

tan 8(sec 2 0 - 2) = 0 

tan 0 = 0 or sec 2 0 = 2 

secO=±.J2 

0=0°, 180° 0 =45°, 135°, 225°, 315° 

0= 0°, 45°,135°,180°,225°,315° 

20-sin20=0 

32. cos 

cos 2 0 - (1- cos 2 0) = 0 

2cos 2 0-1 = 0 

cos 2 0 = 1 

2 

cosO=±{f 

.J2 

cosO=± 

2 

.J2 

If cosO = -,0= 45°,315° 

2 

.J2 

If cosO = --,0= 135°,225° 

2 

0=45°, 135°, 225°, 315° 

33. sinO+cosO = 1 

(sinO + cosO)2 = 1 2 

sin 2 0+2sinOcosO+ cos 2 0 = 1 

1+ 2 sin 0 cos 0 = I 

2sinOcosO =0 

sinOcosO =0 

sin 0 = 0 or cos 0 = 0 

0=0°, 180 0 0 =90°, 270° 

Since the solution was found by squaring an 

equation, we must check that each proposed 

solution is a solution of the original equation . 

Since sin 180° + cos 180° = -1 and 

sin 270° + cos 270° = -1, 180° and 270° are not 

solutions. 

0=0°,90° 

34. sec 0 - tan 0 =1 

_1__ sinO = 1 

cosO cosO 

I-sinO =1 

cosO 

1-sin 0 = cos 0 

1-sin 0 = ±~""I---si-n-2-0 

(1-sinO)2 =(±~I-sin2 or 

1-2sinO+sin 2 0 = I-sin 2 0 

2 sin 2 0 - 2 sin 0 = 0 

2sin O(sin0 -1) =0 . 

sin 0 =0 or sin 0 =1 

0=0° 0=90° 



solution is a solution of the original equation. 

sec 90° - tan 90° is undefined. Thus, 90° is not a 

solution. 

0=0°. 

3~ 3~n20-~nO=2 

3sin 2 0 - sin 0 - 2 = 0 

(3sinO+ 2)(sinO -1) = 0 

3sin 0 + 2 =0 or sin 0 -1 = 0 

sin 0 = - ~ sin 0 = 1 

3 

0= 221.8°, 318.2° 0 = 90° 

0=90°,221.8°,318.2° 

237



.... 

36. 

37. 

38. 

39. 

40. 

2 tan 9 = 1 

3-tan 29 

2 tan 9 = 3 - tan 2 9 

tan 2 9+2 tan9- 3 = 0 

(tan 9 -1)(tan 9 + 3) = 0 

tan 9 -1 = 0 or tan 9 + 3 = 0 

tan 9 = 1 tan 9 =-3 

9 = 45°,225° 9 = -71.6° 

Since the solution -71.6° is not in the interval 

[0°,360°), we must use it as a reference angle to 

find solutions in the interval. 

9 = -71.6° + 180° = 108.4° 

9 = -71.6° + 360° = 288.4° 

9 = 45°, 108.4°, 225°, 288.4° 

sec 2 9 = 2tan9+4 

tan 2 9+ 1= 2tan9+4 

tan 2 9 -2tan9-3 = 0 

(tan 9 - 3)(tan9 + 1)= 0 

tan9-3=0 

or tan9+1=0 

tan 9 = 3 tan 9 = -1 

9 = 71.6°, 251.6° 9 = 135°, 315° 

9= 71.6°,135°,251.6°,315° 

5sec 2 0 = 6secO 

5sec 2 0 - 6secO = 0 

secO(5secO-6) = 0 

sec0 = 0 or sec9-6 = 0 

No solution sec0 = ~ 

5 

o= 33.6°, 326.4° 

9 = 33.6°, 326.4° 

3cot 2 0 = cot 9 

3cot 2 9 - cot 9 = 0 

cot 0(3cot 9-1) = 0 

cot 9 = 0 or 3cot 0 - 1= 0 

0=90°,270° 3cotO =1 

1 

cotO= 3 

9 = 71.6°, 251.6° 

0= 71.6°, 90°,251.6°,270° 

8cos9 = cot9 

cos 9 

8cosO=- 

sinO 

8cosOsin 9 = cosO 

8cosOsinO- cos 9 = 0 

cos 0(8 sin 0 -1) = 0 

cosO= 0 

or 8sinO-l=0 

0=90°,270° 8sin9 = 1 

. 9 1 

sm = 

8 

0= 7.2°, 172.8° 

0= 7.2°,90°,172.8°,270° 

41. 9 sin 2 0 - 6 sin 9 = 1 

9sin 2 0-6sin9-1 = 0 

We use the quadratic formula with a =9, b =~ , 

and c =-1. 

. 6 ±.J36- [4· 9(-1)] 

sm 0 =---'----- 

2·9 

6±m 

= 

18 

1±../2 

=- 3 

. 1+../2 

If sm 0 =--= .80473787, 9 = 53.6°, 126.4°. 

3 

If sinO= 1-../2 =-.13807119, 0=-7.9°. Since 

3 

this solution is not in the interval [0°, 360°), we 

must use it as a reference angle to find angles in· 

the interval. 

0= 180°+1-7.9°1 = 187.9° or 

0= (-7.9°)+360° = 352.1° 

9 = 53.6°, 126.4°, 187.9°, 352.1° 

42. 4cos 2 0+4cosO = 1 

4cos 2 9 +4cosO-l = 0 

We use the quadratic formula with a = 4, b = 4, 

and c =-1. 

-4±~42 -4(4)(-lf 

cosO = --~---- 

2(4) 

-4±.J32 

= 

8 

-4+4../2 

= 

8 

-1±../2 

= 

2 

-1 -../2 . I th 1 hi h i . ibl 

2 IS ess an - w lC IS an tmpossi e 

value for the cosine function . 

-1+../2 

So we use cosO = cosO= .20710678. 

2 

0"" 78.0°,282.0° 

238



43. tan 2 8 + 4 tan8 + 2 = 0 

We use the quadratic formula with a = I, b = 4, 

and c= 2. 

-4±"16-4·1·2 

tan 8 = ----:'--- 

2 ·1 

-4±.J8 

= 

2 

-4±2..fi 

= 

2 

=-2±..fi 

If tan 8 = -2 +..fi = -.5857864, 

8 = 149.6°,329.6°. 

If tan 8 = -2 -..fi = -3.4142136, 

8 = 106.3°,286.3°. 

8 = 106.3°, 149.6°, 286.3°, 329.6° 

44. 3cot 2 8 - 3cot8 -1 = 0 

We use the quadratic formula with a = 3, b = -3, 

and c =-1. 

-(-3)±~(-3)2-4(3)(-1) 

cot 8 = ---'---:...---!....:..-~-....:....:...:-.....:- 

2(3) 

3±.fil 

= 

6 

If cot = 3+.fil = 1.2637626, 8 = 38.4°,218.4°. 

6 

3-.fil 

If cot 8 = = -.26376262, 

6 

8 = 104.8°,284.8°. 

8 = 38.4°, 104.8°,218.4°,284.8° 

45. sin 2 8-2sin8+3 =0 

We use the quadratic formula with a = I, b = -2, 

and c= 3. 

sin 8 = 2 ± ..J4 - (4 ·1 .3) 

2 ·1 

2±.J=8 

= 

2 

Since .J=8 is not a real number, the equation has 

no real number solution. 

46. 2cos 2 8+ 2cos8-1 = 0 


and c =-1. 

_2±~22 -4(2)(-1) 

cos 8 = --"------'-..:....;.,.-..:.... 

2(2) 

- 2 ±.J1i 

= 

4 

-2±2.J3 

= 

4 

-1±.J3 

= 

2 

-1+.J3 ° ° 

If cos 8 = = .36602540,8 = 68.5 ,291.5 . 

2 

cos 8 = -1-.J3 = -1.3660254 is an impossible 

2 

value for the cosine function. 

8 = 68.5°, 291.5° 

47. cot 8 + 2 esc 8 = 3 

cos 8 +_2_=3 

sin8 sin8 

cos8+2 = 3sin8 

(cos8+2)2 = (3sin8)2 

cos 2 8+ 4cos8+4 = 9sin 2 8 

cos 2 8+4cos8+4 = 9(I-oos 2 8) 

cos 2 8+ 4cos8+4 = 9 -9cos 2 8 

lOcos 2 8+ 4cos8-5 = 0 


and c=-5. 

n -4± '4 2 -4(10)(-5) 

cos o = __V"--_---'-~'-:.. 

2(10) 

= -4±M 

20 

-4±6-J6 

= 

20 

-2 ±3..J6 

= 

10 

-2+3../6 

If cos8 = = .53484692, 

10 

8 = 57.7°,302.3°. 

-2-3..J6 

If cos 8 = - .93484692, 

10 

8 = 159.2°. 200.8°. 


equation. we must check that each proposed 

solution is a solution of the original equation. 

302.3° and 200.8° do not check. 

8= 57.7°,159.2° 

239



48. 2 sin 6 = 1 - 2 cos 6 

(2sin6)2 = (1- 2 cos 6)2 

4sin 2 6 = 1-4cos6+4cos 2 6 

4(I-cos 2 6) = 1-4cos6+4cos 2 6 

4-4cos 2 6 = 1-4cos6+4cos 2 6 

0= -3-4cos6+8cos 26 

We use the quadratic formula with a =8, b =-4, 

and c=-3. 

cos6 =b -(-4)±~(-4)2 -4(8)(-3) 

2(8) 

4±M 

= 

16 

4±.w7 

= 

16 

1±..[7 

=- 4 

If cos6 = 1+..[7 =.91143783,6=24.3°,335.7°. 

4 

1-..[7 

If cos6 = --= -.41143783, 

4 . 

6= 114.3°,245.7°. 

Sincethe solution was found by squaringan 


solutionis a solution ofthe original equation. 

24.3°and 245.7° do not check. 

6= 114.3°, 335.7° 

49. 2sin 2 x-sinx-I = 0 

(2sinx+ I)(sinx -I) = 0 

2sinx + 1= 0 or sin x -I = 0 

sin x = _! 

2 

7rr lIrr 

x=6'6 

rr 

x="2 

sin x = 1 

x = - 

rr 

+ 2nrr. 

7rr 

+ 2nrr. --+ 

llrr 2 

nrr.w 

h 

ere n 

. 

IS 

2 6 6 

an integer. 

51. 4cos 2 x-I=O 

cos 2 x=! 

4 

1 

cosx=± 

2 

rr 2rr 4rr 

x=-+2nrr, -+2nrr. -+2nrr. 

3 3 3 

- 5rr + 2 nn, w h ere n . . 

IS an integer. 

3 

52. 2cos 2 x+5cosx+2 =0 

(2cosx+ l)(cosx+2) = 0 

2cosx+I=0 or cosx+2=0 

1 

cosx=- x =-2 

2 

2rr 4rr 

x= 

3 • 

- No solution 

3 

2rr 

x=-+2n1C. "'3+ 

4rr 2 

n1C, 

h . . 

w ere n IS an integer. 

3 

53. cos 2 x+cosx-6=0 

(cosx+3)(cosx-2) =0 

cosx+3=0 or cosx-2=0 

cos x = -3 

No solution. 

cos x = 2 

54. sin 2 x - sin x =0 

sinx(sinx -I) = 0 . 

sinx =0 or sinx-I = 0 

x=O sinx=1 

1C 

x= 2 

1C .. 

x =n1C. "2 + 2n1C •where n is an integer. 

50. 2cos 2 x+cosx=1 

2cos 2 x+cosx-I=O 

(2cosx -I)(cosx + I) = 0 

2cosx-I=0 or cosx+I=O 

1 

cosx=- cosx=-I. 

2 

rr 5rr 

x=- - x=rr 

3' 3 

rr 

x =-+2nrr. rr+ 2nn; -+ 5rr 2 nx; were h n . 

IS an 

3 3 

integer. 

240


55. 59. P =A sin(21¢ + (D) 


rr 

(a) Letf= 261.63, A = .004, and (6=-. 

7 

Graph P =.004 sin[2rr(261.63)t + ~] on the 

interval 0, .005 . 

Forx=r, [ 7T] 

P(r) = .004 sin 27T(261.63)t + "7 

- .005 

(b) 0 =.004 sin[2rr(261.63)t + ~] 

56. The intersection method is shown on the window 0= sin(l643.87t + .45) 

Y2 = i 1 X 

1643.87t + .45 =ns; where n is an integer. 

nx -.45 

t=-- 

1643.87 

Ifn =0, then t =.000274. 

Ifn = I, then t = .00164. 

Ifn =2, then t =.00355. 

Ifn =3, then t =.00546. 

The only solutions for t in the interval 

[0, .005] are .00164 and .00355 . 

(e) 

We must solve the trigonometric equation 

P =0 to determine when P $; O. From the 

graph we can estimate that P$;O on the 

interval [.00164, .00355]: 

(d) P$;O implies that there is a decrease in 

pressure so an eardrum would be vibrating 

outward. 

60. Since fmeasures cycles per second and T 

measures seconds per cycle, 

57.2cosx-l=0 

2cosx =1 

f =..!.. or T =..!... For example, iff=5 cycles per 

1 

T f 

cos x = 

2 sec, then T =-!. sec per cycle. Thus, T is the time 

rr 51r 

5 

x=3"' ""3 

for one complete cycle, which is the period. 

58. 2 sin x -1 = 0 

2sinx = 1 

. 1 

SlOX= 

2 

n 5rr 

x="6' 6" 

241



16D 2 

61. .342DcosO + hcos 2 0 =- 63. e = 20sin( :t-~) 

Vo2 

Vo = 60. D = 80. h = 2 

(a) e =0 

2 16(80)2 

.342(80)cosO+2cos 0= 2 

60 

o= 20sin(:t -~) 

25_6 

2cos 20+27.36cosO-_ =0 Since arcsin 0 = 0, solve the following 

9 equation. 

cos2 0 + 13.68cosO __12_8 = 0 ret _ re =0 

9 4 2 

a = 1, b = 13.68. c = -128/9 ret re 

-= 

-13.68± 1(13.68)2 -4(~) 4 2 

cos 0 = --'V'--- -'---' t =2 sec 

2 

-13.68±15.6215 

= (b) e =1M 

2 

cosO = .97075 or cosO =-14.65075 1M = 20 sin( :t-~) 

-14.65075 does not lead to a solution because it is 

not in the range of cos O. 

If cosO = .97075.0 = 14°. 

.J3 = sin(!!.!.. _ re) 

2 4 2 

1 S· ...[j re I h J: 

mce arcsm - =-. so ve t e I . 

10 lowing 

62. V = cos 21l' t, 0 ~ t ~ . 2 3 

2 

equation. 

(a) V=0.cos21l't=0 ret 1l' re 

---= 

21l't = cos- t 0 

4 2 3 

ret lOre 

21l't =1l' 

-= 

4 12 

2 

10 1 

t=-sec=3- sec 

t=1:.. 3 3 

21l' 

1 

=-sec 

64. s(t) = sin t + 2 cos t 

4 

2+..[3 2 ..[j 

(b) V =.5, cos 21l't= .5 

(a) --=-+ 

2 2 2 

21l't =cos- 1 (.5) 

1l' 

=2 .'!'+ ..[j 

21l't =- 2 2 

3 

s: = 2 co{ ~ ) + sin( ~ ) 

t =-2 

2re 

1 

=- sec 

6 

(c) V = .25. cos 21l't = .25 

One such value is 1l'. 

3 

2ret = cos-1( .25) 

2ret = 1.3181161 

1.3181161 

t=--- 

21l' 

=.21 sec 

One such value is !!... 

4 

242


6.3 Trigonometric Equations II 243 

Section 6.3 

Exercises 

2tr 2tr Str 

1. x=6' 2' 6 

tr 4tr 

x = - 

3' 

tr 

, 

 

3 

2tr 10tr 10tr 

2. x=16' 12' -Str 

5tr 5tr 

x=g' 6'4 

IS0° 630° ' 720° 930° 

3. (}=-- -- -- - 

3' 3 ' 3 ' 3 

e= 60°, 210°, 240°, 310° 

03, 

4. ()=45°·3, 60° 75° ·3, 90°·3 

e= 135°, IS0°, 225°, 270° 

.J3 

5. cos2x= 

2 

Since 0 ~ x < 2tr, 0 ~ 2x < 4tr. 

2x = tr Iltr 13tr 23tr 

6' 6 ' 6' 6 

tr lItr 13tr 23tr 

x= 12 ' 12' 12' 12 

1 

6. cos2x=- 

2 

Since 0 ~ x < 2tr, 0 ~ 2x < 4n: 

2x = 2tr 4tr Str IOtr 

3' 3' 3 ' 3 

tr 2tr 4tr 5tr 

x="3' 3' 3' 3 

7. sin 3x =-1, 

Since 0 ~ x < 21C, 0 ~ 3x < 6tr. 

3x = 3tr 71C 1m 

2' 2 ' 2 

tr 7tr 1I1C 

x=2" 6' li 

8. sin 3x = 0 

Since 0 ~x < 21[, 0 ~ 3x < 6n: 

3x =0, n; 2tr, 3tr, 4tr, 5tr 

x =0 tr 2tr tr 4tr 5tr 

, 3' 3 ' , 3 ' 3 

9. 3tan3x =.J3, 0 s 3x < 6tr 

.J3 

tan3x= 

3 

tr 7tr 13tr 19tr 25tr 31tr 

h=6'6'li'(5'(5'li 


x=18' 18' 18' 18' 18' 18 

10. cot 3x = .J3, 0 < 3x < 6tr 

3x = tr 7tr 13tr 19tr 25tr 31tr 

6' 6' 6' 6 ' (5' li 


x = IS' 18' 18' 18' 18' 18 

11. .,fi cos 2x = -1, 0 s 2x < 4tr 

-1 .,fi 

cos2x=-=- 

.,fi 2 

2x = 3tr 5tr lItr 13tr 

4' 4' 4' 4 

3tr 5tr lItr 13tr 

x=g' g' -S-, "8 

12. 2.J3 sin 2x =.J3, 0 s 2x < 4tr 

. 2 1 

sm x= 2 

2x = tr Str 13tr 17tr 

6 ' 6' 6' 6 

tr 5tr 13tr 17tr 

x=12' 12' 12' 12 

0 x (;:;2 0 x 0 ",. !... < '" 

13. sm 2=-V ~ - sm 2' .::. 2 '" 

oX oX (;:;2 

sm- +sm - =-V ~ 

2 2 

2 sin!... = .J2 

2 . 

o x .,fi 

sm-= 

2 2 

x tr 3tr 

2=4"' 4 

tr 3tr 

x=2" 2 

243



14. 

15. 

16. 

17. 

18. 

 

sin x =sin 2x 

sin x = 2sin xcosx 

sin x - 2sin xcosx =0 

sinx(l-2 cos x) =0 

sin x = 0 or 1-2cos x = 0 

1 

x =0, n cos x = 

2 

n 5n 

x= 3 ' 3 

n 5n 

3 , 3 

x=O, -, n 

tan 4x =0, 0 ~ 4x < 8n 

4x =0, n, 2n, 3n, 4n, 5n, 6n, 7n 

tt n 3n 5n 3n 7n 

x=o, 4' 2' 4' n, 4'2'4 

cos2x -cosx = 0 

(2cos 2 x-l)-cosx=O 

2cos 2 x-cosx-l=O 

(2cosx+ 1)(cosx -1) = 0 

2cosx+l=0 or cosx-l=O 

1 

cos x = -- cosx = 1 

2 

2n 4n 

x=- - x=O 

3' 3 

x=O 2n 4n 

'3 3 

2 

8sec .::.=4 

2 

2 x 1 

sec -= 

2 2 

x ..[i 

sec-=± 

2 2 

No solution. 

. 2 

sm -- 

x 2 

= 

0 

2 

sin 2'::'= 2 

2 

sin'::'=±..[i 

2 

No solution. 

19. 

20. 

. x x 

Stn'2 = cos'2' 

. 2 x 2 x 

Sill -=cos 

2 2 

sin 2 '::' = l-sin 2 '::' 

2 2 

2sin 2'::'=1 

2 

, 2 x 1 

Sill -= 

2 2 

sinI=±~-±~ 

. x ..[i x n 3n n 3n 

If sm-=- -=- - and x=- -. 

22'24' 4' 2'2 

If sm, -x = --, 

..[i th ere are no so lutions unons in m the 

2 2 

interval [0, n). 

n 3n 

x=- 

2' 2 


equation, the proposed solutions must be checked. 

n 

x=-: 

2 

x 

n.J2 

. x . n ..[i 

sm-=sm-= 

2 4 2 

cos-=cos-= 

2 4 2 

3n 

x=-: 

2 

. x . 3n ..[i 

sm-=sm-= 

2 4 2 

x 3n ..[i 

cos-=cos-=- 

2 4 2 

The only solution is !E... 

2 

x x 

sec-=cos 

2 2' 

1 x 

--=cos- 

COS! 2 

2 

cos '::' =1 

2 

cos'::' =±1 

2 

~=O 

2 

x=O 

244



21. 

22. 

23. 

24. 

25. 

26. 

cos2x + cos x = 0, 0 ~ x < 21r 

2cos 2 x-l+cosx=O 

2cos 2 x+cosx -1 = 0 

(2cosx -1)(cosx+ 1)= 0 

2cosx-l=0 or cosx+l=O 

1 

cosx = - cosx =-1 

2 

1r 51r 

x=- - X=1r 

3' 3 

1r 51r 

x="3' x, 3 

. 1 

smxcosx= 

4 

. 1 

2 smxcosx= 

2 

sin2x=..!., 0~2x



32. sin26 = 2cos 2 6 

37. 0°::;26



vx-2 

-3 

41. 

42. 

43. 

In both cases, the value is approximately .7621. 

(a) 

For x = r, 

P( r) = .003 sin 2201rr + 

.003 sin 660lTr + 

3 

.003 sin 1100111 + 

5 

.003 sin 1540111 

'-__--.7~ 

.005 

--J 

44. (a) 3 beats r sec 

Forx = r, 

P(r) = .005 sin 4401rr 

.005sin 446'lTr 

'-------. 

+ 

-.01 

(b) 4 beats 

r sec 

(b) 

The graph is periodic, and the wave has 

"jagged square" tops and bottoms. 

+ 

(c) 

The eardrum is moving outward when P < O. 

This occurs for the time intervals 

.0045 < t < .0091; 

.0136 < t < .0182; 

.0227 < t < .0273. 

.15 

- .01 

(c) The number of beats is equal to the absolute 

value of the difference in the frequencies of 

the two tones . 

247



45. (a) For x = I , 

46. 

P(I) = -tsin [211'(220)1] + 

-lsin(211'(330)1] + 

3 

-lsin (211'(440)1] 

4 

(b) h assumes its least value when 

sm- . 2rr x tak. es on Its . 1 east vue, al w hi IC h IS . - 1. 

365 

. 2rr x 1 

sm--= 

365 

2rr x 3rr +2 . . 

365 = T - rr n, n IS an integer . 

x=e; ±2rrnX~~) 

= 3(365) ± 365n 

4 

= 273.75 ± 365 

-1 x = 273.75 means about 273.8 days after 

March 21, on December 19. 

(b) .0007576, .009847, .01894, .02803 

(c) Let h = 10. 

(c) 110 Hz 35 7. znx 

IO=-+-sm- 

3 3 365 

(d) For x = I, 

P(I) = sin(211'(110)1] + 

30 = 35 + 7 sin 2rr x 

-tsin(21T(220)1] + 

365 

-tsin (21T(330)I] + 

- 5 = 7 sm- · zxx 

365 

1- sin[211'(440)1] 

4 

5 . 2rr x 

--=sm- 

7 365 

- .71428571 = sin(2rr x) 

365 

In quadrant III and IV, sine is negative. 


2rrx . 

-= rr + .79560295 

365 

=3.9371956 

35 7. 2rrx 

x = 365 (3.9371956) 

h=-+-sm- 

2rr . 

3 3 365 =228.7. 

x = 228.7 means 228.7 days after March 21, 

(a) Find x such that h = 14. 

on November 4. 

14= 35 +2 sin 2rrx In quadrant IV, 

3 3 365 

21tt = 2rr - .79560295 

14- 35 =2 sin 2rrx 365 

3 3 365 = 5.4875823 

7 7. 2rr x 

-=-sm- 

x = 365 (5.4875823) 

3 3 365 

2rr 

. 2rr x 1 

= 318.8 

SIO--= 

365 

x = 318.8 means about 318.8 days after 

2rr x rr + 2 . . 

March 21, on February 2. 

-= _ rr n, n IS an integer. 

365 2 

x= (~ ±2rrnX~~) 

= 365 ±365n 

4 

= 91.25 ±365n 

x =91.25 means about 91.3 days after 

March 21, on June 20. 

47. 

i = l max sin21ift 

Let i = 40, lmax =100, f =60. 

40 =l00sin2rr(6O)(t) 

40 =l00sinl20rrt 

.4 =sin 120rrt 

248


6.4 Equatiom Involving Inverse Trigonometric Functions 249 

From a calculator, 

S2. The error in the solution occurs in going from 

1201rt =.4115168 tan 20 2 

--=- to tanO=!. 

.4115168 

t=--2 2 

1201r tan 20 

--¢ tan 0 because tan 20 ¢ 2 tan O. 

t = .0010916 2 

t:: .00 1sec 

Sedion6.4 

48, i =Imax sin 21!ft 

Let i = 50, lmax =100, f = 120. 

Exercises 

50 = 100 sin 2401rt 

1. C; arcsin 0 = 0 

sin 2401rt = 1. 

2 1r 

2. A; arctan 1 = 

2401rt =!E. 

4 

6 

1 

t=- 

1440 

3. C', arcco{-..fi) = 31r 

2 4 

t =.0007 sec 

49. i = lmaxsin 21!ft 

Let i = lmax' f = 60. 

lmax = lmaxsin 21r(6O)t 

1= sin 1201rt 

S. y=5cosx 

I=cosx 

1201rt =!E. 5 

2 

4. C· arcsin(-1.) = -1r 

, 2 6 

x=arccos I 

1201 =1. 5 

2 

1 

t=- 

6. 4y=sinx 

240 x=arcsin 4y 

t =.004 sec 

7. 2y =cot3x 

SO. i =lmaxsin 21!ft 

3x = arccot 2y 

Let i =1.lmax' f= 60. 

1 

2 

x = -arccot2y 

3 

~ lmax = lmax sin 21r(6O)t 

1 

8. 6y=-secx 

1. = sin 1201rt 

2 

2 12y =sec x 

1201rt =!E. 

x = arcsec 12y 

6 

1 9. y=3tan2x 

t= 

720 

I=tan2x 

t = .0014 sec 3 

2x=arctan I 

. 1 1 . 

SIIf • sm-x=--, we must wnte 

3 

22 . 1 y 

x =-arCtan 

1 1 

-x =210 2 3 

0+n·36O" or -x = 330 0+n '360 0 , 

2 2 

where n is any integer. Then multiplying by 2, we 

have x = 420 0 + n . 720 0 or x = 660 0 + n ·7 20° . 

Since n E {... , - 3, - 2. -1, 0, 1, 2, 3, ...}, 

there is no value of n which will give values ofx 

in the interval [0 0,3600 ) . 

249



10. 

11. 

12. 

13. 

14. 

15. 

x 

3 . y = sm 

2 

1:'.=sin~ 

3 2 

z, arcsinl. 

2 3 

x = 2 arcsin 1:'. 

3 

x 

y =6cos 

4 

z= cos~ 

6 4 

z, arccosl. 

4 6 

x = 4 arccosl. 

6 

. X 

y=-sm 

3 

. x 

sm-=-y 

3 

~ =arcsin(-y) 

3 

x = 3arcsin(-y) 

y= -2cos5x 

-1:'.=cos5x 

2 

5x = arccos(- ~) 

x = ~arccos(-~) 

y =3cot5x 

cot5x =l. 

3 

5x = arccotl. 

3 

1 y 

x =-arccot 

5 3 

y = cos(x + 3) 

x + 3 = arccosy 

x = -3+ arccosy 

17. y=sinx-2 

y+2 = sinx 

x = arcsin(y + 2) 

18. y = cotx+ 1 

cotx = y-l 

x = arccot(y -1) 

19. y=2sinx -4 

y+4 = 2sin x 

y+4 . 

--=smx 

2 

20. 

x = arcsin( y ; 4 ) 

y=4+3cosx 

y-4 =3cosx 

y-4 

--=cosx 

3 

x=arcco{y~4) 

4 -I Y 

23. -cos -=1r 

3 4 

-IY 31r 

cos -= 

4 4 

Y 31r 

-=cos 

4 4 

.J2 

y 

-=- 

4 2 

y=-2.J2 

24. 41r+ 4tan- 1 y = 1r 

4 tan-I y = -31r 

-I 31r 

tan y=- 

4 

The range of tan-I y is _!E. < y < 1r . 

2 2 

Since - 31r is not in this interval, the equation 

4 

has no solution. 

25. 2 arcco{ y ~ 1r) = 21r 

16. 

y = tan(2x -1) 

2x - 1= arctan y 

2x = 1+ arctany 

1 

x = -(1 + arctany) 

2 

arccos( y ~ 1r) = 1r 

y-1r 

--=COS1r 

3 

y-1r 

--=-1 

3 

y-1r =-3 

y=1r-3 

250



 

26. arcco{y- ~) =: 

_I . -1 3 

29. cos x =sm 

5 

1r 1r 

y--=cos- 

. -1 3 . 3 

Let sm - = u , so SinU = -. 

3 6 5 5 

.J3 

u is in quadrant I. 

1r 

y=-+-' 

y 

2 3 

3.J3 +21r 

y= 

6 

3 

. 3 

27. arcsin x = arctan ---F-......---...... x 

4 4 

Let arctan- 

3 

= u, so tanu = -, 

3 .. dr I 

U IS In qua ant. 

4 4 

Sketch a triangle and label it. The hypotenuse is 4 

cosu = 

~32+42 5 

=5. 

The equation becomes 

y 

3 

cos-I x = U. or x = cosu. 

4 

x= 5 

_I _1 4 

30. cot x = tan 

--I'-.....__.......--x 

3 

4 

_14 4 

Let tan - = u, so tan u = - . 

3 3 

. 3 3 y 

smu=-=r 

5 

This equation becomes arcsin x = u, or x = sin u. 

3 

x= 

5 

4 

---I'-"-..............---x 

5 3 

28. arctan x = arccos 

13 

5 5 

Let arccos - = u, so cos u =-. 3 

13 13 

cotu = 

4 

Sketch a triangle and label it. The side opposite 

angle u is ~132 - 52 = 12. . -I -11 1r 

31. sm x-tan =- 

y 4 

-----f~::_I---- x 

. -I -11 1r 

sm x=tan - 

4 

. _I 1r 1r 

SIn 

12 4 4 

sin-I x = 0 

12 

tanu= 

5 

The equation becomes arctan x = u, or x =tan u. 

12 

x= 5 

x=-- 

sinO = x 

x=O 

251



, -1 -1 h3 21r 

32. Sin x+tan v-' = 

3 

, -1 1r 21r 

Sin x+-= 

3 3 

, -1 1r 

Sin x= 3 

. 1r 

x 

X=SIn 

3 

sinu =~I_x2 

.J3 

. 1r 1r. 

x=2x = SIn--COSU-COS-SInU 

2 

6 6 

,-./3 

33. arccos x + 2 arcsin - = 1r 

2 

. -./3 

arccos x = 1r- 2 arCSIn 

2 

arccosx =1r- z( ~) 

21r 

arccosx = 1r- 

3 

9x 2 =3-3x 2 

1r 

12x 2 =3 

arccosx= 

3 2 3 1 

x =-= 

1r 

12 4 

x=cos 3 

1 

x= 

2 

, -./3 1r 

34. arccos x + 2 arCSin - = 

2 3 

arccosx+z( ~) =~ 

1r 

arccosx=- 

3 

- ~ is not in the range of arccos x, Therefore, the 

equation has no solution. 

1 -./3 ~ 

2x=-x--(vl-x-) 

2 2 

4x = x--./3~I-x2 

3x =-J3~I- x 2 

(3x)2 =(-J3~I- x 2 )2 

2) 

9x 2 =3(1- x 

1 

x=± 

2 

Check these proposed solutions since they were 

found by squaring an equation. 

1 

x=-: 

2 

il) 

.( I) 1r 1r 1r 

arcsin 2'2 +arcco'\2 ='2+3'*6' 

1 

x=--: 

2 

arcsi .{ 2 (-'2 I)) +arccos (-2-I ) =-'2+'3=6' 

tt 21r 1r 

The only solution is _.!.. 

2 

35• arcsin ' 2 x + arccos x = 1r 

6 

arcsin -2 x = -1r 

- arccos x 

6 

2x = Sin(: - arccos x ) 

Use the identity 

sin(A - B) sin Acos B-cosAsin B. 

2x = sin 1rcos(arccosx)- cos 1rsin(arccosx) 

6 6 

Let U =arccos x. 

x 

cosu=x= 

1 

36• arcsm 

. 2x 

+ arcsm 

. 

x = 

1r 

2 

arcsin 

- 2 

x = - 

1r 

- arcsin 

. 

x 

2 

2x = sin( ~ - arcsin x) 


sin (A - B) = sin A cos B - cosA sin B. 

. 1r ( in r) 1r . ( - ) 

2x = Sin- cos arcsm x - cos - Sin arcsin x 

2 2 

= 1cos(arcsin x) - 0 

= cos(arcsin x) 

252



. . x 

Let arcsm x = U, so Sin U = X = -. 

1 

y 

Let U = tan-I x so tan U = x. 

x 

COSU =~1_x2 

2 

Substitute ~1- x 

2x=~I-x2 

4x 2 = l-x 2 

5x 2 = 1 

for cos(arcsin x). 

2 1 

x = 

5 

..[5 

x=± 

5 

Check these proposed solutions since they were 

found by squaring an equation. 

..[5 

x=-: 

5 

arcsin . (2 ..[5) . (.J5) x 

..[5 

x=--: 

5 

's +arCSIn "'"5 =2' 

arcsin(2· ~) + arCSi{- '7) = - ~ 

· . .J5 

TheonIy so Iution IS -. 

5 

_I -I x 

37. cos x+tan x= 

2 

-I 1r -I 

cos x == - - tan x 

2 

x=co{~ -tan-I x) 


cos(A - B) = cos A cos B + sin A sin B. 

x =cos~cos (tan-I x)+sin 1r sin (tan-I x) 

2 2 

COSU= 

. 

SInU= 

r-1 

vl+x 2 

x 

r-1 

vl+x 2 1r . 1r • 

x =cos-,cosU+SIn- 'SInU 

2 2 

X={b)+lQ 

x 

X=-.=== 

~1+x2 

xx ==0 

~1+x2 

JI- 1 ]=0 

l ~1+x2 

1 

x=Oor 1 

~1+x2 

1 

~1+x2 

~1+x2 

=0 

-1 

- 

=I 

l+x 2 =1 

38. sin-I x+tan- I x=O 

x 2 =0 

x=O 

sin-I x==-tan- I x 

sin(sin-I x) = sin(- tan-I x) 

x=-sin(tan- I x) 

x 

x=---;=== 

~x2 +1 

x2=~ 

x 2 +1 

2(x 2 

x + 1) == x 

2 

x 2(x2 +1)-x 2 =0 

x 2[(x2 +1)-1]=0 

2(x2 

x ) == 0 

x 4 =0 

x==O 

253



39. 41. Y I = (arctan x)3 - x + 2 

-3 

x =4.4622037 

40. 

42. 

x =2.2824135 

43. A =~(AI COSiPI + AZcostPz)2 + (AI siniPI + AZ sin(j)z)2 

an d '" 

(_A.!..Is_i_n..:,.tPI!...+_A.:. 2_si_n...:.tPz~) 

'I' =arctan 

Al cos 41 + Az cos tPz 

Make sure your calculator is in radian mode. 

(a) Let Al =.0012, iPI =.052, Az =.004, and tPz =.61. 

A =~(.OO12cos.052+ .OO4cos.61)2 +(.0012sin.052+ .004-sin .61)2 

=.00506 

'" ( .0012sin .052 + .004sin .61) 

'I' =arctan --------- 

.0012cos.052+ .OO4cos.61 

=.484 

P =A sin(21¢ + 41), let! =220. 

P =.00506sin(440m + .484) 

(b) 

Forx=t, 

P( t) = .00506 sine4401rt + .484) 

P1(t) + P 2 (t ) = .OOI2sin(440m + .052) + 

.004 sin(440m + .61) 

~------. 

The two graphs are the same. 

254



44. A = ~(AI cost/>} + A2COS¢2)2 + (AI sin if>! + A2 sin ¢2)2 

A. 

d 

(AI sin ¢I + A2 sin ¢2 ) 

Al cos¢+ A2 cos¢2 

an 'I' = arctan ~-.........--=------:..= 

Make sure your calculator is in radian mode. 

1r 

(a) Let AI = .0025, if>! = 7' A2 = .001, ¢2 = 6' and!= 300. 

A = ~(AI cos¢! + A2 COS¢2)2 +(AI sin¢1 + A2 sin¢2)2 

= [.0025CO{ ~)+.OOlCO{:)r+ [.0025sin( ~)+.OOlsin(:)r 

"'.0035 

.0025sin(~)+.001Sin(t)] 

¢ = arctan ---+'-i-----7'--'; 

[ .0025 cos(.,) + .001co~f) 

'" .470 

P = Asin(21¢ + ¢) 

P = .0035 sin(6001rt + .47). 

1r 

(b) 

Forx = t, 

per) = .0035 sin(6OOm + .47) 

PI(r) + P2(t) = .0025 Sin(6OO1I1 + f) + 

.001 sin(600111 + f) 

45. (a) 

(b) 

x 

(e) (x+ y)tana = x tan {3 

tana=z 

x tan {3 

tana=- 

x+Y 

tan{3=- 

x+y 

z 

a = arctan(Xlan{3) 

x . a x+y 

x+y 

tana= tanJJ=- 

z 

z 

z tan a =x ztan{3=x+y 

(d) xtan{3=(x+y)tana 

z = x z=- x+y 

tan{3= (x+ y)tana 

tan a tan{3 x 

x x+y 

--=- {3 = arctan[(x + y~tana] 

tana 

tan{3 

255



46. (a) u = arcsinx Divide by ·Ji 

. 1t 1t 

x =smu - - :s; u :s; - 

x 2 -3~+4=O 

, 2 2 ,----- 

3../3 ±~(_3../3)2 -4(1)(4) 

(b) Y x= 

2 

3../3±.JIT 

x= 

2 

x = 4.26 ft or .94 ft 

(ii) Let ()= rr . 

8 

rr -I 4 -I 1 

-=tan --tan 

x x~l-x2 8 x X 

(c) tanu= = 2 

"Jl-x 2 I-x tan-=tan tan --tan 

8 x x 

From part (a), 

[ I-x 8 x 2+4 

rr (-I 4 -I 1) 

(d) x~ ] rr 3x 

u =arctan 2 tan-=--. 

.414=~ 

47. (a) e = Emax sin21ift x 2+4 

_e_ = sin21ift 

E max 

.414x 2 + 1.656= 3x 

.414x 

2 -3x+1.656=O 

21fft = arcsin _e_ 

E max 

3± "J9 - 4(.414)(1.656) 

x= 

1 . e 

2(.414) 

t=-arcsm- 

21r/ E max 3±2.502 

x'" 

.828 

(b) Let E max = 12, e = 5, and!= 100. 

x '" 6.64 ft or .60 ft 

1 . 5 

t= arcsm 

2rr(I00) 12 

n -1 4 _I 1 

() c 17 = tan --tan 

x x 

t = _I-arcsin .4166667 = .00068 sec 

200rr (i) Let x =4. 

n -1 4 -I 

17=tan --tan 

1 

6 

-II -I 1 

7t _I 4 -I 1 6 = tan -tan 

- = tan - - tan - 4 

6 x x 

1t 

6 =--.245 

tan- 

7t 

= tan (-14 tan --tan ~ - 1) 

4 

6 x x 6= .54 

48~ (b)(i) Let 6 =!:.. 4 4 

../3 ~_.l 

x 

_= x 

(ii) Let x =3. 

3 1+A.·.l 

x x -1 4 -I 1 

6 

../3 

=tan --tan 

_=_x_ 

1. 3 3 

3 1+4 

6 = .927- .322 

x 

6=.60 

../3 ~ 

3"= ijA 

x 

../3 3x 

""3= x 2+4 

../3x 2 + 4../3 = 9x 

../3x 2 -9x+4../3 =0 

256



1 . 4m 

49• y = - sm 

3 3 

50. 

() a 

. 4m 

3y=sm 

3 

4m . 3 

3=arcsm y 

4m = 3arcsin 3y 

3 . 

t=-arcsm 3 y 

41r 

(b) If y = .3 radian, 

t=-arcsm 

3 . 

. 9 

41r 

t= .27sec. 

6. y = arcco{-~) 

1 

cosy =-2"' 0:5: y:5: 1r 

2n 

y=3" 

7. y =tan- 1 (-J3) 

M x 1r 

tany =-v3 --


258 Chapter 6 Inverse Trigonometric Functions and T rigonometric Equations 

1 

14. 6 =arccos 

2 

1 

cos 6 = -, 0° ~ 6 ~ 180° 

2 

6=60° 

15. 6 =arCSin(- ~) 

sin 6 = - .J3 , - 90° ~ 6 ~ 90° 

2 

8=~0 

16. 6 =tan- I 0 

tan 6 =0, _90° < 6 < 90° 

8=0° 




17. 6 = arctan 1.7804675 

Enter 1.7804675 ~ ~ 

or~ ~ 1.7804675~ 

Display: 60. 67924514 

6 = 60.67924514° 

18. 6 = sin- I (-.66045320) 

Enter: -.66045320 8 ~ 

or~ ~ c=J .66045320 

~ 

Display: -41.33444556 

6 == -41.33444556° 

.-' 

19. 6 = cos- I .80396577 

Enter: .80396577 ~ §) 

or ~ §) .80396577 ~ 

Display: 36.4895081 

6 = 36.4895081 ° 

20. 6 = coe l 4.5046388 

Enter: 4.5046388 §)~ ~ 

or 8 §) 4.5045388 ~ ~ 

Display: 12.51631252 

6 = 12.51631252° 

21. 6 = arc see 3.4723155 

Enter: 3.4723155 §) ~ §) 

or ~ §J 3.4723155 ~ ~ 

Display: 73.26220613 

6 = 73.26220613° 

22. 6 = csc- 1 7.4890096 

Enter: 7.4890096 §)8 ~ 

or ~ ~ 7.4890096 ~ ~ 

Display: 7.673567973 

6 = 7.673567973° 

25. The domain of the arccot function is (-eo, 00)since 

the range of the cotangent function is (-eo, 00). 

26. The range of the function y =arc sec x as defined 

in this text is [0, ~) u(~ ,n] or [0, n], y'# ~ . 

. ( . _I 1) 

27. sm sm 2 

Le . -I 1 

t (O=sm 2' 

Then si 1 n n 

en sm(O=-, --~(O~-. 

2 2 2 

. ( . 

sm sm 21). -I =sm (0 =21 

29. cos (arccos (-1» 

Let (0 = arccos (-1) 

Then, cos (0 = 1, 0 ~ x ~ n , 

costarccost-J) = CO!! (0 =-1 

30. sin[arcsin(- ~J) =- ~ 

31. arcco{cos 3:)=x 

3n 

cos x =cOS-, 0 ~ x ~ n, 

4 

3n 

so x=-. 

( 4 3n) 3n 

arccos cos4 =4 

32. arcsec(sec n) 

seen =-1 

arcsec(sec n) =arcsec(-1) =n 

258


1C 

sox=-. 

4 

tan-I(tan =) == 

36. cos (arctan 3) 


Let ()= arctan 3. Then tan () = 3 = ~; 0 is in 

1C 1C 1C 

I 

tan x =tan- --< x


260 Chapter 6 Inverse Trigonometric Functions arid Trigonometric Equations 

Let 0 =sin-'(-~} 

Then sin 0 = _.!.; 0 is in quadrant IV. 

3 

The side adjacent to 0 =~32 - (_1)2 

=.J8 =2.fi 

y 

----loo:=--"*""--r- x 

-, 

2.fi 

cosO =- 3 

sec( 2sin-'(-~)) =sec 20 

. 3 TC TC 

Let (0\ =arcsm -, - - ~ co, ~ -. 

5 2 2 

sinco, =~. The side adjacent to COt is 

39. tan(arcsin ~ + arccos %) 

~52 _3 2 =4. 

y 

1 

=- 

cos20 

y 

1 

= 2 

2ef) -1 

---l'-...:...__L....~ x 

1 

=-'6 

9-1 

1 

='7 

2../6 

"9 

9 

= 7 

tan (0, ='4 3 5 

Let (02 =arccos -, 0 s co2 s TC. 

7 

5 

cos(02 =-. 

7 

The side opposite (02 is ~72 - 52 =.,fi4 =2../6 . 

tan co2 =- 5 

tan (0, + tanco2 

Use tan(co, + (02) = . 

1- tan co, tan co2 

3 5) i+ 2 ./6 

tan(arcsin - + arccos - = -71£) 

5 7 1-(iJ\.¥ 

15+8./6 

_ -----zo 

- 20-6'J6 

-W 

15+ 8../6 20 + 6.J6 

= 20- 6.J6 ' 20+ 6../6 

588+250../6 

= 

184 

294+125../6 

= 

92 

260



40. sinuan"! u) 

-I tr tr 

Let s =tan u, - - < S 0, then s is in quadrant I. 

Ifu < 0, then s is in quadrant IV. 

hypotenuse =~u 2 + 1 

y 

~U2 +1 

Then sees = . 

u 

If u > 0, s is in quadrant I. 

If u < 0, s is in quadrant II.· 

The side opposite s, which is positive in both 

quadrants, is ~(u2 + I) - u 2 =I. 

y 

U, u>o 

--++~--_ x 

u, u



 

45. y = arccot x 51. tan 2 2x -1 = 0, 0 ~2x < 41C 

Domain: (-00,00) 

Range: (0, 1C) 

tan 2 2x = 1 

y tan2x = ±1 

1C 51C 

Iftan 2x = 1, 2x =-, 

91C 131C 

y =arccot x 

4 4' 4' 4 

1C 51C 91C 131C 

x=- 

8 ' 8' 8' 8 

31C 71C lI1C 151C 

---+---:-1---+--_ x If tan 2x = -1, 2x =-, -- , 

-1 0 4 4' 4 4 

31C 71C lI1C 151C 

x=- -- , 

8 ' 8 ' 8 8 

46. sin 2 x = 1, 0 ~ x < 21C 1C 31C 51C 71C 91C 1I1C 131C 151C 

x=-, - , - , 

sin x =±1 8 8' 8' 8' 8' 8 8 8 

1C 31C 

x= x x 

2' 2 52. sec-= cos 

2 2' 

47. 2tanx-l = 0 1 x 

--=cos 

2tanx=1 cost 2 

1 

tanx= cos 2 ::'= 1 

2 2 

x = .463647609or x = 3.605240263 

cos::' = ±1 

2 

48. 3sin 2 x -5sinx +2 = 0 x 

(3sinx -2)(sinx -1) = 0 If cos- = 1, 

2 

. 2 . 1 

::'=0 

3 2 

1C 

x = .7297276562, 2.411864997 or x = 

x=O. 

2 

Sin X =- or Sin x = 

If cos::' =-1 

2 ' 

49. tan x = cot x, 0 ~ x < 21C 

::. = 1C which is not in the domain. 

Use the identity cot x = _1_. 

2 ' 

tan x 

The only solution is O. 

1 

tanx=--, tan e O 

tan x 

53. cos2x +cosx =0, 0 ~ x < 21C 

tan 2 x = 1 

2cos 2 x-l+cosx=O 

tan x =±1 

2cos 2 x+cosx-l=O 

1C 51C 

If tanx=l, X=-, 

(2cosx -1)(cosx +1)=0 

4 4 2cosx-l =0 or cosx+l =0 

31C 71C 1 

If tan x =-1, x=- - . cosx =- cosx =-1 

. 4' 4 2 

1C 31C 51C 71C 1C 51C 

x=- - x=1C 

x="4' 4' 4' 4 3' 3 

50. sec 4 2x =4, 0 ~ 2x < 41C 

22x=2 

sec 

sec2x=±.,fi 

2x =1C 31C 51C 71C 91C lI1C 131C 151C 

4' 4' 4' 4' 4' 4' 4'4 

1C 31C 51C 71C 91C lI1C 131C 151C 

x=s' 8' 8' 8' 8' 8' -8-"-8 

1C 51C 

x=- 3' 

1C , 

 

3 

262



-/.. ' 

54. 4sinxcosx =.J3 

2 . .J3 

smxcosx = 2 

sin2x =.J3 0::; 2x < 41r 

2 ' 

A value for 6 will be a solution if 

sin 26 - cos26 = 1. 

6=45°,26=90° 

sin 90° - cos90° = 1-0 = I 

6 =90°, 26 = 180° 

sin 180°- cos180° = 0 - (-I) = I 

6 = 135°, 26 = 270° 

sin2700-cos270° = -1-0* I 

2x = 1r 21r 71r 81r 

3' 3' 3' 3 

1r 1r 71r 41r 

6 = 225°, 26 = 450° 

x= - - - 6' 3' 6' 3 sin4500-cos450° = 1-0 = I 

6 = 270°, 26 = 540° 

55. sin 26+3sin6+2=0,00::;6



62. cos8-cos28 = 2cos8 

cos8 - (2cos 2 8 -1) = 2cos8 

2cos 28+cos8-1 =0 

(2cos8 -1)(cos8 + 1)= 0 

2cos8 -1 = 0 or cos 8 + 1= 0 

cos8 =..!.. cos8 =-1 

2 

8 = 60°, 300° 8 = 180° 

8 = 60°, 180°, 300° 

63. 4y =2sinx 

2y=sinx 

x =arcsin2y 

. 2 

68. arccosx = arcsm-« 

7 

. 2 

Let u =arcsm-. 

7 

Then sinu = ~, u is in quadrant I. 

7 

The side adjacent to u is 

~72 _2 2 =.[45 = J$. 

y 

x 

64. y=3cos 2 

z.=cos~ 

3 2 

z, arccosZ. 

2 3 

x = 2arccosZ. 

3 

65. 2y =tan(3x + 2) 

3x + 2 = arctan2y 

3x = arctan2y - 2 

1 

x =-(arctan2y-2) 

3 

x =(~arctan2Y)-~ 

66. 5y = 4sinx-3 

5y+3 . 

--=SlnX 

4 

. 5y+3 

x=arcsln- 

4 

4 x 

67. -arctan- = TC 

3 2 

X 3TC 

arctan-= 

2 4 

But, by definition, the range of arctan is 

(- ~, ~) . So, this equation has no solution. 

3..[5 

cosu=- 

7 

This equationbecomes arccos x = u, or x = cos u, 

J./5 

x=- 

7 

lITC 

69. arccosx + arctan1= 

12 

lITC 

arccosx = - - arctan 1 

12 

11K TC 

arccos x = - - 

. 12 4 

lITC 3TC 8TC 

arccosx=---= 

12 12 12 

2TC 

arccosx= 

3 

2TC 

cos-=x 

3 

1 

x=- 2 

.( -- 

.J2) 3TC 

70. arccotx = arcsin + 

. 2 4 

TC 3TC 

arccotx = - - + 

4 4 

TC 

arccotx= 

2 

TC 

x=cot-=O 

2 

264



Tr 

71. d = 550+450cos-t 

50 

Tr 

450cos-t = d -550 

50 

Tr d -550 

cos-t= 

50 450 

-t= x arccos (d-550) 

50 450 

t=-arcco 50 {d-550) 

Tr 450 

CI sin 0 

74. -=- 1 

c2 sinO 2 

.752 = sin 01 

sinO 2 

If O 2 =90°, 

.752 = sin01 

sin90° 

sin0 

=- 1 

1 

= sin0 1 

0 1 = 48.8°. 

75. If 0 1 > 48.8°, then O 2 > 90° and the light beam is 

completely underwater. 

76. L = 6077- 31cos20 

(a) Solve the equation for O. 

L - 6077 = -31cos20 

L-6077 

cos 20 =-- 

-31 

20 = cos-I(L-6077) 

-31 

O=-cos 1 _1(L-6077) 

2 -31 

O=-cos 

1 _1(6074-6077) 

2 -31 

0:= 42.2° 

73. a. Let a be the angle to the left of O. 

b. 

5+10 

Then tan(a + 0) =- 

x 

a+O = arctan(~) 

o= arctan(~)-a 

e= arctan( ~) - arctan(;) 

(b) Substitute L = 6108 in the equation from 

part (a). 

1 _1(6108-6077) 

O=-cos 

2 -31 

0=90° 

(c) Substitute L = 6080.2 in the equation from 

part (a). 

O=-cos 

1 _1(6080.2-6077) 

2 -31 

0:= 48.0° 

77. To see the graph, enter the function 

YI = «x > 0) - (x < 0» -(~ - tan-I ~x 2 -1). 

The maximum occurs at x = 8.6602567. 

-"2 7T 

Radian mode 

265



78. (a) 

(e) 

y=csc-1(2../3) -3 

2../3 rr rr 

cscy=-- --:5y:5- y;tO 

3 ' 2 2' 

n 

y=3" 

(b) In both cases, sin- I (.4) =:: .41151685 


1. 

y 

(I. f) 

+--+----'-tf---t----t- x 

(I) y =cot- I (../3) 

t: rr rr 

cot y ='" 3, - - :5Y :5-, y;t 0 

2 2 

rr 

y= 6 

. 2 . 2 

3. (a) Le t arcsm- = u, so smu =-. 

3 3 

Then, co{arcsin ~) =cos u =~ 

y 

Domain: [-I, 1] 

Range: [-;, ;] 

2. (a) 

(b) 

y =arcco{-~) 

y=cos -I(-'2 

I) 

1 

cosy=--,O:5y:5rr 

2 

2rr 

y= 3 

y=sin-t-v;) 

. ../3 rr rr 

smY=-T' -"2:5 y:5"2 

x 

y=- 3 

(b) Let arccosej) = U,so cosu =-j. 

Then Si{2 cos-I -j) =sin 2u 

y 

=2sinucosu 

_ ' -(2~XI) -2- 

3 3 

4~ 

=- 9 

" 

(c) y = tan-I (0) 

rr rr 

tany=O, --


4• Le t x=arcsmu, 

. th 

en smx= 

. u 

1 

tan(arcsinu) = tan x 

y 

u ~ 

=~1-u2 . ~1-u2 

-s> 

-----i'--==.......... x 

5. sin 2 0 = cos 2 0 + 1 

7. cosx =cos2x 

- l-u 2 1 

cosx =- 

2 

2cosx =-1 

21C 41C 

x= - 

3' 3 

8. 


cosx=2cos 2 x-I 

2 cos 2 x cos x-I = 0 

(2cosx + 1)(cosx -1) =0 

2 cos x + 1=0 or cos x -1 = 0 

x =0 21C 41C 

, 3' 3 

2~ sin(~) = 3 

.(0) 3 

sm"2 = 2~ 

x=O 

e . ( 

sin 20-cos20=1 3 ) 

"2=arcsm 2~ 

-cos20= 1 

cos20=-1 

!!- = 60° 120° 

2 ' 

20 = 180° + n36O° 

e = 120°, 240° 

0= 90° + 180 0n 

0=90°, 270° 

9. (a) y =cos(3x) 

(since 0 must be in the interval [0,360°» 

3x = arccosy 

6. csc 2 x = arccosy 

0 - 2 cot 0 = 4 

3 

1+ cot 2 0 - 2 cot 0 = 4 

cot 2 0 - 2 cot- 3 = 0 

(cot 0 -3)(ootO+ 1)= 0 

cotO-3=0 or cotO+ 1=0 

cot 0 =3 

O=tan-I(i) 

0=18.4°, 198.4° 

0=18.4°, 135°, 198.4°, 

cot 0 =-1 

0= tan-I(-l) 

0= 135°, 315° 

315° 

(b) 

y 

cosx = 1 

Let (tJ = arctan(~). Then 

arcsin x = arctan(~) 

arcsin x = (() 

. 4 

x =sm{() = 5 

tan (() = ~ . 

___-+::::....--'-L--_ x 

4 

267



n-(t- ~) =arccos(O) 

1r ( t - 

'31) ='21r + nn, were here n iIS an integer 

1 1 

t--=-+n 

3 2 

1 1 

t=-+-+n 

2 3 

5 

t=-+n 

6 

In the interval [0, 1r), n =O. I, and 2 provide valid 

values for t. 

5 5 5 

t=- -+1 -+2 

6' 6 • 6 

5 11 17 

t=-sec -sec -sec 

6 '6 '6 

268



CHAPTER 7 APPLICATIONS OF 

5. A = 37°, B = 48°, C = 18 m 

TRIGONOMETRY AND VECTORS 

C=1800-A-~ 

Section 7.1 

C = 180°- 37°-48° 

C=95° 


b C 

--=- 

sinB sinC 

1. House: 

b= 18sin48° 

X = (7400-150)(.9) "" 1131.8 ft 

sin95° 

H 6sec4.1 

b=13 m 

-3.5sin4.1 0 a C 

y = (7400 - 150)(3.5)cos 4.1° z 4390.2 ft --=- 

0-3.5sin4.10 

H 6sec4.1 

sinA sinC 

Fire: 

18sin37° 

a=-- 

X = (7400-690)(2.1) "" 2277.5 ft 

sin95° 

F 

a=l1 m 

6see 4.1° + 2.4sin 4.1° 

Y = (7400-690)(-2.4)eos4.1° z-2596.2 ft 6. B=52°, C=29°, a=43em 

F 6sec4.1 0+2.4sin4.10 

A=1800-B-C 

A = 180°-52° -29° =99° 

2. d=~(XF-XH)2+(YF-YH)2 a b 

--=- 

sinA sinB 

= ~(2277.5 -1131.8)2 + (-2596.2 - 4390.2)2 

b = 43sin52° 

z 7079.7 ft 

sin 99° 

b=34 em 

Exercises 

a C 

--=- 

1. C 

sinA sin C 

43 c 

--=- 

2. C,D sin 99° sin 29° 

43sin29° 

3. The measure of angle C is c =-__.--:-" 

sin 99° _ 

180°- (60° + 75°) = 45°. 

C = 21 em 

a c 

--=- 

sinA sinC 

7. A = 27.2°, C = 115.5°, c = 76.0 ft 

a .fi 

B=1800-A-C 

--=-- B = 180°-27.2°-115.5° 

sin60° sin 45° 

B= 37.3° 

.fisin 60° 

a="";""'--- a c 

sin 45° --=- 

sinA sinC 

.fi(~) 76.0 sin 27.2° 

a=---- 

a = .J2 

sin 115.5° 

T 

a =38.5 ft 

a =..[3 b c 

--=- 

sin B sinC 

4. The measure of angle B is b =76.0sin37.3° 

180°- (45° + 105°) = 30°. sin 115.5° 

a b 

--=- 

b =51.0 ft 

sinA sinb 

a 10 

--=- 

sin45° sin 30° 

lOsin45° 

a=-- 

sin 30° 

1O(~) 

a=-r 

a =10.J2 

269


270 Chapter 7 Applications of Trigonometry and Vectors 

8. C=124.1°, B=18.7°,b=94.6m 

0-B-C 

A=180 

A = 180 0 -18.7 0 -124.1 0 = 37.2 0 

a b 

--=- 

sinA sin B 

94.6 sin 37.2 0 

B = 18.5 0 

a b 

--=- 

a=----sinA 

sinB 

sinI8.7° 

a =178 m 

c b 

--=- 

a=239 yd 

sinC sinB 

b c 

94.6 sin 124.1° 

--=- 

c= 

sinB 

sinI8.7° 

c=244 m 

9. A = 68.41 0 , B = 54.23 0 , a = 12.75ft 

0-A-B 

C=180 

C= 180 0-68.410-54.230 

C= 57.36 0 

a b 

--=- 

sinA sin B 

b = 12.75sin54.23° 

sin68.41° 

b =11.13 ft 

a c 

--=- 

sin A sinC 

12.75sin57.36° 

sin68.41° 

c = 11.55 ft 

10. C = 74.08 0 , B = 69.38 0 , c = 45.38 m 

A =180 0-B-C 

A = 180 0 - 69.38 0 -74.08 0 = 36.54 0 

a c 

--=- 

sinA sinC 

45.38sin 36.54 0 

11. A = 87.2 0 , b = 75.9 yd, C = 74.3 0 

0-A-C 

B= 180 

B = 180 0 - 87.2 0 -74.3 0 

12. B = 38 040', 

a = 19.7 em, C =91 040' 

A =180 0-B-C 

= 180 0-38040' -91 040' 

=49 040' 

b a 

--=- 

sin B sinA 

b = 19.7sin 38 0 40' 

sin 49 040' 

b=16.1em 

c=---- 

c a 

--=- 

sinC 

A = 180 0 - 20 0 50' -103 0 10' 

a=-----A =56 sin 74.08 0 00' 

0 

a=28.lOm 

AC AB 

--=- 

b c 

sinB sinC 

--=- 

sin B sinC 

AB = 132sin 103 0 10' 

b = 45.38sin 69.38 0 

sin 20 050' 

sin 74.08 

AB= 361 ft 

0 

b =44.17 m 

BC AC 

--=- 

sinA sinB 

BC = 132sin 56 000' 

75.9sin87.2° 

a=---- 

sin 18.5 

0 

sinC 

75.9sin 74.3 0 

c=---- 

sin 18.5 

0 

c=230 yd 

sinA 

19.7sin 91 040' 

c=---- 

sin 49 

040' 

c=25.8 em 

13. B = 20 0 SO', AC = 132 ft. C = 103 010' 

A =180 0-B-C 

sin 20 050' 

BC= 308 ft 

270



14. A =35.3°, B =52.8°, AC =675 ft 

C=1800-A-B 

=180°- 35.3° -52.8° =91.9° 

BC AC 

--=- 

sinA sinB 

BC= 675 sin 35.3° 

sin 52.8° 

BC = 490 ft 

AB AC 

--=- 

sinC sinB 

AB= 675 sin 91.9° 

sin 52.8° 

AB=847 ft 

15. A =39.70 0,C =30.35°, b =39.74 m 

B= 180 0-A-C 

B =180°- 39.70° - 30.35° 

B = 110.0° (rounded) 

a b 

--=- 

sinA sinB 

39.74 sin 39.70° 

sin 109.95° 

a =27.01 m 

b c 

--=- 

sinB sinC 

39.74 sin 30.35° 

c= 

sin 109.95° 

c=21.36 m 

16. C =71.83°, B =42.57°, a =2.614 em 

A =180 0-B-C 

A =180°- 42.57° - 71.83° =65.60" 

b a 

C=97°34' 

--=- 

sinB sinA 

a b 

b = 2.614sin42.57° 

--=- 

sinA sinB 

sin 65.6QO 

b = 268.7 sin 42° 32' 

b=1.942 em 

sin 39°54' 

c a 

--=- 

b=283.2 m 

sinC sinA 

a c 

2.614sin 71.83° 

--=- 

c=-----sinA 

sinC 

sin 65.60° 

c=2.727 em 

17. B =42.88°, C =102.40°, b =3974 ft 

A = 180°- B-C 

A = 180° - 42.88° -102.40° 

A = 34.72° 

a b 

--=- 

sinA sinB 

3974 sin 34.72° 

a=---- 

sin 42.88° 

a =3326 ft 

b c 

--=- 

sin B sinC 

3974 sin 102.40° 

c=----- 

sin42.88° 

c =5704 ft 

18. A = 18.75°, B =51.53°, c = 2798 yd 

C=1800-A-B 

C =180°-18.75°-51.53° = 109.72° 

a c 

--=- 

sinA sinC 

2798 sin 18.75° 

a=---- 

a=---- 

sin 109.72° 

a = 955.4 yd 

b c 

--=- 

sinB sinC 

b = 2798sin51.53° 

sin 109.72° 

b=2327 yd 

19. A =39° 54', a =268.7 m, B = 42° 32' 

C=1800-A-B 

C = 180°-39°54' -42°32' 

268.7sin97°34' 

c=----- 

sin 39° 54' 

c= 415.2 m 

271



20, C=79°18',c=39.81mm, 26, T 

A =32°57' 

B=1800-A-C 

B =180° - 32°57' -79°18' 

=67°45' 

a c 

--=- 

sinA sinC 

39.81 sin 32°57' 

a= 

sin 79° 18' 

s 

a=22.04mm 

b c 

S = 180° - 32°50' -102°20' 

--=-- = 44°50' 

sinB sinC 

b = 39.81sin67°45' 

RS 582 

= 

sin 79°18' 

sin 32°50' sin 44°50' 

b=37.50 mm 

RS=448 yd 

22, In Example 1, the Pythagorean theorem does not 27. Let C =the transmitter. 

apply because triangle ABC is not a right triangle. 

N N 

Angle A is 32.0°, B =81.8°, therefore C = 66.2°, t t 

One of the angles has to be 90° for ABC to be a I c I 

I 

I 

right triangle. 

I 

I 

147.7' 

1 302S 

24, No. If a is twice as long as b, then sin A =2 sin B, I 

I 

but A is not necessarily twice as large as B. 

A 3.46mi B 

25. c 

Since side AB is on an east-west line. the angle 

between it and any north-south line is 90°. 

A =90° - 47.7° = 42.3° 

B =302.5° - 270° = 32.5° 

C=1800-A-B 

=180° - 42.3° - 32.5° 

C=105.2° 

AC 3.46 

---=--- 

sin 32.5° sin 105.2° 

AC =3.46sin32.5° 

sin 105.2° 

AC= 1.93 mi 

A =180 0-B-C 

A = 180° -112°10'-15°20' 

A =52°30' 

BC AB 

--=- 

sinA sinC 

AB =354 sin 15°20' 

sin 52°30' 

AB= 118 m 

Led d =the distance the ship traveled between 

the two observations; 

L =the location of the lighthouse. 

N 

44.2' 

d 

L 

12.5 

38.8' 

L = 180° - 38.8° - 44.2° 

L=97° 

d 12.5 

--= 

sin 97° sin 44.2° 

d=17.8km 

272



29. 

x 12.0 

32. Label the centers of the atoms A, B, and C . 

C 2+3 

sin 54.8° sin 70.4° 

12.0sin54.8° 

x=------: 

sin 70.4° 

x =10.4 in. 

30. Let M = Mark's location; 

L =Lisa's location; 

T =the tree's location; 

R =a point across the river from Mark so that 

MR is the distance across the river. 

N 

115.45' 

428.3rn 

45.47' 

M - R 

8 a T 

L 

(J = 180° -115.45° = 64.55° 

a =180° -45.47°-64.55° = 69.98° 

f3 =(J =64 .55° Alternate interior angles 

In triangle MTL, 

MT 428.3 

= 

sin 45.47° sin 69.98° 

MT = 324.9645. 

In right triangle MTR, 

MR . R. 

-=SlnfJ 

MT 

MR =sin 64.550 

324.9645 

MR =293.4 m. 

31. Label a in the triangle as shown. 

1.6 

.." 1.6 + 3.6 =-- 

sin38° sina 

. (,Ic;6_+-:;2:.... .7"-)s:..:;i:..:;n.:..38=-° 

sma= 

1.6+3.6 

a=31° 

(J= 180 0-38°-a 

= 180°-38°-31° 

(J=11l0 

A 

a = 2.0+ 3.0 = 5.0 

c = 3.0+4.5 = 7.5 

sinC sinA 

--=- 

c a 

. C 7.5sin 18° 

sm =--- 

5 

C=28° 

B = 180° -18° - 28° = 134° 

b a 

--=- 

sinB sinA 

b 5.0sin134° 1 

----= 2 

sin 18° 

The distance between the centers of atoms A and 

Cis 12. 

33. Let x = the distance to the lighthouse at bearing 

N 37° E. 

y = the distance to the lighthouse at bearing 

N25°E. 

N 

Lighthouse 

2.5 mi 

(J =180° - 37° =143° 

a = 180 0-(J-25° 

= 180° -143° - 25° =12° 

2.5 x 

--=-- 

sina sin 25° 

2.5sin 25° 

x=--- 

sinl2° 

=5.1 mi 

2.5 y 

--=- 

sina 

sin(J 

2.5sin 143° 

y= 

sin 12° 

=7.2mi 

273



34. Let A = the location of the balloon; 

B = the location of the farther town; 

C = the location of the closer town. 

A 

~ 

B 1.5 C 

Angle ABC = 31° and angle ACB = 35° because 

the angles of depression are alternate interior 

angles with the angles of the triangle. 

AngleBAC= 180°-31°-35°= 114° 

sin BAC sin ACB 

---=-- 

1.5 AB 

AB = 1.5sin 35° 

sinl14° 

AB=.94 

sin ABC = AD 

AB 

AD = .94(sin 31°) 

AD=.49 

The balloon is .49 mi above the ground . 

35. c 

A 

Angle C is equal to the difference between the 

angles ofelevation. 

C = B - A = 52.7430° - 52.6997° = .0433° 

The distance BC to the moon can be determined 

using the law of sines. 

BC AB 

--=- 

sinA sinC 

BC= ABsinA 

sinC 

398 sin 52.6997° 

= 

sin.0433° 

== 418,930 km 

Ifone finds distance AC, then 

AC= ABsinB 

sinC 

398sin(180° - 52.7430°) 

= 

sin .0433° 

==419,171 km. 

In either case the distance is approximately 

419,000 km compared to the actual value of 

406,oookm. 

36. 

We must find the length of CD. 

Angle BAC is equal to 35° - 30° = 5°. 

Side AB = 5000 ft. Since triangle ABC is a right 

triangle, 

5000 

cos5°=- 

AC 

AC= 5000 

cos5° 

== 5019 ft. 

The angular coverage of the lens is 60°, so angle 

CAD = 60°. From geometry, angle ACB = 85°, 

angle ACD = 95°, and angle ADC =25°. We now 

know three angles and one side in triangle ACD 

and can use the law of sines to solve for the 

length of CD. 

CD AC 

---=-- 

sin 60° sin 25° 

CD =5019sin60° 

sin 25° 

CD == 10,285 ft 

The photograph would cover a horizontal distance 

. '85 ft 10,285 1 95 . 

of approximately 10,2 or ---==. rm, 

5280 

37. A 

Determine the length of CB which is the sum of 

CD and DB. Triangle ACE is isosceles so angles 

ACE and AEC are both equal to 47°. It follows 

that angle ACD is equal to 47° - 5° =42°. Angles 

CAD and DAB both equal 43° since the 

photograph is taken with no tilt and AD is the 

bisector of angle CAB. The length of AD is 

3500 ft. Since the measures of the angles in 

triangle ABC total 180°, angle ABC equals 

180° - (86° + 42°) = 52° . 

D 

274



Using the law of sines, 

CD= 3500 sin 43° "" 3567.3 ft 

sin 42° 

and 

DB = 3500sin 43° "" 3029.1 ft. 

sin52° 

The ground distance in the resulting photograph 

will equal the length of segment CB. 

CB=CD+DB 

=3567.3+3029.1 

CB = 6596.4 ft 

38. A 

c 

In the figure, angle CAB is 72°. 

Triangle ACE is isosceles so angles ACE and 

AEC are both equal to 54°. It follows that angle 

ACB is equal to 54° - 5° = 49° . Angles CAD and 

DAB both equal 36° since the photograph is taken 

with no tilt and AD is the bisector of angle CAB 

with length 3500 ft. Since the measures of the 

angles in triangle ABC total 180°, angle ABC 

equals 180° - (72° + 49°) = 59°. 

Using the law of sines, 

CD = 3500 sin 36° ""2725 .9 ft 

sin 49° 

and 

DB= 3500 sin 36° "" 2400.1 ft. 

sin 59° 

The ground distance in the resulting photograph 

will equal the length of segment CB. 

CB=CD+DB 

= 2725.9 + 2400.1 

CB = 5126 ft 

The coverage is less with a longer focal length. 

Cameras that have shorter focal lengths have a 

greater ground coverage in aerial photographs. 

39. Using K =!bh, 

2 

K =!(l)(.J3) 

2 

.J3 

= 

2 

E 

Using K = !absin C, 

2 

K = t(.J3)(l) sin 90° 

=!(.J3)(l)(I) 

2 

.J3 

= 

2 

40. Using K=!bh, 

2 

K=!(2)(.J3) 

2 

=.J3. 


2 

K = ! (2)(2)sin 60° 

2 

. = t(2)(2{ ~) 

=.J3. 

41. Using K =!bh, 

2 

K = !(l)(.J2) 

2 

.J2 

= 

2 

Using K =! ab sin C, 

2 

K = !(2)(I)sin45° 

2 

=~(2)(l)( v;) 

.J2 

= 

2 

42. Using K=!bh, 

2 

K =!(2)(1) 

2 

=1. 


2 

K = !(2)(.J2)(sin45°) 

2 

= ~(2)(.J2{ v;) 

=1. 

275


276 Chapter 7 Appl i ~~ations of Trigonometry and Vectors 

43. A = 42.5°, b = 13.6 rn, e =10.1 m 

Angle A is included between sides band e. 

50. Area =.!. (52.1)(21.3) sin 42.2° 

2 

2 

Area =.!. be sin A 

= 373 m 

2 

=.!.(13.6)(10.1) sin 42.5° 

51. The function y = sin x is increasing from y = 0 to 

2 

2 

Y =1 on the interval 0 s x s ~ . 

=46.4 m 

2 

44. C = 72.2°, b = 43.8 ft, a = 35.1 ft 

a b 

Angle C is included between sides a and b. 

53. --=- 

sin A sinB 

Area =.!. ab sin C 

Solve for b. 

2 

asinB =b 

= .!.(35.1)(43.8) sin 72.2° 

sin A 

2 

b=asinB 

= 732 ft2 

sin A 

45. B = 124.5°, a = 30.4 ern, e = 28.4 em 

. asinB 

Angle B is included between sides a and e. 

54. From Exercise 51, b =--. 

sin A 

Area =.!. ac sin B 

sinB 

sinB 

2 

If--< 1, then b =a .--< a ·1 or a. 

sin A 

sin A 

= .!.(30.4)(28.4) sin 124.5° 

So, b



58. (b) If A =18° and B =36° Another possible value for B is 

Section 7.2 

Exercises 

1. A 

2. D 

A. =[5 sin 18°sin 36°]R2 

sin(18 + 36)° 

2 

'" 1.12257R 

(c) i. 11.4 in. x (10 + 13) in. =8.77 in. 2 

ii. A =5J5 sin 18°sin 36° }.308)2 

'l sin(l8 0+360 ) 

A =5.32 in. 2 

iii. red 

3. The vertical distance from the point (3, 4) to the 

x-axis is 4. 

(a) Ifh is more than 4, two triangles can be 

drawn. But h must be less than 5 for both 

triangles to be on the positive x-axis. 

So, 4 < h 5, then only one triangle is possible on 

the positive x-axis. 

(c) If h < 4, then no triangle is possible, since the 

side of length h would not reach the x-axis. 

4. (a) Since the side must be drawn to the positive 

z-axis, no value ofh would produce two 

triangles. 

(b) Since the distance from the point (-3, 4) to 

the origin is 5, any value of h greater than 5 

would produce exactly one triangle. 

(c) Likewise, any value of h less than or equal to 

5 would produce no triangle. 

180° - 31.2 0 =148.8°. This measure, combined 

with the angle of 95°, is too large, however. 

Therefore, only one triangle is possible. 

6. b =60, a = 82, B =100° 

b a 

--=- 

sinB sinA 

60 82 

=- 

sin 100° sin A 

. A 82 sin 100° 

Sin =--- 

60 

sin A = 1.346 

Since this is impossible, no triangle can be drawn 

with these parts. 

7. a = 31, b = 26, B = 48° 

a b 

--=- 

sinA sin B 

31 26 

--=-- 

sinA sin 48° 

. A 31sin48° 

Sin =--- 

26 

sin A = .886 

A =62.4° 

Another possible value for A is 

180 0 - 62.4° = 117.6°. Therefore, two triangles 

are possible. 

8. a = 35, b = 30, A = 40° 

a b 

--=- 

sinA sinB 

35 30 

---=- 

sin 40° sinB 

. B 30sin40° 

Sin =--- 

35 

sinB =.551 

B-=33.4° 

Another possible value for B is 

180° - 33.4° =146.6°, but this is too large to be 

in a triangle that also has a 40° angle. Therefore, 

only one triangle is possible. 

9. a = 50, b = 61, A = 58° 

a b 

--=- 

S. a =50, b =26, A =95° 

sinA sinB 

a b 

50 61 

--=- 

---=- 

sinA sinB 

sin 58° sin B 

50 26 

. B 61sin58° 

---=- 

sm =--- 

sin 95° sinB 

50 

. . 26 sin 95° 

sin B -= 1.03 

smB=--- 

50 

sinB '" .518 

B = 31.2° 

sin B > 1 is impossible. Therefore, no triangle is 

possible for the given parts. 

277



10. B=54°,c=28,b=23 

b c 

--=- 

sinB sinC 

23 28 

--=- 

sin 54° sinC 

. C 28sin54° 

Sin =--- 

23 

sinC:::: .985 

C::::80° 

Another possible value for C is 

180° - 80° =100°. Therefore, two triangles are 

possible. 

b 

sin B 

a 

sinA 

11. --=- 

2 -J6 

--=- 

sinB sin 60° 

. B 2 sin 60° 

sm = -J6 

2(4) 

= -J6 

~ 

= 

2 

B=45° 

Another possible value for B is 180° - 45 = 135°, 

but this is too large since A = 60°. Therefore, 

B = 45°. 

12. A =45°, a=~, b=3 

b a 

--=- 

sinB sinA 

3 ~ 

--=- 

sinB sin 45° 

. B 3sin45° 

Sin = ~ 

~4) 

=~ 

1 

= 

2 

B=3O° 

Another possible value for B is 

180° - 30° =150°, but this is too large since 

A = 45°. Therefore, B = 30°. 

In Exercises 13-30, the number of possible triangles is 

found based on the given conditions. Remember that, for 

example, if angle B and sides a and b are given such that 

a> b > a sin B, then two triangles are possible. 

13. A = 29.7°, b =41.5 ft, a = 27.2 ft 

b > a > b sin A tells us that there are two possible 

triangles. 

sin B sinA 

--=- 

b 

a 

· B 41.5sin 29.7° 

Sin =---- 

27.2 

sin B = .75593878 

Bt=49.1° 

C) = 180° - A - B) 

= 180°-29.7-49.1° 

= 101.2° 

~ = 180° - Bt = 130.9° 

Cz = 19.4° 

14. B = 48.2° , a =890 em, b = 697 ern 

a > b > a sin B tells us that there are two possible 

triangles. 

sinA sinB 

--=- 

a b 

sin A = 890sin 48.2° = .95189905 

697 

A) =72.2° 

C) = 180°- B-A) 

= 180° - 48.2° - 72.2° 

=59.6° 

A 2 = 180°-A) =107.8° 

C 2 = 180°- B-Az 

= 180° - 48.2° -107.8° 

=24.0° 

15. C =41°20', b = 25.9 m, C = 38.4 m 

c > b tells us that there is only one triangle. 

sinB sinC 

--=- 

b c 

· B 25.9sin 41°20' 

Sin =----- 

38.4 

sin B =.44545209 

B =26°30' 

Note that B,* 153°30' because 

153°30'+C> 180°. 

A = 180°-26°30' -41°20' 

A = 112°10' 

16. B =48° 50', a = 3850 in., b = 4730 in. 

Since b > a, only one triangle exists. 

sinA sinB 

--=- 

a b 

· A 3850sin48°50' 

510 =----- 

4730 

= .61274255 

A =37°50' 

278



A ~ 142° 10' since 142 010' + B> 180 0 

C=1800-A-B 

= 180° - 37° 50' - 48°50' 

=93°20' 

17. B=74.3°,a=859m,b=783m 

sinA sinB 

--=- 

a b 

· A 859 sin 74.3° 

Sin =---- 

783 

sin A =1.0561331 

sin A > 1 is impossible. 

Note that b < a sin B. 

. No such triangle exists. 

18. C = 82.2°, a = 10.9 km, C = 7.62 km 

sinA sinC 

--=- 

a C 

· A 10.9 sin 82.2° 

Sin =---- 

7.62 

= 1.4172115 

sin A > 1 is impossible. 

Note that c < a sin C. 

No such triangle exists. 

19. A = 142.13°, b = 5.432 ft. a = 7.297 ft 

a > b tells us that there is one triangle. 

sinB sinA 

--=- 

b a 

· B 5.432 sin 142.13° 

Sin =----- 

7.297 

sin B = .45697580 

B =27.19° 

B must be acute since A is obtuse. 

C = 180°-142.13° -27.19° 

C=1O.68° 

20. B = 113.72°. a = 189.6 yd. b = 243.8 yd 

Since b > a, only one triangle exists. 

sinA sin B 

--=- 

a b 

189.6sin 113.72° 

sin A 

243.8 

= .71198940 

A =45.40° 

A is acute since B is obtuse. 

C=1800-A-B 

= 1800- 45.40° -113.72° 

=20.88° 

21. A = 42.5°, a = 15.6 ft, b = 8.14 ft 

a > b tells us that there is only one triangle. 

sinB sinA 

--=- 

b a 

· B bsinA 

Sin =- a 

8.14 sin 42.5° 

= 

15.6 

sin B =.35251951 

B=20.6° 

B ¢.159.4° since 159.4°+ A > 180°. 

C =180° - 42.5° - 20.6° 

=116.9° 

c a 

--=- 

sinC sinA 

asinC 

c=- 

sin A 

15.6sin 116.9° 

= 

sin42'so 

c= 20.6 ft 

22. C = 52 .3°, a = 32.5 yd, c = 59.8 yd 

Since c > a, only one triangle exists. 

sinA sinC 

--=- 

a c 

· A 32.5 sin 52.3° 

Sin = 

59.8 

= .43001279 

A = 25.5° 

A ~ 154.5° since 154.5° of: C > 180°. 

B = 180° - 25.5° - 52.3° 

=102.2° 

b c 

--=- 

sin B sinC 

b = 59.8sin102.2° 

sin 52.3° 

= 73.9 yd 

23. B = 72.2°, b = 78.3 m, c = 145 m 

sinC sinB 

--=- 

c b 

· C csinB 

Sin =- b 

145 sin 72.2° 

= 

78.3 

sin C = 1.7632026 

sin C > 1 is impossible. Note that c sin B > b. 


279



24. C = 68.5°, C = 258 em, b = 386 em 

sin B sinC 

--=- 

b C 

. B 386 sin 68.5° 

sm =---- 

258 

= 1.39202008 

sin B > 1 is impossible. Note that c 


25. a = 38°40', a = 9.72 Ian, b = 11.8 Ian 

b > a > b sin A tells us that there are two possible 

triangles. 

sin B sinA 

--=- 

b a 

. B bsinA 

sm =- a 

11.8sin 38° 40' 

= 

9.72 

sin B = .75848811 

~ =49°20' 

C l = 180° - 38° 40' - 49° 20' 

CI =92°00' 

_c_l_=_a_ 

sinCI 

sin A 

asinC I 

ci = sin A 

9.72 sin 92°00' 

= sin 38°40' 

ci = 15.5 Ian 

~ =130°40' 

C 2 =180°-38°40'-130°40' 

C 2 = 10°40' 

~=_a_ 

sinC 2 sin A 

asinC 2 

c2 = sinA 

9.72 sin 10° 40' 

= sin38°4O' 

C2 =2.88 km 

26. C = 29° 50' , a = 8.61 m, c = 5.21 m 

sinA sinC 

--=- 

a c 

.." 

a > c> a sin C tells us that there are two possible 

triangles. 

sin A = 8.61sin29°50' = .82212894 

5.21 

Al = 55°20' 

~ =180°-55°20'-29°50' 

=94°50' 

-!L=_c_ 

sin ~ sinC 

~ = 5.21sin 94° 50' 

sin29°50' 

= 10.4 m 

A 2 = 180° - AI 

= 124°40' 

~ = 180° -124°40' -29°50' 

=25°30' 

-.!!L=_c_ 

sin~ sinC 

b _ 5.21sin25°30' 

2 - sin 29°50' 

=4.51m 

27. A = 96.80°, b = 3.589 ft, a = 5.818 ft 

a > b tells us that there is only one possible 

triangle. 

sinB sinA 

--=- 

b a 

. B bsinA 

sm =- a 

3.589 sin 96.80° 

= 

5.818 

sin B = .61253922 

B=37.77° 

B must be acute since A is obtuse. 

C= 180°-96.80°-37.77° 

C= 45.43° 

c a 

--=- 

sinC sinA 

asinC 

c=- 

sin A 

5.818sin45.43° 

= 

sin 96.80° 

c=4.174ft 

28. C = 88.70°, b = 56.87 yd, c = 112.4 yd 

Since c > b, only one triangle exists. 

sinB sinC 

--=- 

b c 

. B 56.87 sin 88.70° 

sm =----- 

112.4 

= .50583062 

B= 30.39° 

B~ 149.61° since 149.61°+C > 180°. 

A = 180° - 30.39° - 88.70° 

= 60.91° 

a c 

--=- 

sinA 

sin C 

112.4sin 60.91 ° 

a=----- 

sin 88.70° 

= 98.25 yd 

280


7.2 The Ambiguous Case ofthe Law of Sines 281 

29. B = 39.68°, a = 29.81 m, b = 23.76 m 

a > b > a sin B tells us that there are two possible 

triangles. 

sinA sin B 

--=- 

a b 

. A osinB 

sm =- 

b 

29.81sin 39.68° 

= 

23.76 

sin A = .80 108002 

Al = 53.23° 

C I= 180° - 53.23° - 39.68° 

C I =87.09° 

b ci 

--=- 

sinB sinq 

bsinq 

CI = sinB 

23.76 sin 87.09° 

sin 39.68° 

ci = 37.16 m 

A2 =126.77° 

C2 = 180° -126.77° - 39.68° 

C2 =13.55° 

b c2 

--=- 

sinB sinC2 

bsinC 2 

C2 = sinB 

23.76sin 13.55° 

= sin 39.68° 

c2 =8.719 m 

30. A = 51.20°, C = 7986 em, a = 7208 em 

c> a > c sin A tells us that there are two triangles. 

sinC sinA 

--=- 

c a 

. C 7986 sin 51.20° 

sm =---- 

7208 

=.86345630 

C I =59.71° 

Bt =180°-51.20°-59.71° 

=69.09° 

...l!L=_o_ 

sin Bt sinA 

.. ~ = 7208 sin 69.09° 

sin51.20° 

=8640 em 

C 2 =180 0-q 

=120.29° 

~ = 180° - 51.20° -120.29° 

=8.51° 

--.!!L=_a_ 

sin B 2 sinA 

~ = 7208 sin 8.51 ° 

sin 51.20° 

=1369 em 

31. a = .$,c = 2...[5, A = 30° 

a c 

--=- 

sinA sinC 

. C 2.$sin 30° 

sm = ...[5 

sinC= 1 

C=90° 

This is a right triangle. 

35. Let A = 38° 50', a = 21.9, b = 78.3. 

b sin A = 78.3 sin 38° 50' 

=49.1 

Thus, 21.9 < 49.1. That is, a 

The piece of property cannot exist with the given 

data. 

36. Let C = 28° 10', so c = 21.2. 

Let a = 26.5. Then B is the obtuse angle in the 

figure. 

a sin C = 26.5 sin 28° 10' 

= 12.509006 

Thus, a > c > a sin C. 

This fact tells us that there are two possible 

triangles with the given datil: one has an acute 

angle B and the other has an obtuse angle B as 

shown. 

sinA sinC 

--=- 

a c 

. A 26.5 sin 28° 10' 

sm = 

21.2 

=.59004745 

Al = 36°10' 

Bt = 180° - 28° 10' - 36° 10' 

= 115°40' 

-.lL=_c_ 

sinBt sinC 

b = 21.2 sin 115° 40' 

I sin 28°10' 

= 40.5 yd 

A 2 = 143°50' 

~ = 180° - 28° 10'-143°50' 

=8°00' 

~ c 

--=- 

sin~ sinC 

b _ 21.2sin8°00' 

2 - sin 28° 10' 

=6.25 yd 

281



.... 

a + b sin A + sin B Connections (page 305) 

37. Prove that --= . 

b sin B 

Start with the law of sines. 

1 

a b 2 

--=- 

sinA sin B = .!.K2 . 3 - 5(-1)+(-1)(0) - 3·4+4·5 - O· 2~ 

2 

( 

sin A X-!!-) = (sin A )(-/!-) 

b smA b smB = .!.K6+5 -12+20~ 

a sinA 

2 

-=- 

b sinB =.!.~9\ 

2 

a 

-+ 

1 

=--+ 

sinA 1 

= 9.5 square units 

b sinB 

a b sinA sinB 

-+-=--+- 

1. A =- kX\Y2 - y\x2 + x2Y3 - Y2x3 + x3Y\ - Y3x\ ~ 

b b sin B sinB 2. a=~(2-(-I»2+(5-3)2 

a+b sinA+sinB 

--=---- = ~32 +2 2 

b 

sinB 

= ..[13 '" 3.60555 

a - b sin A - sin B b = ~(2-4)2 +(5 _0)2 

38. Provet h at --= . 

a+b sinA+sinB 

Start with the law of sines. 

= ~(_2)2 +5 2 

a b =.J29 '" 5.38516 

--=- 

sinA sinB 

c = ~(-1-4)2 +(3-0)2 

bsinA 

a=--. 

sin B =~(_5)2 +3 2 

Substitute for a in the expression 

=.J34 '" 5.83095 

(a-b)+(a+b). 

1 

a-b ~-b 

s=-(a+b+c) 

2 

-=~ 

a+b sin B +b = .!.(..[13 +.J29 +.J34) '" 7.41083 

bsinA-bsinB 

_ sIDB 

2 

- bsjn t\+bsin B 

A =..j's("""s---a-)(-s-----,-,b)--,(s:---c....,...) 

sIDB 

sin A -sinB '" ~(7 .41083)(3.80528)(2.02567)(1.57988) 

= 

sin A +sinB 

'" 9.5 square units 

Section 7.3 

3. By the law of cosines, 

c 2 =a 2 +b 2 -2abcosC 

Connections (page 303) 2 

2ab cos C =a 2 + b 2 - c 

1. In a triangle, the ratio of the tangent of half the a 2 +b 2 _c 2 

cosC=---- 

difference between two angles to the tangent of 

2ab 

half the sum of those angles is equal to the ratio 

Using the lengths from part 2, we have: 

of the difference between the sides opposite those C 13+29-34 

angles and the sum of those sides. cos = 2..[13. Ji9 

2. In a triangle, the sum of two sides divided by the 

third side is equal to the cosine of half the 

difference between those two sides and the sine of 

half the angle opposite the third side. 

'" 0.20601 

C '" 78.1 0 

A = .!.ab sin C 

2 

= .!...[13 ..J29 sin 78.1 0 

2 

'" 9.5 square units 

282



Exercises 

1. (a) law of cosines 

c 2 = a 2 + b 2 - 2ab cos C 

2abcosC=a 2 +b 2 _c 2 

a 2 +b 2 _c 2 

cosC=---- 

2ab 

(342)2 + (116)2 - (401)2 

=-'-----'---'----'--"'-----'- 

2(342)(116) 

"" -.38290 

C "" 112.5° 

(b) law of cosines 

2 2 

+ b 2 c = a - 2ab cos C 

= (12.2)2 +(13.1)2 -2(l2.2)(13.l)cos20.2° 

"" 20.47 

c""4.52 

(c) law of sines 

b a 

--=- 

sinB sinA 

b 

a 

--= 

sin B sin(180 -(B+ C» 

b a 

--= 

sin B sin(B + C) 

b= asinB 

sin(B+ C) 

(21.13) sin(48° 13') 

= sin(48° 13' + 81° 42') 

=20.54 

(d) Neither is applicable. 

2. Subtract the sum of the two given angles from 

180° to obtain the third angle. 

3. a 2 = 3 2 + 8 2 - 2(3)(8) cos 60° 

=9+64-4s(~) 

=73-24 

=49 

a=7 

4. a 2 = 1 2 + (4.J2i - 2(1)(4.J2) cos 45° 

=1+32- 8 4~) 

=33-8 

=25 

a=5 

b 2 +c 2 _a 2 

5. cosO=--- 

2bc 

1 2 + (../3)2 _1 2 

= -2""';'(-I)('../3F'"3) 

1+3-1 

= 2../3 

3 

- 2../3 

../3 

= 2 

0=30° 

b 2 +c 2 _a 2 

6. cosO =--- 

2bc 

3 2 +5 2 _7 2 

= 

2(3)(5) 

9+25-49 

= 

30 

1 

=- 

2 

0=120° 

7. C=28.3°,b=5.71 in.,a=4.21 in. 

2 2+b2-2abcosC 

c =a 

c = ~(4.21)2 +(5.71)2 -2(4.21)(5.71)cos28.3° 

c = 2.83 in. 

Keep all digits of ...[;2 iii the calculator for use in 

the next calculation. If2.83 is used, the answer 

will vary slightly due to round-off error. Find 

angle A next, since it is the smaller angle and 

must be acute. 

. A asinC 4.21sin28.3° - 

SIn = -c-= ...[;2 

sin A = .70581857 

A =44.9° 

B = 180° - 44.9° - 28.3° 

B= 106.8° 

8. A = 41.4°, b =2.78 yd, c =3.92 yd 

a 2 =b 2 + c 2 - 2bccosA 

= (2.78)2 + (3.92)2 - 2(2.78)(3.92)(cos 4 1.4°) 

= 6.7459792 • 

a = 2.60 yd 

Keep all the digits of,J;;2 in your calculator for 

use in the next calculation. If2.60 is used, the 

answer will vary slightly due to round-off error. 

Find angle B next, since it is the smaller angle and 


283



sinB sin41.4° 

2.78 = J:i 

B=45.1° 

C = 180° - 41.4° -45.1° 

C=93.5° 

9. C =45.6°, b =8.94 rn, a =7.23 m 

c 2 =a 2 +b 2 -2abcosC 

c = ~(7.23)2 + (8.94)2 - 2(7.23)(8.94) cos 45.6° 

c=6.46 m 

Find angle A next, since it is the smaller angle and 


. asinC 7.23sin45.6° 

smA=--= G 

c ve 2 

sin A =.79946437 

A=53.1° 

B =180° -53.1°-45.6° 

B =81.3° 

10. A = 67.3°, b = 37.9 km, e = 40.8 krn 

2 = b 

2 2 

a + c - 2bc cos A 

= (37.9)2 + (40.8)2 - 2(37.9)(4O.8)(cos67.3°) 

= 1907.581537 

a =43.7 km 


must be acute . 

sin B sin 67.3° 

37.9 = ..r;;z 

B=53.2° 

C= 180°-67.3°-53.2° 

C=59.5° 

11. A = soo 40' b = 143 cm, c = 89.6 em 

a 2 = b 2 + e 2 - 2bc cos A 

a = ~1432 + (89.6)2 - 2(143)(89.6) cos 80° 40' 

a= 156 em 

Find angle C next, since it is the smaller angle 

and must be acute. 

csinA 89.6sin80040' 

sin C = --= r=:; 

a va 2 

sin C = .56692713 

C=34°30' 

B = 1800- 80° 40' - 34°30' 

B=64°50' 

12. C = 72° 40', a = 327 ft, b = 251 ft 

c 2 =a 2 +b 2 -2abcosC 

= 327 2 +251 2 -2(327)(251)cos72°40' 

c =348 ft 



sin B 

sin 72° 40' 

--- 

251 .J 

B=43°30' 

A =180 0-B-C 

A = 63°50' 

13. B = 74.80°, a = 8.919 in., e = 6.427 in. 

b 2 =a 2 +c 2 -2accosB 

b = ~(8.919)2 + (6.427)2 - 2(8.919)(6.427) cos 74.80° 

b=9.529 in. 

Find angle C next, since it is the smaller angle 

and must be acute. 

Jbi 

c sin B 6.427 sin 74.80° 

sinC=-b-= 

sin C = .65089219 

C=4O.61° 

A = 180°-74.80° -40.61° 

A =64.59° 

14. C = 59.70°, a = 3.725 mi, b = 4.698 mi 

c 2 =a 2 +b 2 -2abcosC 

= (3.725)2 +(4.698)2 - 2(3.725)(4.698) cos 59.70° 

c=4.276 mi 



sin A sin 59.70° 

3.725 = V 

A =48.77° 

B = 180° -48.77° -59.70° 

= 71.53° 

IS. A = 112.8°, b = 6.28 m, c = 12.2 m 

a 2 =b 2 +c 2 -2bccosA 

a = ~(6.28)2 + (12.2)2 - 2(6.28)(12.2)cos 112.8° 

a = 15.7 m 

Angle A is obtuse, so both Band C are acute. Find 

either angle next. 

b sin A 6.28 sin 112.8° 

sinB=--= r 

a va 2 

sin B = .36787456 

B=21.6° 

C = 180° -112.8° - 21.6° 

C=45.6° 

284



16. 8 =168.2°, a = 15.1 em, C =19.2 em 

b 2 = a 2 + c 2 - 2ac cos 8 

= 15.1 2 + 19.2 2 - 2(l5.1)(l9.2)cos 168.2° 

b=34.1 em 



sin A sin 168.2° 

--= 

15.1 fbi 

A =5.2° 

C =180° - 5.2° -168.2° 

=6.6° 

17. a =3.0 ft, b =5.0 ft, C = 6.0 ft 

Angle C is the largest, so find it first. 

c 2 =a 2 +b 2 -2abcosC 

3.0 

222 

+5.0 -6.0 

cos C=----- 

2(3.0)(5.0) 

cos C = -.06666667 

C=94° 

. 8 bsinC 

SIO =- c 

sin8 = .83147942 

8=56° 

A = 180° - 56° - 94° 

A=300 

18. a = 4.0 ft, b = 5.0 ft, c = 8.0 ft 

Find the largest angle first, in case it is obtuse. 

Since c is the longest side, angle C will be the 

largest angle . 

2 2 + b 

2 

c = a - 2ab cos C 

a 2+b2_c2 

cosC=--- 

2ab 

= 16+25-64 =-.575 

40 

C= 125° 

sin 8 sin 125° 

--= 

5.0 8.0 

8=31° 

A = 180° -31°-125° 

A=24° 

19. a =9.3 em, b = 5.7 ern, c =8.2 em 

Angle A is the largest, so find it first. 

a 2 = b 2 + c 2 - 2bc cos A 

5.7 

222 

+8.2 -9.3 

cos A =----- 

2(5.7)(8.2) 

cosA = .14163457 

A =82° 

. 8 bsinA 

SIO =- 

a 

sin B = .60672455 

B=37° 

C= 180°-82°-37° 

C=61° 

20. a = 28 ft, b = 47 ft, c =58 ft 


28 2 + 47 2 - 58 2 

cos C = ----- 

2(28)(47) 

= -.14095745 

C=98° 

sinA sinC 

--=- 

28 58 

A =29° 

B = 180° -29° -98° 

=53° 

21. a = 42.9 m, b = 37.6 m, c = 62.7 m 


c 2 = a 2 + b 2 - 2ab cos C 

42.9 

222 

+37.6 -62.7 

cos C = 

2(42.9)(37.6) 

cos C = -.20988940 

C= 102.1° 

. B bsinC 

SID =- c 

sin 8 = .58632321 

B=35.9° 

A = 180° - 35.9° -102.1° 

A =42.0° 

22. a =187 yd, b = 214 yd, c = 325 yd 


187 2 +214 2 - 325 2 

cosC= 

2(187)(214) 

= - .3106 1022 

C = 108.1° 

sinB sinC 

--=- 

214 325 

B=38.7° 

A = 180° - 38.7° -108.1° 

= 33.2° 

23. A8 = 1240 ft, AC = 876 ft, 8C = 918 ft 

AB=c,AC=b,BC=a 


c 2 =a 2 +b 2 -2abcosC 

a 2 +b 2 _c 2 

cosC=--- 

2ab 

cosC = .04507765 

C= 87.4° 

285


286 Chapter 7 Apr. ieatlons of Trigonometry and Vectors 

. B bsin C 

SID =- 

c 

sin B = .70573350 

B=44.9° 

A = 180° - 44.9° - 87.4° 

A = 47.7° 

24. AB=298 m,AC= 421 m, BC= 324 m 

Since AC is the longest side and is opposite 

angle B, angle B will be the largest angle. Find 

angle B first. 

324 2 +298 2 -421 2 

cos B= ------ 

2(324)(298) 

cos B = .08564815 

B=85.1° 

sinA 

sinB 

--=- 

324 421 

A = 50.1° 

C= 180 0-A 

- B = 44.8° 

27. Findx. 

x 2 = 25 2 + 25 2 - 2(25)(25)cos52° 

=480 

x = 22 ft 

28. Let B be the harbor, AB is the course of one ship, 

BC is the course of the other ship. 

Then b = the distance between the ships. 

A 

B 402mi C 

b 2 = 402 2 + 402 2 - 2(402)(402)cos 135° 40' 

b=745 mi 

29. Find AC, or b, in the following triangle. 

N 

t 

25. Find AB, or c, in the following triangle. 

A 

B 

c 2 =a 2 +b 2 -2abcosC 

c = ~2862 + 350 2 - 2(286)(350) cos 46 .3° 

c=257 

AB=257m 

26. Find the diagonals, BD and AC, of the following 

parallelogram. 

B 

C 

Icm/ 122 . / 

4D;L 

c 

A 6.0cm D 

BD 2 = AB 2 + AD 2 - 2(AB)(AD)cos A 

= 16+36-48(.52991926) 

= 26.5638755 

BD=5.2cm 

AC 2 = AB 2 + BC 2 - 2(AB)(BC) cos B 

= 16 + 36 - 48(-.52991926) 

= 77.4361245 

AC=8.8 em 

The lengths of the diagonals are 5.2 em and 

8.8 em. 

A 450km B 

Angle 1 =180°-128°40'=51°20' 

Angle 1 = Angle 2 

Angle B = 90° - Angle 2 = 38°40' 

b 2 =a 2 +c 2 -2ac~osB 

b = ~3592 + 450 2 - 2(359)(450)cos38° 40' 

b=281 km 

30. Let d = the length of the rope from the top of the 

tower to the bottom of the hill . The situation is 

illustrated in the fOllOrng fig",e. 

459.0 ft 

1 

12.47· 

a = 35.98° -12.47° 

=23.51° 

0= 90° -12.47° = 77.53° 

f3 = 180°-77.53° 

= 102.47° 

d 459 

----=-- 

sin 102.47° sin 23.51 ° 

d =1123 ft 

286



31. AB 2 = 10 2 + 10 2 - 2(10)(10)cos 128° 

AB 2 =323 

AB= 18 ft 

32. Let A =the angle between the beam and the 45-ft 

cable. 

45 2 + 90 2 - 60 2 

cos A =----- 

2(45)(90) 

A=36° 

Let B =the angle between the beam and the 

60-ft cable. 

90 2 + 60 2 - 45 2 

cos B =----- 

2(90)(60) 

B=26° 

33. Let x = the distance from the tracking station to 

the satellite at 12:03 P.M. 

Notice that the distance from the center of the 

earth to the satellite is 6400 + 1600 = 8000 km. 

Let 9 =the angle made by the movement of the 

satellite from noon to 12:03 P.M. 

21t radians 9 radians 

= 

2 hours 3 minutes 

21t 9 

--= 

2(60) 3 

9 = 61t = ~ radian 

120 20 

x 2 = 6400 2 + 8000 2 - 2(64OO)(8000)cos~ 

20 

x=2000 

The distance between the satellite and the 

tracking station is 2000 km. 

34. Using the law of cosines, 

c 2 = (14.0)2 + (13.0)2 - 2(l4.0)(l3.0)cos 70° 

"" 240.5047 

c "" 15.5 ft 

The length of the property line is approximately 

18.0 + 15.5 + 14.0 = 47.5 feet 

35. Let A = the man's location; 

B = the factory whistle heard at 3 sec after 5:00; 

C = the factory whistle heard at 6 sec after 5:00. 

B 

c~a 

~ 

Abe 

Since sound travels at 344 m per sec, and the man 

hears the whistles in 3 sec and 6 sec, the factories 

are 

c = 3(344) = 1032 m 

and b = 6(344) = 2064 m 

from the man. 

a 2 = 1032 2 + 2064 2 - 2(1032)(2064) cos 42.2° 

a = 1470 

The factories are 1470 m apart. 

36. Let x =the distance from the plane to the 

mountain when the second bearing is taken. 

course of 

plane 

7.92km 

9 = 180° - 32.7° = 147.3° 

Using the law of sines gives 

7.92 x 

= 

sin 147.3° sin 24.1 ° 

5.99= x. 

The plane is 5.99 km from the mountain. 

37. Let a be the length of the segment from (0, 0) to 

(6,8). Use the distance formula. 

a = ~(6 - 0)2 + (8 - 0)2 

= ~62 +8 2 

= ,.J36+64 

=,.J100 = 10 

Let b be the length of the segment from (0, 0) to 

(4,3). 

b = ~r-(4---0-)2-+-(-3_-0-)-2 

= ~42 +3 2 

=,.J16+9 

=.../25 =5 

Let c be the length of the segment from (4, 3) to 

(6,8). ,------ 

c=~(6-4)2 +(8-3)2 

= ~22 +5 2 

=,.J4+25 =..fi9 

a 2 +b 2 _c 2 

cos 9 = .:....-....:.....::...----=- 

2ab 

10 2 +5 2 _(..fi9)2 

cos 9 =..:...-......:....=--~.=.=~ 

2(10)(5) 

100+25 -29 

cos 9 = ------=..:.... 

100 

cos9= .96 

9"" 16.26° 

38. Let a be the length of the segment from (0, 0) to 

(8, 6). Use the distance formula. 

a = ~(8-0)2 +(6-0)2 

= ~82 +6 2 

= ,.J64+ 36 

= ,.Jloo = 10 

287



.... 

Let b be the length of the segment from (0, 0) to 

(12,5). 

b = ~r-(1-2_-0-)-2+-(-5---0)-2 

=~122+52 

=.J144+25 

= .J169 =13 

Let c be the length of the segment from (8, 6) to 

(12,5). 

c = ~'-(l-2_-8-)-=-2+-(-5---6)-=-2 

=~42 +(_1)2 

=.Jfi 

a 2 +b 2 _c 2 

cos(}=--- 

2ab 

10 2 +13 2 _(.Jfi)2 

cos() = ----- 

2(10)(13) 

e 100+169-17 

cos =---- 

260 

cos(} = .9692308 

()= 14.25° 

39. Using K = .!.bh, 

2 

K =.!.(16)(3.J3) 

2 

= 24.J3 = 41.57. 

To use Heron's Formula, first find the 

semi perimeter. 

1 

s='2(a+b+c) 

1 

=-(6+14+16) 

2 

1 

='2(36) =18 

Now find the area of the triangle. 

K =..js(s a)(s b)(s c) 

= ..j18(18 6)(18 -14)(18 -16) 

= ..j18(12)(4)(2) 

= .J1728 '" 41.57 

Both formulas give the same area. 

40. Using K = .!.bh, 

2 

K = .!.(10)(M) 

2 

= 15.J3 

=25.98. 

To use Heron's Formula, first find the 

semiperimeter. 

1 

s=-(a+b+c) 

2 

1 

=-(10+6+14) 

2 

=.!. (30) 

2 

=15 

Now find the area of the triangle. 

K =..js(s - a)(s - b)(s - c) 

= ..j15(15-10)(15 - 6)(15 -14) 

= ..j15(5)(9)(1) 

= .J675 

= 25.98 

Both formulas give the same result. 

41. a = 12 m, b = 16 m, c = 25 m 

1 

s = -(12 + 16+ 25) 

2 

s=26.5 

area =..js(s-a)(s-b)(s-c) 

=..j(26.5)(14.5)(10.5)(1.5) 

=78m 2 

42. a = 22 in., b = 45 in., c = 31 in. 

1 

s=-(a+b+c) 

2 

1 

= -(22 +45+31) 

2 

=49 

The area is 

K = .Jr-s(-s---a-)(-s---b)-(s---c-) 

= ..j49(49 - 22)(49 - 45)(49 - 31) 

= ..j49(27)(4)(18) 

= 310 in. 2 

43. a = 154 em, b = 179 em, c = 183 em 

1 

s=-(154+179+183) 

2 

s=258 

area =..js(s-a)(s-b)(s-c) 

=..j(258)(104)(79)(75) 

=12,600 cm 2 

288



44. a = 25.4 yd, b = 38.2 yd, c = 19.8 yd 50. Find the area of the Bermuda Triangle. 

1 

s=-(a+b+c) s = .!. (850 + 925 + 1300) 

2 2 

= 41.7 s = 1537.5 

The area is 

area = ..jr-15-3-7.-5(-6-87-.5-)-(6-12-.5-)-(23-7-.5-) 

K = ..jr-s(-s---a-)(-s--b)-(s---c-) = 392,000 

= ~41.7(16 .3)(3.5)(21.9) 

The area of the Bermuda Triangle is 392,000 mi 2 . 

= 228 yd 2 . 

51. Perimeter: 9 + 10 + 17 = 36 feet, so the semi 

45. a = 76.3 ft, b = 109 ft, c = 98.8 ft perimeter is .!..36 = 18 feet. 

1 

s = -(76.3 + 109 + 98.8) 

2 

s = 142.05 

area =..jr s(":'"""s---a":'""")("-s-_7 b)"-(s---c--:-) 

2 

Use Heron's Formula to find the area. 

Area: =~s(s-a)(s-b)(s-c) 

= ~(142 .05)(65 .75)(33.05)(43 .25) = ~18(9)(8)(1) 

= ~18(18 - 9)(18 -10)(18-17) 

= 3650 ft2 = .J1296 =36 ft 

Since the perimeter and area both equal 36 feet. 

46. a = 15.89 in., b = 21.74 in., c = 10.92 in. the triangle is a perfect triangle. 

1 

s=-(a+b+c) 

2 1 

52. (a) s=-(a+b+c) 

=24.275 2 

The area is 

1 

K = ..jr-"s(-s-_-a-)(s-_---,-,.b)-(s-_-c 7) 

=-(11+13+20) 

2 

1 

= 83.01 in. 2 =-(44) =22 

2 

47. AB = 22.47928 mi, AC = 28.14276 mi, Area =.Js(s-a)(s-b)(s-c) 

A = 58.56989° 

= .J22(22 - 11)(22- 13)(22 - 20) 

BC2 = AC2 + AB2 - 2(AC)(AB) cos A 

= ~22(11)(9)(2)" 

Use a calculator and substitute. 

BC 2 = 637.5539346 

BC = 25.24983 mi 

= .J4356 

= 66 which is an integer 

48. AB = 22.47928 mi, BC = 25.24983 mi, 1 

(b) s=-(a+b+c) 

A = 58.56989° 

2 

sinC sin A 1 

= -(13+ 14+ 15) 

22.47928 25.24983 2 

sin C = .75965065 

1 

C= 49.4334° 

="2(42) =21 

B= 180 0 - A - C 

~----- 

Area =~s(s-a)(s-b)(s-c) 

= 71.99670° 

= ~21(21-13)(21-14)(21-15) 

49. Find the area of the region. = ~21(8)(7)(6) 

1 

s = -(75+68+ 85) 

= .J7056 

2 

= 84 which is an integer 

s=114 

~----- 

area = ~(114)(39)(46)(29) 

1 

(c) s=-(a+b+c) 

=2435.3571 m 2 2 

(area in m 2 ) 

1 

Number of cans needed = "':""""2::---"";" =- (7 + 15 + 20) 

(m percan) 

2 

1 

2435.3571 = 32.471428 cans 

= -(42) = 21 

2 

75 

She will need to open 33 cans. 

289



Area ="'/s(s-a)(s-b)(s-c) 

a 2 =b 2 +c 2 -2bccosA 

= "'/21(21-7)(21-15)(21- 20) 2bccosA = b 2 +c 2 _a 2 

= .../21(14)(6)(1) b 2 +c 2 _a 2 

cosA=--- 

=.../1764 2bc 

= 42 which is an integer (4)2 +(5)2 _(6)2 

= 

2(4)(5) 

53. Applying the law of cosines when a = 3. b = 4. 

and c = 10 gives 

c 2 =a 2 +b 2 -2abcosC 

10 2 =3 2 +4 2 -2(3)(4)cosC 

75 = -24cosC 

-3.125 = cos C. 

Since -1 ~ cos C ~ 1. a triangle cannot have sides 

3,4, and 10. 

= 

16+25 -36 

40 

1 

= 

8 

Does cos A = 2 cos 2 B-1 

1 (3)2 

'8.1 2 '4 -1 

.1 2(~)-~ 

55. (a) Using the law of sines 

16 16 

sinA sinC 

2 1 

--=- 

~-=a 

c - 16 8 

sin C = E- sin A 

a 

=~sin60° 

13 

Since the identity holds, A is twice the size of B. 

Recall that cos 2A = 2 cos 2 A-I. 

57. Using the law of cosines, 

=0.99926 

a 2 =b 2 +c 2 -2bccosA 

c = 87.8° or 92.2° c 2 - 2bc cos A + (b 2 _a 2 ) = 0 

... 

(b) By the law of cosines, c 2 - 2(24.9)(cos55.3°) c + (24.9 2 - 22.8 2) = 0 

c 2 = a 2 + b 2 - 2ab cos C c 2 - 28.35c + 100.17 = 0 

2abcosC= a 2 + b 2 _c 2 

Apply the quadratic formula 

a 2 +b 2 _c 2 

cosC=---- 

2ab 

28.35 ± ~(-28.35)2 - 4(1)(100.17) 

c =---"----'---.;..;...-- 

2(1) 

(13)2 + (7)2 _ (15)2 

28.35 ± .../803.72 - 400.68 

= 

= 

2(13)(7) 

2 

=-.03846 

28.35 ±20.08 

= 

C=92.2° 2 

28.35 + 20.08 

(c) With the law of cosines, we are required to ci = 2 

find the inverse cosine of a negative number 

=24.2 ft 

therefore, we know angle C is greater than 

28.35 - 20.08 

90 0 • 

c2 = 2 

56. Using the law of cosines, = 4.14 ft 

b 2 = a 2 + c 2 - 2ac cos B 

2accosB=a 2 +c 2 _ b 2 

a 2 +c 2 _b 2 

cosB=--- 

2ac 

(6)2 + (5)2 _ (4)2 

= 

2(6)(5) 

36+25-16 

58. Prove that if a and b are equal sides of an 

isosceles triangle. then c 2 = 2a 2 (l cos C). 

By the law of cosines, 

c 2 =a 2 + b 2 - 2abcos C. 

Since a = b, 

c 2 = a 2 +a 2 -2a 2 cosC 

c 2 = 2a 2 - 2a 2 cos C 

= 

60 c 2 = 2a 2 (1- cos C). 

3 

= 4 

290



59. Given that point D is on side AB of triangle ABC, 

such that CD bisects angle C; thus, 

angle ACD =angle DCB. 

AD b 

Show that - =-. 

DB a 

A 

63. b 2 + c 2 > b 2 and b 2 + c 2 > c 2 . If a 2 > b 2 + c 2 , 

then a 2 > b 2 from which a > b and a > c 

because a, b, and c are all nonnegative. 

64. Because A is obtuse it is the largest angle, so the 

longest side should be a, not c. 

Section 7.4 

c 

B 

Let 9 =the measure of angle ACD; 

a = the measure of angle ADC. 

Then 9 =the measure of angle DCB and angle 

BDC =180 - a. 

By the laws of sines, 

sin9 sina 

--=- 

AD b 

. 9 ADsina 

SIO =-- 

b 

and 

sin9 

--= 

DB 

sin 9 

ADsina 

b 

= 

sin(1800-a) 

a 

DBsin(l80° - a) 

a 

DBsin(1800-a) 

a 

Since sin ex. =sin (1800 _ u), AD = DB . 

b a 

Mu tip I' b th sid b b AD b 

DB DB a 

I . ymg 0 SI es y -,we get - =- . 

60. Let a = 2, b =2-/3 ,A = 30°, B = 60°. 

. tan!(A-B) a-b 

Venfy - 

tan!(A+B) a+b 

tan!(A-B) _ tan!(300-600) 

tan t (A + B) - tan! (30° + 60°) 

tan(-15°) 

= tan 45° 

=-.26794919 

a - b = 2 - 2-/3 = -1.46 =- .26794919 

a+b 2 +2-/3 5.46 

(Note: - .26794919 =-2 + -/3.) 

61. Since A is obtuse, 90° < A < 180°. The cosine of a 

quadrant II angle is negative. 

Exercises 

3. Equal vectors have the same magnitude and 

direction: m = p, n =r. 

4. Opposite vectors have the same magnitude but 

opposite direction . Opposite vectors are m and q, 

p and q, nand s, rand s. 

S. One vector is a positive scalar multiple of another 

if the two vectors point in the same direction; they 

may have different magnitudes. 

m = Ip, m = 2t, n = lr, p = 2t or 

1 1 

P =1m t =- m, r =In t =- P 

'2 ' 2 

6. One vector is a negative scalar multiple of another 

if the two vectors point in the opposite direction; 

they may have different magnitudes. 

m = -lq, p =-lq, r= -Is, n = -Ls, q =-2t 

7. 

-b 

8. 

9. 

10. 

2 

62. In a =b 2 + c 2 - 2bccos A, cos A is negative, so 

2 

a = b 2 + c 2 plus a positive quantity. 

2 

Thus, a 2 > b 2 +c . 

291



11. 19. 

12. 20. 

13. , ,,,,,, 

21. 

, 

14. 

22. 

15. 

23. 

16. - - - ; , 

~ 

" , 

d '0 , 

24. 

b 

17. 

18. 

25. 

26. 

27. 

28. 

a + (b + c) = (a + b) + c 

Yes, vector addition is associative. 

c-s d =d+ c 

Yes, vector addition is commutative. 

lui = 12, Iwi = 20, 8 = 27° 

~ 

2 ~--=- --=a. 

20 ----~.- 

lui =8, !wI =12, 8 = 20° 

- 

2~ ~---- 

-~) 

12 ...-- 

292



29. lui = 20, IwI = 30, 8 = 30° 34. a =70°, /vi = 150 

~ 

- - - 

~-:3) 

30 - 

30. lui = 27, Iwl= 50, 8 = 12° 

27 #/_ J2" = 

50 ~ __ SZ) 

y 

31. lui = 50, Iwl = 70, 8 = 40° x 

50 

........ Ix! = 150cos70° = 51 

. . 

IYl = 150sin 70° =140 

.- . 

In Exercises 32-39, x is the horizontal component of v, 35. a =50°, Ivl = 26 

and y is the vertical component of v. 

32. a = 20°, Iv! = 50 

~ 

e::4CJY 

x 

cos200=M 

cos 500 = IxI 

50 26 

Ixl = 50 cos 20° = 47 IxI =26 cos 50° 

=17 

sin200=M 

50 sin500=M 

~I = 50sin20° = 17 26 

IYl = 26sin 50° 

33. a = 38°, Ivl = 12 =20 

y 

x 

36. a =35°50', Iv! =47.8 

y 

x 

cos38o=M 

12 

Ixl =12cos 38° 

Ixl =47.8cos 35° 50' =38.8 

=9.5 

111 =47.8sin35°50' ~ 28.0 

sin38°= M 

12 

~I= 12sin38° 

=7.4 

x 

y 

293



37. a = 27°30', Ivl = 15.4 41. Forces of 250 newtons and 450 newtons, forming 

an angle of 85° 

250 

y 

38. 

x 

cos 27° 30' = ~ 

15.4 

Ixl = 15.4cos27°30' 

=13.7 

sin 27° 30' = J!L 

15.4 

IYI = 15.4 sin 27° 30' 

=7.11 

a = 128.5°, Iv! = 198 

250 

a = 180°-85° = 95° 

1v1 2 = (250)2 +(450i 

= 284,610 

Ivi = 530 newtons 

2(250)(450) cos 95° 

y 

42. 

Forces of 19 newtons and 32 newtons, forming an 

angle of 118° 

180° - 118° = 62° 

-------- 

62· 

\ 

\ 

\ 

\ 

\ 

x 

8 = 180°-128.5° = 51.5° 

cos51.5°=~ 

198 

Ixi = 198cos51.5° 

=123 

sin 51.5° = J!L 

198 

IYI = 198sin 51.5° 

=155 

43. 

32newtons 

Ivl 2 = 19 2 + 32 2 - 2(19)(32)cos 62° 

= 814.1226 

Ivi = 29 newtons 

Forces of 17.9 Ib and 25.8 Ib, forming an angle of 

105.5° 

17.9 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

39. a = 146.3°,1v1 = 238 

y 

x 

8 = 180°-146.3° = 33.7° 

Ixl =238cos33.7° = 198 

IYI =238sin 33.7° =132 

17.9 

a =180° -105.5° 

=74.5° 

Ivl 2 =(25.8)2 + (17.9)2 - 2(25.8)(17.9)cos 74.5° 

= 739.218 

IvI= 27.2 Ib 

294


7,4 Vectors 295 

44, Forces of 75.6 Ib and 98.2 Ib forming an angle of 

82°50' 

180°- 82°50' = 97° 10' 

75.61b 

97'10' 

y 

----fL-=--~ x 

98.21b 

Ivf = (75.6)2 + (98.2)2 - 2(75.6)(98.2) cos 97°10' 

= 17.210.957 

Iv! = 1311b 

45, Forces of 116 Ib and 139 lb. forming an angle of 

140°50' 

, ,,,, 

48. Magnitude: ~(-4)2 + (4.J3) 2 =.J64 = 8 

Angle: (J = arctan( 4':;) 

= arctan(-.J3) (in quadrant II since 

x 0) 

= 120 0 

y 

116 , ,,, 

140'50' " 

_-,"-'---+-...J..-__ X 

-4 

49. Magnitude: ~(8..J2)2 +(-8..J2) 2 = ~256 = 16 

. (-8..J2) 

Angle: (J = arctan s..J2 

a = 180°-140°50' = 39° 10' 

1v!2 = (139)2 + (116)2 - 2(139)(116)cos 39°10' 

=7774.74 

!vi = 88.2 Ib 

= arctan(-l) (in quadrant IV 

since x > 0 and y < 0) 

= 315° 

y 

46, Forces of 37.8 lb and 53.7 lb, forming an angle of 

68.5° 

180°- 68.5° = 111.5° 

--------- 

111.5' 

53.71b 

1v1 2 = (37.8)2 + (53.7)2 - 2(37.8)(53.7) cos 111.5° 

= 5800.422 

!vi = 76.2 Ib 

47, Magnitude: ~1 2 + 1 2 =..J2 

50. Magnitude: ~(.J3)2 + (_1)2 =.J4 = 2 

Angle: (J = arctan( ':A) 

= arctan(- .J3) (in quadrant IV since 

3 x > 0 and y < 0) 

=330° 

Angle: (J = arc tan(!) (in quadrant I since 

1 x > 0 and y > 0) 

=45° 

295



y 

54. Magnitude: ~02 +(-12)2 =..J144 =12 

Angle: e=arctan( - ~2) 

(Division by zero makes this undefined so the 

--+--k-...l...-'--.,_ .r 

-I 

only possible values are (} = 90° or (} = 270° .) 

Since y < 0, (} = 270°, 

y 

51. Magnitude: ~(l5)2 + (_8)2 = ·./289 = 17 270' 

---+--+---'--_ x 

Angle: e = arctan (-8)- (i m qua d rant IV smce ' 

15 x>Oandy



57. y 60. y 

--I--4.-L,,-,- x 

7890 

a = Ivlcos 8 b = Ivlsin 8 

=198cos 128°30' = 198sin128°30' 

= 198cosI28.5° = 198sin128.5° 

== -123.258 == 154.956 

v = (-123.258.154.956) 

58. y 

a = Ivlcos8 b = Ivlsin 8 

= 7890cos302°40' = 7890sin 302°40' 

== 7890cos 302.67° == 7890sin 302.67° 

== 4258.633 == -6641.999 

v = (4258.633. -6641.999) 

61. u + v = (-2.5)+(4.3) 

=(-2+4.5+3) 

=(2.8) 

62. u - v = (-2.5)-(4,3) 

=(-2-4.5-3) 

= (-6.2) 

59. 

a = /vi cos 8 b = /vi sin8 

= 238cos 146°10' = 238sin 146°10' 

= 238cos 146.167° =238sin146.167° 

== -197.697 == 132.513 

v = (-197.697.132.513) 

y 

63. 

64. 

v -u =(4.3) -(-2.5) 

=(4-(-2).3-5) 

=(6. -2) 

5v =5{4, 3) 

= (5 ·4, 5 '3) 

=(20,15) 

--rl'--T---'-- x 

65. -5v = -5(4. 3} 

=(-5 .4. - 5 .3) 

=(-20. -15) 

. e 

a = /vi cos 8 

b = Ivlsin8 

= 69.1cos251°20' = 69.1sin251°20' 

= 69.1cos 251.33° = 69.1cos251.33° 

==-22.116 ==-65.465 

v = (-22.116. -65.465) 

66. 

67. 

3u+ 6v = 3(-2.5).+6(4. 3} 

=(3·-2,3.5)+(6·4.6·3) 

= (-6. 15)+ (24. 18) 

= (-6+ 24.15 + 18) 

=(18,33) 

(-5.8)=-5i+8j 

68. (6,-3)=6i-3j 

69. (2.0}=2i+Oj=2i 

70. (0. -4)= 0i-4j =-4j 

297



.... 

71. a = 8cos45° b = 8sin45 

( (1,7) ·(I,I) J 

= 8 . .,fi = 8. .,fi 82. ()= arccos 

2 2 ~12 +7 2 ' ~ 1 2 +1 2 

=4.,fi =4.,fi 

(4.,fi, 4.J2) = 4.,fii + 4.,fij 

( 1+7 ) 

= arccos .J5O ..,fi 

72. a = 3cos 210° b = 3sin21O° = arccos( 1~) 

-~ 1 

=3·- =3·- 

2 2 

:= 36.87° 

-3~ 3 

:;::- 

= 

2 2 83. ()= arccos 

[ (I, 2) (-6, 3) J 

~12+22 ' ~(-6) 2 + 3 2 

(_Wi _~)=_3~i_~j 

2 ' 2 2 2 

= arcco { .J5.J45 -6+6) 

73. a = .6cos115° b = .6sin 115° = arccos(O) 

:= -.254 :=.544 

=90° 

(-.254, .544) = -.254i+.544j 

74. a = .9cos 208° b = .9sin208° 84. ()= arcco 

{ (4, 0}.(2, 2) J 

:= -.795 =-.423 ~42 +0 2 .~22 +2 2 

(-.795, -.423)=-.795i-.423j 

75. (6, -1) ·(2, 5) = 12 - 5 = 7 

76. (-3,8)·(7, -5)=-21-40=-61 

77. (2, - 3)·(6,5) = 12 -15 =-3 

{ 8+0 ) 

= arcco ..Jl6 ..J8 

=arcco{81) 

=arcco{*J 

78. (1,2)'(3, -1)=3-2=1 =arcco{ ~) 

79. (4,0) ·(5, -9)=20+0=20 

SO. (2,4) .(0, -1)=0-4=-4 85. 

For Exercises 81-86, use the formula ()= arcco {u IuIIvI .v). 

=45° 

()= arcco 

{ (3, 4).(0, I) J 

~32 +4 2 . ~0 2 +1 2 

{ 0+4 ) 

= arcco .,fi5 ...[i 

=arcco{~) 

81. ()= arcco 

{ (2, I) (-3, 1) J '" 36.87° 

~22 +1 2 '~(_3)2 +1 2 { (-5, 12). (3, 2) J 

{ -6+1 ) 86. () = arcco 

= arcco .J5..JW ~(_5)2 +12 2 .~32 +2 2 

=arcco{5~) 

169 13 

=arcco{~) =arcco{ ~) 

1313 

(-15+24) 

= arccos ..Jl69.Jl3 

=arcco{- ~) 

= arcco{ 9.Jl3) 

169 

= 135° := 78.93° 

298



87. (3u)' v = (3(-2, 1)). (3,4) 97. Draw a line parallel to the x-axis and the vector 

u + v (shown as a dashed line). 

= (--6,3)' (3,4) y 

= -18+ 12 

=--6 

88. u.(v-w)=(-2, 1).(3, 4)-(-5, 12)) 

=(-2,1)'(3-(-5),4-12) 

----r.:::+-:--L-._ x 

=(-2,1)'(8, -8) 

80' a A 

=-16-8 3 

=-24 Since 0 1 = 110°, its supplementary angle is 70°. 

Further since O 2 = 260° , a = 260° - 180° = 80°. 

89. u,v-u'w=(-2, 1)'(3,4)-(-2,1)'(-5, 12) 

Then the angle eBA becomes 

= (--6+ 4) -(10+ 12) 

180 - (80 + 70) = 180 - 150 = 30°. Using the law 

=-2-22 of cosines, the magnitude of u + v is 

=-24 

2+c2-2accosB 

In+vl 2 =a 

90. u ·(3v)=(-2, 1).(3(3, 4)) = (3)2 + (12)2 - 2(3)(12)cos 30° 

=90.65 

= (~2, 1).(9, 12) 

= -18+ 12 In+ vi = ../90.6461709 

=--6 "" 9.52082827. 

91. (1, 2) ·(--6, 3) = --6 + 6 Using the law of sines, 

=0 sinA sin B 

--=- 

The vectors are orthogonal. a b 

. A asinB 

SIO =- 

92. (3,4)'(6,8)= 18+32 b 

=50 

The vectors are not orthogonal. 

A =arcsl .{asinB) - 

b 

.{ 3sin 300) 

A =arcs) 

93. (1, 0).(..fi,0)=..fi +0 9.52 

=..fi 


= 9.0646784° 

The direction angle of U + v is 

110 + 9.0646784 = 119.0646784°. 

94. (1, 1). (1, -1) = 1-1 

98. al =lui cos 0 1 ~ = luisin 0 1 

=0 =12 cos 110° =12sin 110° 

The vectors are orthogonal. = -4.10424172 "" 11.27631145 

(-4.10424172, 11.27631145) 

95. (...[5,-2),(-5,2...[5) = -5...[5 -4...[5 

=-9.,J5 99. a2 =IvicoS 02 ~ =lvl sin 02 

The vectors are not orthogonal. =3cos26O° = 3sin260° 

96. (-4,3).(8, -6)=-32-18 

=-50 


= -5.20944533 = -2.954423259 

(-.520944533, - 2.954423259) 

100. a =a1 + a2 

= -4.10424172 - .520944533 

= -4.625186253 

b=~+~ 

= 11.27631145 - 2.954423259 

= 8.321888191 

(-4.625186253,8.321888191) 

299



101. Magnitude: ~(-4.625186253)2 +(8.321888191)2 

= 9.52082827 

A 1 8 ( 8.321888191 ) 

ng e: =arccos -4.625186253 

=119.0646784° 

(in quadrant II since y > 0 and x < 0) 

102. (a) They are the same. 

Section 7.5 

Exercises 

1. Let a = the angle between the forces. 

-, 

(J ',423 

, 

-, 



3. Let (J = the angle that the hill makes with the 6. Let r = the vertical component of the person 

horizontal. 

exerting a 114-lb force and s = the vertical 

component of the rson exerting a ISQ-lb force . 

25 

The 80-lb downward force has a 25-lb component 

parallel to the hill. The two right triangles are 

similar and have congruent angles. 

. (J 25 

sm = 

80 

(J = 18° 

4. Let If1 = the force required to keep the car parked 

on hill. 

62.4· 

The weight of the box is the sum of the 

magnitudes of the vertical components of the two 

vectors representing the forces exerted by the two 

people. 

Irl = 114 sin 54.9° = 93.27 

~ = ISOsin 62.4° = 132.93 

Weight of box = Irl + ~ 

= 2261b 

7. Let Ixl = the weight of the boat. 

r 

J!L=sinISo 

3000 

If1 = 3000sin 15° = 780 Ib 

5. Find the force needed to pull a 6O-ton monolith 

along the causeway. 

2.3· 

500 =sinI8 0 

500lb 

Ixl 

500 

Ixl =sin 180 

Ixl= 1600 Ib 

The force needed to pull 60 tons is equal to the 

magnitude of x, the component parallel to the 

causeway. 

sin2.3o=M 

60 

IxI = 60sin 2.3° 

=2.4 

The force needed is 2.4 tons . 

301


302 Chapter 7 Applications of Trigonometry and VectOI'S 

The equilibrant has a magnitude of 2640 lb and 

makes an angle of 167.2° with the 1480-lb force. 

The weight is 2640 lb at an angle of 167.2° with 

the 1480-lb force. 

9, Find the weight of the crate and the tension on the 

horizontal rope. 

v 

x 

-, 

'\. 

'\. 

-, 

-, 

-, 

w 

b 

v has horizontal component x and vertical 

component y. The resultant v + h also has vertical 

component y. The resultant balances the weight 

of the crate, so its vertical component is the 

equilibrant of the crate's weight. 

IwI ;;:: ~I ;;:: 89.6 sin 46° 20' 

Iwl;;::64.8Ib 

Since the crate is not moving side-to-side, h, the 

horizontal tension on the rope, is the opposite of 

x . 

1hI;;:: Ixl;;:: 89.6cos46°20' 

IhI;;:: 61.91b 

The weight of the crate is 64.8 lb: the tension is 

61.9 lb. 

10, Draw a vector diagram showing the three forces 

acting at a point in equilibrium. Then, arrange the 

forces to form a triangle. The angles between the 

forces are the supplements of the angles of the 

triangle. 

760lb 

The angle between the 1220-lb and 980-lb forces 

is 

180° - a= 141.5°. 

(3) f3 =180° - a - 0 

=53.4° 

The angle between the 760-lb and 1220-lb forces 

is 

180° - f3 = 126.6°. 

II, Find the magnitude of the second force and of the 

resultant. Use the parallelogram rule; v is the 

resultant and x is the second force. 

a 

I 

-71-----v 

I 

I 

I 

I 

I 

78'50' 

I 

41'10' 

176 

I 

a =180° -78°50' = 101°10' 

the angle between the second force and the 

resultant is f3 =78° 50' - 41°10' = 37° 40' . 

Use the law of sines to find Ivl. 

Iv! 176 

sin a = sinf3 

Ivi = 176 sin 101° 10' 

sin37°40' . 

Ivi =2831b 

Use the law of sines to find Ixl. 

IxI = 176 

sin41°10 sin 37°40' 

Ixl= 176sin41°IO' 

sin 37°40' 

Ixl;;:: 190 lb 

12. Let It1 ;;:: the second force and Irl =the magnitude 

of the resultant. 

9801~7601b 

~ 

12201b 

980 2 +760 2 -1220 2 

(1) cosO;;:: 2(980)(760) 

0;;:: 88.1° 

The angle between the 760-lb and 980-lb forces is 

180°-0 =91.9° 

sina sinO 

(2) 760 =1220 

a =38.5° 

/ 

/ 

r / 

/ 

/ 

/ 

/ 

~2'IO' / 

(J/'/ 

",,-~~ 

32'40' 

0= 180° - 42° 10' =137°50' 

f3 =42° 10' - 32° 40';;:: 9° 30' 

Irl 28.7 

--=-- 

sin 0 sin 9° 30' 

Irl =117 Ib 

302



Irl 

sin 32° 40' sin 137° 50' 

IfI = 93.9 Ib 

The magnitude of the resultant is 117 lb; the 

second force is 93.9 lb. 

13. Let v = the ground speed vector of the plane. 

Find the actual bearing of the plane. 

N 

A 25 

Angle A = 266.6° -175.3° = 91.3° 

Use the law of cosines to find Ivl. 

1v1 2 = 25 2 + 650 2 - 2(25)(650) cos 91.3° 

=423,862 

/vi = 651 

Use the law of sines to find the drift angle, B, the 

angle that v makes with the airspeed vector of the 

plane. 

sinB 

sinA 

25 651 

sin B = .038392573 

B=2.2° 

The bearing is 175.3° - 2.2° = 173.10. 

14. Let Ixl = the airspeed and IdI = the ground speed. 

N 

-+oe:;;;.-.::;.....--~~E 

x 

(J =90°-74.9° = 15.1° 

M=cotI5.1° 

.::, 

42 

Ixl = 42 cot 15.1° 

=156mph 

1!L=cscI5.1° 

42 

1



18. Let c = the ground speed vector. 

N 

20. Let v = the ground speed vector . 

N 

B 

v 192 

By alternate interior angles, 

angle A = 57° 40' . 

Use the law of sines to find B, the drift angle. 

sinA sinB 

--=- 

168 27.1 

sin B = .13629 

B = 7°50' 

Bearing = 57° 40' + B = 65° 30' 

C= 180°- A- B = 114°30' 

~= 27.1 

sinC sinB 

lei = 181 mph 

The bearing is 65° 30'; the ground speed is 

181 mph. 

19. Let v = the airspeed vector. 

The ground speed must be 

400 

-=160 mph. 

2.5 

N 

II 

B 

Angle A = 328° - 180° = 148° 


Ivl 2 = 11 2 + 160 2 - 2(l1)(160)cosI48° 

Ivl2 = 28,706 

Ivl = 170 

The airspeed must be 170 mph. 

Use the law of sines to find B, the drift angle. 

sin B sin 148° 

--= 

11 170 

. B 11sin 148° 

Sin =--- 

170 

B=2.0° 

The bearing must be 360° - 2.0° = 358°. 

-- 

A 23 C 

Angle C = 180° - 78° = 102° 

1v1 2 =23 2 +192 2 -2(23)(192)cosl02° 

Ivl= 198 mph 

sin B sin 102° 

--= 

23 198 

B=6.5° 

Bearing = 180° + B = 186.5° 

21. Find the ground speed and resulting bearing. 

N 

245' 

Angle A = 245° -174° = 71° 


Ivl2 = 30 2 + 240 2 - 2(30)(240) cos 71° 

1v1 2 "= 53,812 

Ivl"=230 

The ground speed is 230 Ian per hr. 

Use the law of sines to find angle B. 

sinB sin71° 

--=- 

30 230 

sinB=.1233 

B=7° 

The resulting bearing is 174° _7° = 167°. 

22. Let A =the point where the ship changes to a 

bearing of 62° and C = the point where it changes 

to a bearing of 115°. 

N 

N 

F 

D 

304



Angle CAB = 90° - 62° = 28° 

Angle FCA =180° - angle DAC 

=180°-62° =118° 

AngleACB =360°- angle FCB- angle FCA 

=360° -115°-118° 

=127° 

Angle CBA =180°- angleACB- angle CAB 

=180° -127° - 28° =25° 

Using the law of sines, we have 

sin 127° sin 28 

=- 

50 CB 

CB=29.4 

sin 127° sin 25° 

=- 

50 AC 

AC=26.5. 

The ship traveled 26.5 + 29.4 =55.9 mi. To avoid the iceberg, the ship had to travel 55.9 - 50 =5.9 mi. 

23. At what time did the pilot turn The plane will fly 2.6 hr before it runs out of fuel. In 2.6 hr, the carrier will travel 

(2.6)(32) = 83.2 mi and the plane will travel a total of (2.6)(520) = 1352 mi. Suppose it travels x mi on its initial 

bearing; then it travels 1352 - x mi after having turned. 

338· 

Use the law of cosines to get an equation in x. 

(1352 - x)2 =x 2 +(83.2)2 - 2(x)(83 .2)cos 52° 

l, 827,904 - 2704x + x 2 =x 2 + 6922.24 - 102.45x 

r, 820,982 =2601.55x 

7oo=x 

To travel 700 mi at 520 mph requires 700 =1.35 hr, or 1 hr and 21 min. 

520 

The pilot turned at 1 hr 21 min after 2 P.M., or at 3:21 P.M. 

24. (a) Relative to the banks. the motorboat will be traveling 

2 

x +3 2 =7 2 

x 2 =49-9 

x =.J4O "" 6.32 mph 

(b) At 6.32 mph, the motorboat travels 

6.32miles 5280 feet 1 hour 1 minute 

---_. 

._-- 

hour 1 mile 60 minutes 60 seconds 

=9.276 ft I sec 

Crossing a 132-foot wide river would take 

1 second 

132 feet · '" 14.23 seconds 

9.276 feet 

305



() C SID 

'() = 3 

7 

() =arcsin(%) 

'" 25.38 0 

25. (a) First change 10.34" to radians in order to use the length of arc formula. 

0 

10.34" ._1__ • .....!!..- '"5.013 x 10- 5 radian . 

3600" 180 0 

In one year Barnard's Star will move in the tangential direction a distance of 

s = r(} =(35 x 1OI2X5.013 xJO- 5 ) 

= 1,754,550,000 mi. 

' S 1,754,550,000 . . 

Inone second B amard s tar moves '" 56 nu tangentially, Thus, vt '" 56 mi I sec . 

60 ·60 ·24 ·365 

(b) The magnitude of v is given by 

1v!2 =~l +~l 

Iv! =~672 + 56 2 

'" 87 mi I sec. 

26. m =-cot 0 

From the figure, we see that L passes through the point (beosO, bsinO). 

Using this information and the general form of the line: y - bsinO =-cotO(x -boosO). 

(Other answers are possible.) 

27. At the end point of v, x =a. Using the formula from Exercise 26: 

y- bsinO =-cotO(x - b cos 0) 

y - bsinO =-cotO(a -bcosO) 

y = -a cot 0+ bcosOcotO+ bsin 0 

The coordinates are (a,-a cotO + bcosOcotO + bsin 0). 

28. Iv! = ~(XI - 0)2 + (YI - 0)2 

=~(a _0)2 + (-a cot 0 + bcosOcotO + bsinO _0)2 

=~a2 + a 2 cot 2 ()- 2abcot 2 OcosO - 2abeot Osin 0 + b 2 cos 2 Beot 2 B+ 2b 2 cosOsin BcotO + b 2 sin 2 B 

=~a2 + a 2 cot 2 B- 2abcot 2 OcosO- 2abcosO+ b 2 cos 2 Ocot 2 0+ 2b 2 eos 2 0+ b 2 sin 2 B 

= 

a 2 sin 2 0+a 2 cos 2 0-2abcos 3 B-2abcosBsin 2 0+b 2 cos 4 B+2b 2 cos 2 Bsin 2 0+b 2 sin 4 0 

sin 20 

~a2(sin2 0 + cos 2 0) - 2abeos 8(cos 2 0 + sin 2 0) + b 2(eos4 0 + eos 2 Osin 2 B+sin 2 0) 

sin 0 

" 

=~a2 .1-2ab cos 0·1 + b 2(cos2 B+ sin 2 B)2 

sinO 

~a2 _ 2abcosB+ b 2 .1 

= 2 

sinO 

~a2 -2abcosB+b 2 ~a2 +b 2 -2abcosO 

= = 

sinB 

sin B 

306



29. Applying the law of cosines, we see that the 

quantity 

a 2 +b 2 -2abcosO 

is exactly the segment connecting the endpoint of 

a to the endpoint of b. 


1. Find b, given C = 74.2°, c = 96.3 m, B =39.5°. 

Use the law of sines to find b. 

b c 

--=- 

sinB sinC 

b= csinB 

sinC 

96.3 sin 39.5° 

= sin 74.2° 

=63.7 m 

2. Find B, given A = 129.7°, a = 127 ft, 

b= 69.8 ft. 

Use the law of sines to find B. 

sinB sinA 

--=- 

b a 

. B 69.8sinI29.7° 

Sm = 

127 

= .42286684 

Since angle A is obtuse, angle B is acute. 

B=25.0° 

3. Find B, given C = 51.3°, c = 68.3 m, b =58.2 m. 

Use the law of sines to find B. 

sinB sinC 

--=- 

b c 

. B bsinC 

Sm =- 

c 

58.2sin51.3° 

68.3 

sin B = .66502269 

B =41.7° 

Angle B cannot be obtuse, since b < c , B < C, and 

C is acute. 

4. Find b, given a =165 m, A = 100.2°, B = 25.0°. 

b a 

--=- 

sinB sin A 

b = 165sin25.0° 

sin 100.2° 

b=70.9m 

S. Find A, given B = 39° 50' , b = 268 m, a =340 m. 

a > b > a sin B tells us that two triangles are 

possible and A can have two values. 

Use the law of sines to find A. 

sinA 

sinB 

--=- 

a b 

. A asinB 

Sm =- 

b 

340 sin 39° 50' 

268 

sin A = .81264638 

A = 54°20' or A = 125° 40' 

6. Find A, given C=79°20', c=97.4mm, 

a=75.3 mm. 

Use the law of sines to find A. 

sinA sinC 

--=- 

a c 

. A 75.3sin 79°20' 

Sm = 

97.4 

=.75974194 

Angle A cannot be obtuse, since a < c and Cis 

acute. 

A = 49°30' 

9. a = 10, B = 30° 

(a) The value of b that forms a right triangle 

would yield exactly one value for A. That is, 

b = 10 sin 30° = 5. Also any value of b 

greater than or equal to 10 would yield a 

unique value for A. 

(b) Any value of b between 5 and 10, would 

yield two possible values for A. 

(c) If b is less than 5, then no value for A is 

possible. 

10. a = 10, B =150° 

(a) any value of b greater than 10 would yield 

exactly one possible value for A. 

(b) No values of b would yield two values for A, 

since A has to be an acute angle . 

(c) 

Any value of b less than or equal to 10 would 

yield no possible values for A. 

11. Find A, given a = 86.14 in., b =253.2 in., 

c = 241.9 in. 

Use the law of cosines to find A. 

a 2 = b 2 +c 2 - 2bc cos A 

b2+c2_a2 

cos A = 

2bc 

253.2 2 +241.9 2 -86.14 2 

= 

2(253.2)(241.9) 

cos A = .94046923 

A = 19.87° or 19° 52' 

307



12. Find b. given B = 120.7°. a = 127 ft, C = 69.8 ft. 

Use the law of cosines to find b. 

b 2 =a 2 +c 2 -2accos B 

= 127 2 + 69.8 2 - 2(127X69.8)cos 120.7° 

b=173ft 

13. Find a. given A = 51 ° 20'. C = 68.3 m. b = 58.2 m; 

Use the law of cosines to find a. 

a 2 =b 2+c2 -2bccosA 

a = ~(58.2)2 + (68.3)2 -2(58.2X68.3)cos51° 20' 

a=55.5 m 

14. Find B. given a = 14.8 rn,b = 19.7 m, c = 31.8 m. 

b 

Use the law of cosines to find B. 

a 2 +c 2 _b 2 

cosB=---- 

2ac 

14.8 2 + 31.8 2 -19.7 2 

= 

2(14.8X31.8) 

B = 265° or 26° 30' 

15. Find a, given A = 60°. b = Scm, c = 21 em. 

Use the law of cosines to find a. 

2 = b 

2 

a + c 2 - 2bc cos A 

a = ~(5}2 + (21)2 - 2(5X21)cos 60° 

a =19cm 

16. Find A. given a = 13 ft, b = 17 ft, c = 8 ft. 

b 2 +c 2 _ a 2 

cosA=--- 

2bc 

17 2 +8 2 _13 2 

2(17X8) 

A =47° 

17. Solve thetriangle, given A = 25.2 0, a = 6.92 yd. 

b =4.82 yd. 

a > b tells us that only one triangle is possible. 

Use thelaw of sines to find B. 

sinB smA 

--=- 

b a 

. B bsinA 

sm · =- 

a 

. B 4.82 sin 25.2° 

sm =---- 

6.92 

B=I7.3° 

(Angle B cannot be obtuse, since b < a, B < A, 

and A is acute.) 

C =18O"'-A- B= 137.5° 

Use the law of sines to find c. 

c a 

--=- 

sinC sinA 

asinC 

c=- 

sin A 

6.92 sin 137.5° 

= 

sin25.2° 

= 11.0 yd 

18. Solve the triangle, given A = 61.7°, a = 78.9 m, 

b= 86.4 m. 

Since b > a> b sin A, there are two triangles . 

sin B sin A 

--=- 

a 

. B 86.4 sin 61.7° 

sm =---- 

78 .9 

=.96417292 

Bt = 74.6° 

C 1 =180 0 - A - B = 43.7° 

_c_l_=_a_ 

sinC I sin A 

78.9sin43.7° 

cl=---- 

sin 61.7° 

=61.9 m 

Bz = 180° - Bt = 105.4° 

C2 = 180 0-A-Bz = 12.9° 

~=_a_ 

sinC 2 

sinA 

78.9sin 12:9° 

c2 = sin 61.7° 

=20.0m 

19. Solve the triangle, given a = 27.6 em. b = 19.8 em, 

C=42° 30'. 

Using the law of cosines, we have 

2 2 

+ b 2 

c = a - 2ab cos C 

c 2 = 27.6 2 + 19.8 2 -2(27.6XI9.8)cos42° 30' 

c= 18.7 em. 

Note: Keep all values of...[ in your calculator 

for use in the next calculation. If 18.7 is used, the 

answer will vary slightly due to round-off error. 

Now use the law of sines to find B so there is no 

ambiguity. 

c b 

--=- 

sinC sinB 

18.65436576 19.8 

=- 

sin 42° 30' sin B 

. B 19.8sin42° 30' 

SID =---- 

18.65436576 

B=45° 50'. 

308



Since the sum of the angles of a triangle is 180°, 24. Given a = 43 m, b = 32 m, c = 51 m, find the area. 

A = 180°- B-C 

Use Heron's formula to find the area. 

= 180° - 45° 50' - 42° 30' 

1 

s=-(a+b+c) 

A =91° 40'. 

2 

1 

= -(43+ 32 + 51) 

20. Solve the triangle, given a = 94.6 yd, b =123 yd, 

2 

c = 109 yd. 

=63 

Using the law of cosines, we have 

123 2 + 109 2 -94.6 2 

Area = ..j's('-s---a:-)(-s-----,-,b):-(s---c-) 

cosA =------ 

= ,J63(63 - 43)(63 - 32)(63 - 51) 

2(123)(109) 

A =47.7°. 

= 680m 2 

Again, using the law of cosines, we have 

94.6 2 + 109 2 -123 2 

25. Substitute a = 7, b = 7.J3, c =14, A = 30°, 

cos B = --..,.......--..,-..,.......--,-- 

2(94.6XI09) 

a+b cos.L(A- B) 

B = 60° and C = 90° into __ = _...=2:....:-._..: 

B= 73.9°. 

, c sintC 

Since the sum of the angles of a triangle is 180°, 

in order to verify the accuracy of the information. 

C=1800-A-B 

a+b 7+7.[3 

= 180° - 47.7° -73.9° 

--=-- 

c 14 

= 58.4°. 

7(1+ .J3) 1+ .J3 

= =:- 

21. Given b = 840.6 m, c = 715.9 m, A = 149.3°, find 

14 2 

the area. 

cOS!(A-B) _ cos-t(300-600) 

Angle A is included between sides band c. 

sin!C - sin-t(900) 

area = !bcsin A 

2 

= cost(-300) 

= !(840.6)(715.9)sin 149.3° 

sin 45° 

2 

cos 15° 

area = 153,600 m 

2 

=-- 

sin 45° 

.,fi +../6 . 

22. Given a = 6.90 ft, b = 10.2 ft, C= 35° 10', find the 

=-.r- =__ 

1+.J3 

area. Angle C is included between sides a and b. 

if 2 

Area =!absinC 

(Use half-angle formulas to find cos 15°.) 

2 

. . 1 1+..[3 

Each expression IS equa to - --. 

= ! (6.90XIO.2) sin 35° 10' 

2 

2 

Area = 20.3 ft2 

26. Substitute a = 7, b = 7.J3, c = 14, A = 30°, 

23. Given a = .913 km, b = .816 km, c = .582 km, find 

a - b sin.L(A - B) 

B = 60°, and C = 90° into --= 2.L in 

the area. 

c 

cos~C 

1 

s=-(a+b+c) 

order to verify the accuracy of the information. 

2 

a-b 7-7-./3 

=----'- 

= !(.913+ .816+.582) 

c 14 

2 

= 1.1555 

7(1- -./3) 1- -./3 

= =- 

area = ~(1.1555)(.2425)(.3395X.5735) 

14 2 

area = .234 km 2 

309



sint(A-B) sint(300-600) 

Use the law of sines to find AC = b. 

b c 

coste - cos t (90°) --=- 

sinB sinC 

= sint(-300) 

b = csinB 

cos 45° 

sinC 

-sin 15° 

= 

7.0sin68° 

cos 45° = 

sin 30° 

=13 

The tree is 13 meters tall. 

=__4___ :fi=:il 1- .J3 

..fi - 2 

2 

(Use half-angle formulas to find -sin 15°.) 

.. I 1-.J3 

30. Let c = the length of the tunnel. 

A 

E ac h expression IS equa to --. 

2 

B 

1:1. Since B = 58.4° and C= 27.9°, 

A = 180 - (58.4 + 27.9) = 93.7° 

Using the law of sines: 

C 

125 AB 

Use the law of cosines to find c. 

--=- 

sinA sinC 

c 2 = 3800 2 + 2900 2 - 2(3800X2900) cos 110° 

AB = (125)sin e c 2 = 30,388,124 

sin A c=5500 

_ (125Xsin27.9°) 

The tunnel is 5500 meters long. 

- sin(93.7°) 

31. Let h =the height of tree. 

=58.6 

The canyon 58.6 feet across. 

28. The angle opposite the 8.0 ft flagpole is: 

180 - (l15 + 22) = 43° 


14.3' 

8 x 

--= 

8 = 27.2° -14.3° = 12.9° 

sin 43° sin 115° a = 90° - 27.2° = 62.8° 

8sin115° 

x=-- 

h 212 

sin43° 

--=- 

sin8 sina 

=11 h= 212sinI2.9° =53.2 

The brace is 11 feet long. 

sin 62.8° 

The height of the tree is 53.2 ft. 

29. Let Ae= the height of the tree. 

I C 

I 

I 

I 

8~ 

I 

I 

.- 

A I 

B 

o 

Angle A = 90° - 8.0° = 82° 

Angle e = 180° - B A = 30° 

310



32" Sketch a triangle showing the situation as follows. 

N 

N 

rock 

B 

35. Let d = the distance between the submarine and 

the battleship. 

17"30' 

24"\0' 

45"20' 

A l'--------.,~ 

15.2 C 

308"40' 

Angle A = 90° - 45° 20' = 44° 40' 

Angle C = 308° 40' - 270° = 38° 40' 

Angle B = 180° -A - C= 96° 40' 

Use the law of sines to find BC = a. 

a b 

--=- 

sinA 

sin B 

bsinA 

a=- 

sinB 

15.2sin44° 40' 

= 

sin 96° 40' 

a = 10.8 

The distance between them is 10.8 miles. 

33. Let x = the distance between the boats. 

In 3 hours, the first boat travels 3(36.2) =108.6 km 

and the second travels 3(45.6) = 136.8 km. 

\08.6 

x 

= 

sin 6° 40' sin 155° 50' 

d = 1450 ft 

The distance between them is 1450 ft. 

36" To find the angles of the triangle formed by the 

ship's positions with the lighthouse, we find the 

supplementary angle to 55°: 

180° - 55° = 125° 

and the third angle in the triangle: 

180° - (125° + 30°) = 25° 

Lighthouse 

Use the law of cosines to find x. 

x 2 = (108.6)2 + (136.8)2 

- 2(108.6X136.8)cos 54° 10' 

=13,113.359 

x = 115 km 

They are 115 km apart. 

34. Sketch the parallelogram. The figure has equal 

opposite sides and an les. 

12 em 147" 

---=-- 

sin 25° sin 30° 

2sin30° 

x= sin 25° 

= 2.4 mi 

The ship is 2.4 miles from the lighthouse. 

15cm 

Let d 1 and d2 be the diagonals of the 

parallelogram. 

d l 

2 =12 

2 

+15 2 - 2(12)(15)cos33° 

d l =8.2 em 

d 2 

2 = 12 2 + 15 2 - 2(12)(15)cos 147° 

d 2 =26 cm 

The diagonals are 8.2 cm and 26 em. 

311



37. Let A = home plate, B = first base, C = second 

base, D =third base, P =the pitcher's rubber. 

Draw ACthrough P, draw PB and PD. 

C 

D £..__.!+---:;7'B 

A 

In triangle ABC, angle B = 90°, 

angle A = angle C = 45°. 

AC= ~902 +90 2 =90../2 

PC =90../2- 60.5 = 66.8 ft 

In triangle APB, angle A = 45°. 

PB 2 =Ap 2 +AB 2 -2(AP)(AB)cos45° 

= 60.5 2 + 90 2 - 2(60.5)(90)cos 45° 

PB= 63.7 ft 

Since triangles APB and APD are congruent, 

PB = PD = 63.7 ft. 

The distance to both first and third base is 63.7 ft 

and the distance to second base is 66.8 ft. 

38. Let x = the distance between the ends of the two 

equal sides. 

JC 

246.75 ~ 

ft 246.75 ft 

x 2 = (246.75)2 +(246.75)2 

- 2(246.75)(246.75) cos 125° 12' 

x =438.14 

The distance between the ends of the two equal 

sides is 438 .14 feet. 

39. Use the distance formula to find the distances 

between the points. 

Distance between (-8, 6) and (0, 0) : 

~(-8-0)2 +(6-0)2 = 10 

Distance between (-8, 6) and (3, 4): 

~(-8 - 3)2 + (6 - 4)2 = ../125 = 11.2 

Distance between (3, 4) and (0, 0): 

~(3-0)2 +(4-0)2 = 5 

1 

s = -(10+ 11.2 +5) 

2 .. 

s =13.1 

Area = ../--(1-3.-1)-(3-.1)-(1-.9-)(-8.-1)= 25 

The area of the triangle is 25 square units. 

40. Divide the quadrilateral into two triangles. 

The area of the quadrilateral is the sum of the 

areas of the two triangles. 

Area = ! (65)(130) sin 80° + ! (120)(120) sin 70° 

2 2 

=11,000 ft 2 

41. a=7, b=7,c=6 

1 

s =- (7 + 7 + 6) =10 

2 

area = ~r-10(--;-3-:-:-X3-:-:)(---:-4) 

= ../360 

= 18.97 m 2 

18.97 m 2 ... 7.5 m 2 /can ""2.5 cans 

He must buy 3 cans. 

42. If C = 90°, then 

c 2 =a 2 + b 2 - 2abcos90° 

=a 2 + b 2 - 2ab(0) 

=a 2 +b 2 

This is the Pythagorean Theorem. 

43. a+ b 

44. a-b 

45. a + 3c 

/ 

/ \ 

/ \ 

/ 

/ \ 

\ 

\ 

\ 

\ 

\ 

46. (a) true 

(b) false 

312



47. Force~ol.!.SJ~ and 23 Ib, forming an angle of 87° 52. horizontal: 

x = Ivlcos 0 =964 cos IS4° 20' 

'" 964 cos IS4.333° =869 

vertical: 

23 

y = Ivlsin 0 =964sin IS4° 20' 

'" 964 sin IS4.333° = 418 

53. u = (21, -20) 

15 

0= 180°-87° = 93° lui = ~212 + (-20 2) 

Use the law of cosines to find Ivl . 

= ..J841 =29 

Ivl 2 =IS 2 +23 2 - 2(IS)(23)cos 93° a 21 

2 

cosO=-= 

Ivl =790 lui 29 

Ivi =281b 0 =43.6° 

But u is in the 4th quadrant, so 

48. Forces of 142 newtons and 21S newtons, forming 0= 360°-43.6° = 316.4°. 

an angle of 112° 

180° -------- 

- 112° = 68° 54. u = (-9, 12) 

68' \ 

\ lui =~(_9)2 + 12 2 

\ 

\ = ..J22S =IS 

\. 8 9 

cosO=-=- 

215 newtons lui IS 

1u1 2 = 215 2 + 142 2 - 2(215)(142)cos68° 

0=126.9° 

lui = 209 newtons 

55. (8) u,v=(6, 2}'(3, - 2) 

49. Forces of 475 Ib and 586 lb, forming an angle of =18-4 

78° 20' =14 

9 

(b) lui = ~62 + 2 2 =.,f40 = 2.JW 

Ivl = ~32 +(_2)2 =.,fl3 

u·v 

cosO = IuIlvl 

586lb 

o= 180° - 78° 20' = 101° 40' {u.v) 

o= arcco lui Ivl 

1u1 2 = 475 2 + 586 2 - 2(475)(586) cos IO!° 40' 

lui = 826 Ib -cos-t( 14 ) 

- 2-JlO .:.rrr 

50. horizontal: = 52.13° 

50 ·../2 -J2 

x = IvlcosO = 50cos45° =--= 25 2 56. (8) u -v =(2,,[3 , 2) .(5,5,,[3) 

2 

vertical: =10,,[3+ 10,,[3 

y = IvIsin 0 =50sin 45° = 50 ·../2 = 25../2 

2 

51. horizontal: x = Ivlcos0 = 69.2 cos 75° = 17.9 

vertical: y = Ivlsin 0 = 69.2sin75° = 66.8 

= 20,,[3 

313



2 

(b) lui= ~(2.J3)2 + 2 =.J16 = 4 

Ivl = ~52 +(5.J3t = ../100 = 10 

u·v 

cosO =IuJIvI 

61. Let 0 = the angle that the hill makes with the 

horizontal. 

D 

o= arcco{ 2~~J . 

=cos-t~J 

=30° 

57. lui =~(-4)2 + 3 2 =..fi5 =5 

u=;1 =¥=(-~,~) 

58. ~JI = ~52 + 12 2 = ../169 = 13 

u=~= (5,12) =(2- g) 

lui 13 13' 13 

59. Let x = the resultant vector. 

N 

100 

'/ 

/ 

/ 

/ 

x //200 

---~// 

100 

a = 180° -45° = 135° 

1x1 2 = 200 2 + 100 2 - 2(200)(100)cos 135° 

Ix! = 280 

Using the law of cosines again, we get 

200 2 -100 2 - 280 2 

cosO=-----.,.....- 

-2(100)(280) 

0=30.4° 

The force is 280 newtons at an angle of 30.4° 

with the first rope. 

60. Let Ixl = the resultant force. 

~I~).._~ 

.....................x 10. 

18 

0= 180° -15° _10° = 155° 

1x1 2 = 12 2 +18 2 - 2(12)(18)cos155° 

Ixl=29 

The magnitude of the resultant force on Jessie and 

the sled is 29 lb. 

C 

B 186 

The downward force of 2800 Ib has component 

AC perpendicular to the hill and component CB 

parallel to the hill. AD represents the force of 186 

Ib that keeps the car from rolling down the hill. 

Since vectors AD and BC are equal, lRel = 186. 

Angle B = angle EAB because they are alternate 

interior angles so the two right triangles are 

similar. Hence, angle 0 = angle BAC. 

SID 

. B'AC 186 

J'\ =- 

2800 

HAC =3° 50' =0 

The angle that the hill makes with the horizontal 

is 3° 50' 

62. Let v = the ground speed vector. 

N 

310' 

a = 212° -180° = 32° 

f3 =50° 

because they are alternate interior angles. 

Angle opposite 520 =a + f3 = 82° 

sinO sin 82° 

--=- 

37 520 

0=4° 

Bearing = 360° - 50° - 0 = 306° 

Angle opposite v = 180° - 82° _4° = 94° 

---.!!L=~ 

sin 94° sin 82° 

Ivl= 524 mph 

The pilot should fly on a bearing of 306°. Her 

actual speed is 524 mph. 

314



63. Let v =the resultant vector. 

!--...130· 

! " 7 

I 

I "' "' 

-, 

I 

I "' "' 

-, 

I 

I 

-, 

"' 

! A 7 v 

Angle A =180° - (130° - 90°) =140° 

Use the law of cosines to find the magnitude of 

the resultant v. 

Ivl 2 =15 2 + 7 2 - 2(15}{7)cos 140° 

Ivl =21 

The resulting speed is 21 kph. 

Use the law of sines to find a. 

sina sin 140° 

--= 

7 21 

. 7sin140° 

sma=-- 

21 

a=12° 

The bearing is 130° - 12° =118°. 

64. Component of C along track is ICI cos 8. 

Component of weight along track is !wi sin 8 . 

We get 

IQcos8 = !wisin 8 

or tan 8 = /Cl 

!wi 

.aut 

=_7_ 

65. 

1 

= 

2 

8 = tan- I .5 = 27° 

32m 

2 

v 

= 

32r 

40 

= 2 

32(100) 

In each of the triangles ABP and PBC we know 

two angles and one side. Solve each triangle using 

the law of sines. 

AB= 92.13sin63° 4' 25" 

sin 2° 22' 47" 

"" 1978.28 ft 

BC = 92.13sin 74° 19' 49" 

sin 5° 13' 11" 

"" 975.05 ft 


1. Using the law of sines: 

sin 25.2° sin B 

=- 

6.92 4.82 

. B (sin 25.2°)(4.82) 

sm 

=-'-----'-'--~ 

6.92 

. _1( Sin 25.2 0 )(4.82») 

B =sm 

6.92 

== 17.3° 

Using the fact that the angles of a triangle sum to 

180°: 

C= 180-(A+ B) 

=180-(25.2+17.3) 

= 137.5° 

The angle C is 137.5°. 

2. Using the law of cosines: 

c 2 =a 2 + b 2 - 2abcosC. 

= (75)2 + (130)2 - 2(75)(130) cos 118° 

"" 31679 

c=..}31679 

== 180 km 

c is approximately 180 km long. 

3. Using the law of cosines: 

b 2 =a 2 +c 2 -2accosB 

2ac cos B =a 2 + c 2 - b 2 

a 2 +c 2 _ b 2 

cosB=--- 

2ac 

2 2 

_(a 2 +c _b ) 

B=cos 1 

2ac 

_1(17.3)2 + (29.8)2 - (22.6)2) 

=cos 

2(17.3)(29.8) 

== 49.0° 

B is approximately 49.0° . 

315


316 Chapter 7 Applications or Trigonometry and Vectors 

4. A=.!.absinc 

2 

= .!.(75)(130)sin 118° 

2 

=4300 km 2 

(Note : Since c was found in Exercise 2, Heron's 

formula can also be used.) 

The area of the triangle is approximately 

43ookm 2 . 

5. Since B > 90°, b must be the longest side of the 

triangle. 

(8) b » 10 

(b) none 

(c) b s 10 

6. The semi-perimeter s is: I 

s=.!.(a+b+c) 

2 

= .!.(22 + 26 + 40) 

2 

=44 

Using Heron's formula 

A =~s(s-a)(s-b)(s-c) 

= ~44(44 - 22)(44 - 26)(44 -40) 

=.../69696 

= 264 square units 

The area of the triangle is 264 square units. 

7. In any triangle, the sum of the lengths of any two 

sides must be greater than the length of the 

remaining side. 

8. 1v1=~(-6)2+82 =.../100=10 

11. u ·v=(-I, 3)·(2, -6) 

=-2-18 

=-20 

12. Find angle C: 

C=180-(47° 20'+24° 50') 

= 180-(72° 10') 

=107° 50' 

Use this information and the law of sines to find 

AC: 

8.4 AC 

----=--=---"":' 

sin 107° 50' sin47° 20' 

AC= 8.4 sin 47.333° 

sin 107.833° 

== 6.5 mi 

Drop a perpendicular line from C to segment AB: 

C 

B ~ A 

8.4mi 

then sin 24.833° = ~ 

6.5 

h = (6.5)(sin 24.833°) 

z2.7 mi 

The balloon is 2.7 miles off the ground. 

13. By the Pythagorean Theorem, we know that the 

distance from the home plate to second base is 

~602 + 60 2 =roJ2 ft 

Since cos 0 = iJ!Ji' 

o=cos-l(~)=cos-t ~) 

=cos-t~) 

tanO=.! 

x 

O=tan-l(~) 

1 

= tan- ( ~) 

z 126.9° 

The magnitude /vi is 10 and 0 = 126.9°. 

9. u+v={-I, 3}+{2, -6} 

=(-1+2), (3+(-6»)) 

={1, - 3} 

10. -3v = -3{2, - 6) 

=(-6, 18) 

Using the law of cosines: 

x 2 =a 2 +b 2 -2abcosO 

=(46)2 +(60)2 -2(46X6O)cos45 0 

z 1813 

x =.../1813 

z 43 ft 

The distance from the pitcher'S mound to third 

base is 43 feet. 

316


14. horizontal : x =IvlcosO=569 cos 127.5° "" -346 

vertical: y = IvjsinO = 569 sin 127.5° "" 451 

The vector is (-346, 451). 

15. Transmitter 

C 

west -..L...,:~-------+~...L..+- east 

Consider the figure. 

Since the bearing is 48° from A, the angle A in 

6.ABC must be 90° - 48° = 42° . Since the bearing 

is 302° from B, the angle B in 6.ABC must be 

302° - 270° =32°. The angles of a triangle sum to 

180°, so 

C = 180 0-(A + B) 

= 180° -(42° + 32°) 

=106° 


b c 

--=- 

sinB sinC 

b 3.46 

--= 

sin 32° sin 106° 

b = (3.46)(sin 32°) 

sin 106° 

=1.91 mi 

The distance from A to the transmitter is 

1.91 miles. 

.... . 


317


318 Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations 

CHAPTER 8 COMPLEX NUMBERS, 

POLAR EQUATIONS, AND 

PARAMETRIC EQUATIONS 

Section 8.1 


1. A beginning algebra student might have difficulty 

with the concept, because no real number 

(positive, negative, or zero) has a square that is 

negative. The student might wonder "How can a 

square be negative" 

2. i 4 = 1 since (i 2 )2 = (_1)2 = 1; 

.3 . , .3 .2 - 1 - 

I = -I, sInce I = I - I = - -I = -I 

Exercises 

1. B; .J:::s =R .$ =i$ 

2. C; i.J:::s = i -R .$ = i 2 $ = -$ 

3. D; -i..J=5 = (-i).J=i .$ = (-i)(i).J5 

=_i 2 $ =-(-1)$ =$ 

4. A; -J=5 =-.J=i ...[5 =-i$ 

5. (a) Imaginary Numbers 

(b) Rational Numbers 

2=-16 

6. x 

x=±..J-16 

x=±i.J16 

x=±4i 

The solutions are 4i and -4i. 

7. y2 =-36 

y=±..J-36 

y=±i../36 

y=±6i 

The solutions are 6i and -6i. 

8. z2 +12 =0 

z2 =-12 

z=±..J-12 

z=±i..Jli 

z= ±i(2..[3) 

z=±2i..[3 

The solutions are 2i..[3 and -2i..[3. 

9. w 2 +48=0 

w 2 =-48 

w =±..J-48 

w =±i.,J48 

w=±i4..[3 

w =±4i..[3 

The solutions are 4i..[3 and - 4i..[3. 

10. 3x 2 +4x+2 =0 

Use the quadratic formula with 

a =3, b =4, and c =2. 

_4±~42 -4.3-2 

x =---=---- 

2 ·3 

-4±..J16-24 

x= 

6 

-4±H 

x= 

6 

-4±i../8 

x= 

6 

2(-2 ±i...fi) 

x= 

6 

-2±i...fi 

x= 

3 

· 2 ...fi· 2...fi. 

The so I unons are --+-1 and ----I. 

3 3 3 3 

11. 2k 2 + 3k =-2 

2k 2 +3k+2 =0 


a =2, b = 3 , and c =2. 

-3±~32 -4·2·2 

k = -----:~--- 

2·2 

-3±..J9-16 

k =......:...:;:....:...:...-....:.. 

4 

-3±H 

k =......:...----:. 

4 

k= -3±i.J7 

4 

- 3.J7. d 3 ..fi . 

Th e so I unons are --+-1 an ----I. 

4 4 4 4 

318



12. m 2-6m+14=0 15. 9a 2 + 7 =6a 


9a 2 -6a+7 =0 

a = I, b = - 6, and c = 14. 

Use the quadratic formula with a =9, b = 6, and 

-(-6)±~(-6i -4 ·1·14 

c=7. 

m= 

2 ·1 -(-6) ± ~(_6)2 - 4·9·7 

6±.J-20 

a= 

m=-"""';"" 

2·9 

2 6±..J=2l6 

6+ i.,fiO 

a= 

m= 

18 

2 

6(1 ±i../6) 

2(3± i../5) 

a= 

m =---'-----' 

18 

2 

m =3±i../5 

The solutions are 3 +i../5 and 3 - i../5 . 

l±i../6 

a=- 

3 

Th I · 1../6. d l../6. 

e so utions are -+-1 an ---I. 

3 3 3 3 

13. p2+ 4p+ 11=0 

Use the quadratic formula with 16. m 2 +1 =-m 

a = I, b =4 , and c = II. 

m 2 +m+1 =0 

-4±~42-4 ·1 ·11 Use the quadratic formula with a = I , b =I, 

p 

-4+.J-28 

p=--'-- 

2 

2(-2 ± i.,fi) 

p= 

2 

p=-2±i.,fi 

2 ·1 and c= 1. 

m= 

-l±~12_4.1 .1 

2 ·1 

-1±.J=3 

m=-"""";' 

2 

-l±i$ 

The solutions are -2 + i.,fi and - 2 - i.,fi . m= 

2 

14. 4z 2 =4z - 7 

4z 2 -4z+7= 0 


a =4, b =- 4, and c = 7. 

-(-4)±~(-4)2 -4 ·4·7 

z= 

. 1../3' ·d 

The solutions are --+-1 an 

2 2 

17. i =2y-2 

I $. ----I. 

2 2 

i-2y+2=0 

Use the quadratic formula with a = I, b = -2, and 

c=2. 

2·4 

4±.J=% 

z= 

-(-2)±~(-2)2-4·1 ·2 

8 y= 2 ·1 

4±i..j% 

z= 

8 

4(1 ±i../6) 

z= 

8 

I ±i../6 

z=- 

2 

· 1../6. d 

Th e so I unons are -+-1 an 

2 2 

l ../6 . 

---I. 

2 2 

2±R 

y= 

2 

2(1 ±i) 

Y=-2 

y= I±i 

The solutions are I + i and I - i. 

19. .J=3..J=3 =i../3· i../3 

=-3 

20. -Pi .-Pi =i..fi .i..fi 

=i 2 (..fi)2 

= (-1)·2 =-2 

319



320



-~ 

47. (2 - i)(2 + i) =4 + 2; - 2i - i 2 

=4-(-1) 

53. 

4 +; 4+i 6-2i 

--=-_._ 

6+2i 6+2; 6-2; 

=5 or 5+0; 24-2i-2;2 

= 

Note that the product of conjugates is the 36-4;2 

difference of squares. 

24 - 2; - 2(-1) 

(2 - i)(2 +i) =2 2 = 

_;2 

36-4(-1) 

=4 -(-I) 26-2i 

=- 

=5 40 

13-; 

48. (5+ 4;)(5 - 4i) = 52 - (4;)2 

= 25-(-16) 

=41+0ior41 

49. 5 5 O-i 

-=-_._ 

0+; O-i 

5 -i 

=_. 

; -i 

SO. 

51. 

52. 

54. 

=- 

20 

13 1 . 

=---1 

20 20 

3-2; (3-2i) (5-3i) 

-=~---=- ~---=- 

5+3; (5+3;) (5-3i) 

15-19;+6;2 

=-----:::- 

25 -9;2 

-5i 9-19i 

=~ 

=- 

-I 34 

-5; 

9 19. 

=- 

=---1 

-(-1) 34 34 

-5; 

= 

1 57. ;12 = (i 4 )3 

=-5i =1 3 

orO- 5; 

=1 

-4 -4 0-; 

-=-_.- 

58. ;9 =;8. i 

= (i4)2 .; 

O+i O-i 

(-4)(-i) 

= 

= 1 2 'i 

(i)(-i) 

= l·i 

4i =; 

=~ 

-I 

4; 

= 

1 

= 0+4i or 4i 

12-8; 12- 8; 5-i 

--=-_. 

5+i 5+i 5-i 

60 -52i+ 8i 2 

= 

25- ;2 

52-52; 

= 

26 

,. . 

59. ;18 = i16.;2 

=(i4)4'(_1) 

= 1 4 .(- 1) 

=1·(-1) 

=-1 

60. 

·99 .96.3 

1 = 1 '1 

= (i4)24 .;3 

= (1)24 .(-;) 

=2-2; = 1(-i) 

=-; 

-13-141 -13-14i 8-3; 

= 

·-3 

8+3; . 8+3; 8-3i 61. 

.-4 ·.1 

1 = 1 '1 

-104 -73; + 42;2 

= (;4)-1 . ; 

= 

64-9i 2 

-146-73i 

= 1-1.; 

= 

73 = 1·; 

=-2-i =i 

321



.... 

62. z 

·-5 

= z 

·-8 

·z 

·3 69. 1= 7 + 5;, E = 28 + 54; 

= (;4)-2 .(_;) 

E=IZ 

E 

= 1- 2 .(-i) Z= Ī 

= 1·(-;) 28+54; 

= -; = 

7+5; 

28+54; 7-5; 

.-10 

63. 

·- 12 .2 

z =z · z = 

7+5; 7-5; 

= (;4)-3 .(- 1) 196-140; + 378; - 270;2 

= 

= r 3 .(- 1) 49-25;2 

=1 ·(-1) 466+238; 

= 

=-1 74 

233 119. 

;-40 = (;4)-10 

=-+-z 

64. 37 37 

= 1- 10 70. E = 35 + 55;, Z = 6 + 4; 

=1 

E=IZ 

35 + 55; = I( 6 + 4;) 

65. 1+; +;2 +;3 

35+55; 

= 1+; +(-1)+(-;) 1= 6+4; 

=0 

35+55; 6-4; 

= 

6+4; 6-4; 

66. 1+; +;2 +;3 + ... + ;100 

210 -140; + 330; - 220;2 

=(1+;+;2 +;3)+(;4 +;5 +;6 +;7) = 

36-16;2 

+(i8 +;9 +;10 +;11)+ ... 430+ 190; 

+(i96 + ;97+ ;98+ ;99) + ;100 = 

52 

Notice that each set of 4 terms within parentheses 215 95. 

=-+-z 

can be simplified to 26 26 

1+; +;2 +;3 

71. From the figure in the text. the light bulbs have 

= 1+; +(-1)+(-;).= O. 

resistances of 50 and 60 ohms and the motors 

Then ;100 = (i4)25 = 1 25 = 1. 

have reactances of 15 and 17 ohms, respectively. 

Therefore, 

Thus, the total impedance is given by 

1+ i + ;2 + ;3 + ... + ;100 Z = 50+ 60+ 15; + 17; 

= 110+ 32; 

= 0+0+0+...+0+ 1 

=1. 72. From Exercise 71, a = 110 and b = 32. Then 

b 

67. 1= 8 + 6;, Z =6 + 3; tan6=a 

E=IZ 32 

= (8 + 6;)(6 + 3;) 

tan6= 

110 

= 48+24;+ 36; + 18;2 6"" 16.22°. 

= 48 + 60; + 18(-1) 

=30+60; 

68. 1= 10 + 6;, Z = 8 + 5; 

E=IZ 

E = (1O+6;)(8+5i) 

= 80+50;+48; + 30;2 

= 80+98;+30(-1) 

=50+98; 

322



73. (~+ ~.J 

74. 

=( ~)' +2~~ i+(~ i)' 

2 4. 2 .2 

=-+-1+-1 

4 4 4 

1 . 1 

=-+1+-(-1) 

2 2 

1 . 1 . 

=-+1--=1 

2 2 

SİDce 

root of i. 

(..fi..fi .)2 ..J2 ..fi .. 

-+-1 =I,-+-Ilsasquare 

2 2 2 2 

Section 8.2 


1. (6, -2}-(-4, -3) 

=(6-(-4), -2-(-3)} =(10, I} 

or 10+ i 

y 

-4-3; -4 

2. Answers will vary. 

Exercises 

lO+i 

1. The modulus of a complex number represents the 

magnitude (or length) of the vector representing it 

in the complex plane. 

2. It is the angle formed by the vector and the 

positive horizontal axis. 

3. 

y 

-2+3; 

3 

+-t-+-t-='-+-t-+~ JC 

-2 

4. 

y 

5 

=i 

t+++++n'H-H-H~JC 

.J3 1. . be f' 

Th ere ti ore - - + -I IS a cu root 0 I. 

2 2 

75. The step ..[::t...[::t= ~( -1)(-1) is invalid, since 

the rule ..r;;:b = ~ ..,Jb does not apply when 

botha and b are negative. 

77. b = 0 

5. 

y 

323


324 Chapter 8 Complex Numbers and Polar Equations 

6. 

y 11. 

y 

++++++d,rt-++-+4~ x 

-8 

7. 

6-5; 

y 12. 

y 

x 0 

2 

x 

8. 

-4 

y 13 1-4; 

14. -4-; 

6 

15. 4 -3;, -1+2; 

x 

(4 - 3;)+ (-1 + 2;) =3


25. 2(cos 45° + i sin 45°) 

=2(~ + i~)=~ +;~ 


34. 6 cis 135° 

= 6(cos 135° + i sin 

={-~+~i) 

26. 4(cos 60° +; sin 60°) =-~+(~); 

=4(~+i~) 

= 2 +2;../3 

27. 10(cos 90° + i sin 90°) 

= 10(0+ i) 

=O+lOi= lOi 

28. 8(cos 270° + i sin 270°) 

=-3~+3;~ 

35. -Iicis 180° 

135°) 

=~(cos 1800+; sin 180°) 

= ..{i(-1 +;·0) 

= -..{i +0; 

=-..{i 

= 8(0 -Ii) = -8; 36. .../3 cis 315° 

29. 4(cos 240° +; sin 240°) =.../3(cos 315°+i sin 315°) 

={-~-i~) =../3[~ +{ ~)] 

= -2-2;.../3 

30. 2(cos 330° +; sin 330°) 

=2(~ -~;) 

37. 

.J6 ~ . 

=---1 

2 2 

3-3; =x+ yi 

Find rand () 

x=3.y=-3 

=.../3-; 

r=~x2 +y2 

~ 

31. cis 30° 

= (cos 30° + i sin 30°) 

32. 

33. 

../3 1. 

=-+-1 

2 2 

3 cis 150° 

= 3(cos 150° + i sin 150°) 

={- ~+~;) 

3..J3 3 . 

=--+-1 

2 2 

5 cis 300° 

= 5(cos 300° +; sin 300°) 

={~+~)] 

5 5.../3 . 

=---1 

2 2 

= ~32 +(_3)2 

=.J18 

=~ 

y -3 

tan (}=-=-=-1 

x 3 

3 - 3; is in quadrant IV, so () = 315°. 

3-3; = ~(cos 315°+; sin 315°) 

38. -2 +2;.../3 

Find rand e. 

Since x = -2 and y = 2../3, 

r=~x2+i 

=~(_2)2 +(2../3)2 

= ~4+ 12 =.Jl6 

=4. 

tan () = r = 2../3 =-.J3 

x -2 

325


326 

Chapter 8 Complex Numbers, Polar Equations, and Parametric Equations 

Since - 2 + 2;.J3 is in quadrant II, 6 = 120°. 

Thus, 

43. -5-5; 

x=-5,y=-5 

-2 + 2;.J3 =4(cos 120°+; sin 120°). r = ~(_5)2 +(_5)2 

39. 

40. 

41. 

42. 

-3-3;.J3 =.J50 =5../2 

-5 

x=-3, y=-3.J3 

tan6=-=1 

-5 

r= ~(_3)2 + (_3.J3)2 = 6 -5 - 5; is in quadrant III, so 6 = 225°. 

-3.J3 .J3 -5 - 5; = 5../2(cos 225°+; sin 225°) 

tan 6=--= 3 

-3 

-3-3;.J3is in quadrant III, so 6 = 240°. 44. -J2 +;../2 

-3 - 3;.J3= 6(cos 240° +; sin 240°) x =-J2, y =../2 

r=~(-J2)2 +(../2)2 =2 

1+;.J3 

x=l, y=..J3 

r= ~12 + (.J3)2 =.J4 =2 

../2 

tan6=~=-1 

-J2 +;../2 is in quadrant II, so 6 = 135°. 

-J2 +;../2 = 2(cos 135°+; sin 135°) 

tan 6=1..=.J3 =.J3 

x 1 · 45. 2+2; 

Since 1+;.J3 is in quadrant I, 6 = 60°. 

x=2,y=2 

1+;.J3 = 2(cos 60° +; sin 60°) r = ~22 + 2 2 =.J8 = 2../2 

.J3 -t 

2 

tan 6 =-= 1 

2 

x=.J3, y=-1 2 + 2; is in the quadrant I, so 6 = 45°. 

r=~(.J3)2 +(_1)2 =2 

2+2i =2../2(cos 45~+; sin 45°) 

-1 .J3 

tan6=13=-3 

.J3 -; is in quadrant IV, so 6 = 330°. 

46. -J3+; 

x=-J3,y=1 

.J3-; = 2(cos 330° +; sin 330°) r=~(-J3)2+12 =2 

1 .J3 

4.J3 +4; tan6==J3=3 

x=4.J3,y=4 -J3+ i is in quadrant II, so6 = 150°. 

r=~(4.J3)2+42 =...[64 =8 -J3+; = 2(cos 150°+; sin 150°) 

4 

tan6=4J3 47. -4=-4+0; 

4 3 

x=-4,y=0 

1 .J3 

=.J3=3 r=~(-4)2 +0 2 =4 

Since 4.J3 + 4; is in the quadrant I, 6 = 30°. 

4.J3 + 4; = 8(cos 30° + i sin 30°) 

-4 + 0; is on the negative x-axis, so 6 = 180°. 

-4 = 4(cos1800+;sinI800) 

48. 5; =0+5; 

x=0,y=5 

r=~02 +5 2 =5 

0+5; is on the positive y-axis, so 6 =90°. 

5; = 5(cos 90° + i sin 90°) 

326


49. -2;=0 - 2; 

x=O,y=-2 

r = ~02 +(_2)2 = 2 

0-2; is on the negative y-axis so, 0 = 270°. 

-2; = 2(cos 270° +; sin 270°) 

50. 7 =7+0; 

x=7,y=0 

r=~72 +0 2 =7 

7 + 0; is on the positive x-axis, so 0 = 0°. 

7 = 7(cos 0° +; sin 0°) 

51. 2+3; 

x=2,y=3 

r=~22+32=..Jf3 

tan8=~ 

2 

2 + 3; is in quadrant I, so 8 = 56.31°. 

2 + 3; = ..Jf3(cos 56.31° +; sin 56.31°) 

52. cos 35°+ i sin 35° 

= .81915204 + .57357644; 

53. 3(cos250° +; sin 250°) 

= 3[-.34202014+ (-.93969262);] 

= -1.0260604 - 2.8190779; 

54. -4+; 

x=-4,y =1 

r =~(_4)2 + 1 =.Jfi 

1 1 

tan8=-=- 

-4 4 

-4+; is in the quadrant II, so 0 = 165.96°. 

-4 +; = .Jfi(cos 165.96° +; sin 165.96°) 

55. 12; = 0 + 12; 

x=0,y=12 

r=~02+122=12 

o+ 12; is on the positive y-axis, so 0 = 90°. 

12; = 12(cos 90° +; sin 90°) 

56. 3 cis 180° 

= 3(cos 180°+; sin 180°) 

= 3(-1+0;) 

= -3+0; or -3 

8.2 Trigonometric (polar) Form of Complex Numbers 327 

57. 3+5; 

x=3,y=5 

r = ~32 + 52 =.J34 

5 

tan 0= 3 

3+5; is in the quadrant I, so 8= 59.04°. 

3+ 5; = .J34(cos 59.04°+; sin 59.04°) 

58. cis 110.5° 

= cos 110.5° + i sin 110.5° 

= -.35020738 + .93667219; 

59. Since the modulus represents the magnitude of the 

vector in the complex plane, r =1 would 

represent a circle of radius one centered at the 

origin. 

60. When graphing x + y; in the plane as (x,y) , if x 

and y are equal, we are graphing points in the 

form (x, x). These points make up the line y =x. 

61. Since the real part of Z = x + yi is 1, the graph of 

1+ yi would be the vertical line x = 1. 

62. When graphing x + y; in the plane as (x,y) , since 

the imaginary part is I, the points are of the form 

(x, 1). These points constitute the horizontal line 

y =1. 

63. z =-.2; 

z2 -1 =(-.2;)2 -1 

=-.04-1 

=-1.04 

The modulus is 1.04. 

(z2 _1)2 -1 = (-1.04)2 -1 = .0816 

The modulus is .0816. 

[(z2 _1)2 _1]2 -1 = (.0816)2 -1 

=-.9933 

The modulus is about .9933. 

The moduli do not exceed 2. Therefore, z is in the 

Julia set. 

64. (b) Let zl =a+b;andz2 =a-b;. 

zr -1 =(a + bi)2 -1 

= (a 2 - b 2 -1)+(2ab); 

=c+d; 

z~ -1 =(a - b;)2 -1 

= (a 2 - b 2 -1) - (2ab); 

=c-d; 

The results are again complex conjugates of 

each other. At each iteration, the resulting 

values from zl and z2 will always be 

complex conjugates. 

327



(c) We must show that if (a, b) is on the graph of 

the Julia set, then (a, -b) is also on its graph . 

Notice that the results after the first 

calculation are complex conjugates of each 

other. By part (a), they have the same 

modulus. If the computation is then repeated 

with c + di and c - di from part (b), they will 

again produce complex conjugates with the 

same modulus. 

Either both numbers will be larger than 2 in 

modulus or both will be less than or equal to 

2 in modulus. Thus, either both (a, b) and 

(a, -b) are on the graph of the Julia set 

determined by z2 - 1 or both are not. 

(d) By part (c), if (a, b) is on its graph then, 

a, -b) is also on its graph. This means that 

the graph is symmetric with respect to the 

x-axis. 

(e) We must show that if (a, b) is in the Julia set, 

then (-a, b) is also in the set. Test 

zi =a + bi and Z2 =-a+ bi. 

Zl2 -1 =(a + bi)2_ 1 

= (a 2 - b 2 -1) +(2ab)i 

=c+di 

zl-1 =(-a + bi)2 -1 

= (a 2 - b 2 -1) - (2ab)i 

=c-di 

Again they are complex conjugates which 

have the same modulus. By the same 

reasoning as above, the graph of the Julia set 

is also symmetric with respect to the y-axis. 

67. (a+bi)-(c+di)=e+Oi, so,b-d=Oorb=d. 

Therefore, the terminal points of the vectors 

corresponding to a + bi and c + di lie on a 

horizontal line. (B) 

Section 8.3 

Exercises 

1. multiply, add 

2. divide; subtract 

3. [3cos 60° + i sin 60°)]· [2(cos 90° + i sin 90°)] 

=6(cos 150° + isin 150°) 

=6(- ~ +i±) 

=-M+3i · 

4. [4(cos 30° + i sin 30°)] 

· [5(cos 120° + i sin 120°)] 

=20(cos 150° + i sin 150°] 

=2{-~ +±i) 

=-1M+lOi 

5. [2(cos45° + isin45°)] 

· 2[(cos225 + isin 225°)] 

=4(cos 270° + i sin 270°) 

=4(0-i) 

= 0-4i or -4i 

6. [8(cos3000+isin3000)] 

·[5(cos 120° + isin 120°)] 

=40(cos420° + i sin 420°) 

=40(cos 60° + i sin 60°) 

=41J.. + i .,[3) 

'\2 . 2 

=20 + 20i.,[3 

68. The modulus of a complex number is equal to the 

length of the vector representing it in the complex 

plane. Therefore, the only way for the modulus of 

the sum of two complex numbers to be equal to 

the sum of their moduli is for the two vectors 

representing the numbers to have the same 

direction. (C) 

7. 

[4(cos6O° + i sin 60°)]· [6(cos 330° + i sin 330°)] 

=24(cos 390°+ i sin 390°)] 

=24(cos 30° + i sin 30°) 

=24( ~ +i±) 

69. The difference of the vectors is equal to the sum 

of one of the vectors plus the opposite of the other 

vector. Ifthis sum is represented by two parallel 

vectors the sum of their moduli will equal to 

modulus of their sum. In order for this sum to be 

represented by two parallel vectors their 

difference must be represented by two vectors 

with opposite directions. (A) 

=12.,[3 + 12i 

8. [8(cos2100+ isin2100)] 

. 2(cos330° + isin 330°)] 

=16(cos 540° + i sin 540°) 

=16(cos 180° +isinI800) 

= 16(-1+0·i) 

=-16+0i or -16 

328


8.3 The Product and Quotient Theorems 329 

9. [5 cis 90°][3 cis 45°] 16. 24(cos 150° + isin 150°) 

= 15 cis 135° 2(cos 30° + i sin 30°) 

== 15(cos 135° + i sin 135°) = 12(cos1200 + isin 120°) 

=1{-'7 + '7 i) 

=1{-~+i~) 

15.fi 15.fi. =-6+6i../3 

=---+--1 

2 2 

17. 3cis 305° 

10. [6 cis 120°][5 cis(-300)] 9 cis 65° 

= 30 cis 90° 

=.!. cis 240° 

= 30(cos 90° + i sin 90") 3 

= 30(0+ Ii) 

=.!. (cos 240° + i sin 240°) 

=0+30i 

Of 30i 

3 

=~(-~- ~ i) 

11. [../3 cis 45°][ .J3 cis 225°] 

= 3 cis 270° 1 ../3. 

=----1 

= 3(cos 270° + i sin 270°) 6 6 

=3(O-i) 

=0-3i or -3i 12 cis 293° 

18. 

6cis 23° 

12. [.fi cis 3000][.J2 cis 270°] =2 cis 270° 

= 2(cos 270° + i sin 270°) 

= 2 cis 570° 

=2(0-lrl 

=2 cis 210° = 0-2i or -2i 

= 2(cos 210° + isin2100) 

19. numerator: r =~82 + 0 2 = 8 

=2(-;: -~i) 

tan 0 =~ 

=-./3-t 

13. 4(cos 120° + i sin 120°) 

8 

0=0° 

denominator: r =~(../3)2 + 1 2 =2 

2(cos 150° + i sin 150°) tan 0 =13 

=2[cos(-300) + i sin(-300)] 

0=30° 

8 _ 8cisOo _ 4 . ( 31\0) 

=2(~ -~i) --- - CIS - v 

../3+i 2 cis 30° 

=../3-i 

=4[cos(-300) + isin(-300)] 

IO(cos225° + isin225°) 

14. =4[~ +{-~)] 

5(cos 45° + i sin 45°) 

=2(cosI80° + isin 180°) 

=2../3 -2i 

=2(-1 +O·i) 

=-2+0i 

or -2 

15. 16(cos3000+ isin 300°) 

8(cos 60° + i sin 60°) 

=2(cos 240° + i sin 240°) 

=2(-~- ~ i) 

=-I-i../3 

1 

329



20. numerator: r = ~02 + 2 2 = 2 

tan 0 = ~ is undefined 

o 

0=90° 

denominator: r = ~r-(_-1-)2-+-(-./3-3-)-2 = 2 

2i _ 

---;= 

-1-i..[3 

tan 0 = --J3 

-1 

o= 60° + 180°= 240° 

2cis90° _ . ( 1500) 

- CIS 

2 cis 240° 

= cos(-1500)+ isin(-1500) 

..[3 1. 

=----1 

2 2 

21. numerator: r = ~02 + (_1)2 = 1 

tan 0 = -1 is undefined 

o 

0= 90° + 180° = 270° 

denominator: r = ~1 2 + 1 2 =.J2 

tanO=! 

1 

0=45° 

~ = cis 270° =.J2 cis 2250 

l+i .J2cis 45° 2 

22. numerator: r = ~12 + 0 2 = 1 

=.J2 (cos225° + i sin 225°) 

2 

=.J2 (_ .J2-i .J2) 

222 

1 1. 

=----1 

2 2 

tan0 =Q 

1 

0=0° 

denominator: r = ~22 + (_2)2 = 2.J2 

-2 

tanO= 

2 

0=-45° 

... 1 cis0° .J2 . 450 

--= =-CIS 

2 - 2i 2.J2 cis(-45°) 4 

=.J2 (cos45° + i sin 45°) 

4 

=.J2(.J2 +i.fi)=.!.+!.i 

4 2 2 4 4 

330


8.3 The Product and Quotient Theorems 331 

23. numerator: r = ~(2.,J6)2 +(-2../2)2 = 4../2 

-2../2 1 

tan9=--=- 

2.,J6 ../3 

9 =-30° 

denominator: r = ~r-(../2-2-)-2-+-(-./6-6-)-2 = 2../2 

-./6 

tan 9 = .J2 =-J3 

9=-60° 

2.,J6 -2i.J2 = 4..[2cis(-300) = 2 cis 30° 

.J2 - i.,J6 2../2 cis (-600) 

= 2(cos 30° + i sin 30°) 

=2( ~ +i~) 

=../3+i 

24. numerator: r = ~4 2 + 4 2 = 4../2 

4 

tan9=-=1 

4 

9=45° 

denominator: r = ~'22=-+-(-_-2)-=-2 = 2../2 

-2 

tan9=-=-1 

2 

9 = -45° 

4 + 4i = 4../2 cis 45° = 2 cis 900 

2-2i 2../2 cis (-45°) 

= 2(cos 90° + i sin 90°) 

= 2(0+i) 

=2i 

25. [2.5(cos35°+ isin35°)] .[3.0(cos500 + isin500)] = 7.5(cos85° + isin85°) 

= .65366807 + 7.4714602i 

26. [4.6(cosI2°+ isin12°)] ·[2.O(cos13° + isin 13°)] = 9.2(cos 25° + i sin 25°) 

= 8.3380316 + 3.8880880i 

27. (12cis 18.5°)(3cis 12.5°) = 36cis 31° 

= 36(cos 31°+isin31°) 

= 30.858023+ 18.54137li 

28. (4 cis 19.25°)(7 cis 41.75°) = 28 cis 61° 

= 28(cos61 ° + isin61°) 

= 13.574669 + 24.489352i 

45(cosl27°+ isinl27°) 2( 840 . , 840) 

29• = cos +Ism 

22.5(cos43° + i sin 43°) 

= .20905693 + 1.9890438i 

30 30(cos 130°+ isin 130°) 3( 1090 . , 1090) 

• = cos +Ism 

IO(cos21° + isin 210) 

= -.97670446 + 2.8365557i 

331



36. 2 cis 0° = 2(cos 0° + i sin 0°) 

31. [2 cis 5;r= [2 cis 5;12 cis 5;] = 2(1+ ;(0» 

=2(1) 

. IOlC 

=4 CIS- =2 

9 The result here is the same as the result in 

Exercise 33. 

= 4(cos l~lC + isin l~lC) 

= -3.7587705 -1.3680806i w -1+i 

37. -=- 

z 

-1-i 

-1+i -1+i 

=-_._ 

-1-; -1+i 

l-i - i+ i 2 

=--.....,.... 

1-;2 

-2i 

= 

2 

=-i 

In Exercises 33-39, w = -1 + i and z = -1 - i. 

33. w . z =(-1 + ;)(-1 - i) 

=(-1)(-1) + (-I)(-i) 

+(i)(-I) +(i)(-i) 

=1+;-i-i 2 

=1-(-1) =2 

39. cis (-90°) 

34. w=-I+i =-i 

r = ~r--(--1)-=-2+-1--=-2 =.J2 

= cos(-90") + i sin(-900) 

=O+i(-I) 

The result here is the same as the result in 

Exercise 37. 

1 

tan(}=-=-1 

-1 43. E = 8(cos 20° + i sin 20°), R = 6, 

() =135° X L=3 

w =.,fi cis 135° 

z=-I-i 

r = ~,-(_-1)~2-+-(_-1)~2 =../2 

I=E,Z=R+XLi 

Z 

Write Z = 6 + 3i in trigonometric form. 

-1 X= 6,y = 3, so r =~62 +3 2 =.J45. 

tan(}=-=1 

-1 3 1 

tan(} = - = - so () = 26.6° 

(}=225° 

6 2' 

z = .J2 cis 225° 

Z = .J45 cis 26 .6° 

1= 8 cis 20° 

35. w· z = (../2 cis 135°)(../2 cis 225°) J45 cis 26.6° 

= (.J2)(../2) cis (135° + 225°) = 1.19 cis (-6.6°) 

= 2 cis 360° .... 1.18-.14; 

=2cisO° 

332


44. E=12(cos25°+;sin25°), R=3, Section 8.4 

XL =4, and Xc == 6 

1. [3(cos300+isin300)]3 

1= E 

R+(X L -Xc); 


= 3 3 [cos(3)(300)+;sin(3)(300)] 

12 cis 25° = 27(cos 90° + ;sin 90°) 

= 

3+(4-6); 

=0+27; or 27; 

12 cis 25° 

= 2. [2(cos 135° + ; sin 135°)]4 

3-2; 

=2 4[cos(4 .135°)+isin(4 · 135°)] 

Write 3 - 2; in trigonometric form. x =3, y =-2. = 16(cos 540° + isin 540°) 

== 16(-1+0 ·;) 

so r =~32 +(_2)2 =.Jl3. 

=-16 + 0; or - 16 

2 

tan 0 =--, so 8 = 326.31°. 

3 3. (cos 45° +; sin 45°)8 

Continuing to find the current, we have 

= [cos(45°)(8) +; sin(8)(45°)] 

1= 12 cis 25° 

.Jl3 cis 326.31° 

=1+0; or 1 

=3.3282 cis (-301.31°) 

"'" 1.7+ 2.8;. 

4. [2(cos 120° +;sin 120°)]3 

= 8(cos 360° + i sin 360°) 

45. Since zl =50 + 25; and Z2 =60 + 20;, it follows =8(1+0·i) 

that 

= 8+0; or 8 

1 1 50 - 25i 50 - 25; 

-= = 

ZI 50+25; 50-25i 3125 

5. [3 cis 100°]3 

2 1. = 3 3 cis (3 ·100°) 

=---1; 

125 125 

=27 cis 300° 

1 1 60 - 20; 60 - 20; = 27(cos 300° + i sin 300°) 

-= = 

Z2 60 + 20; 60 - 20; 4000 

3 1. 

=27(~- ~ i) 

=---1. 

200 200 27 27.J3 . 

' - - - - 1 

11[2 I'J[3 I'J 2 2 

ZI + Z2 = 125 - 125 1 + 200 - 200 1 

31 13 . 6. [3 cis 40°]3 

=-----1 

1000 1000 = 3 3 cis (3.40°) 

1 

Z=- =27 cis 120° 

= 27(cosI20° + isinl200) 

*+* 

1 

- 1660 -1&10; 

=27(-~+ ~;) 

1000 31+13; 27 27.J3. 

= =--+--1 

31-13; 31+ 13; 2 2 

31,000+ 13,oooi 

== 

1130 

3100 1300. 

=--+--, 

113 113 

"'" 27.43 + 11.5; 

46. tan 0 = J..!2. 

27.43 

8 = tan- l ...!..!.2- == 22.75° 

27.43 

333



7. (...[3 + i)5 

x =...[3, Y = 1 

r=~(...[3)2+12 =2 

tan 8 = JJ so 8 = 30°. 

(...[3 + i)5 

= [2(cos30° + isin 30°)]5 

= 2 5(cos 150°+ i sin 150~ 

=32(- ~ +i~) 

= -16...[3+ 16i 

8. (2..{i - 2i..{i) 6 

x= 2..{i, Y = -2..{i 

r = ~(2..{i)2 + (-2..{i) 2 = .J8+8 = 4 

-2..{i 

tan 8 =--zJ2" =-1 so 8=-45°. 

2 2 

(2..{i -2i..{i)6 

= [4(cos(-45°) + i sin(-45o)]6 

= 4 6 (cos(-270, + isin(-2700)] 

=4 6(0+I·i) 

= 0+4 6i 

= 0+4096i 

or 4096i 

9. (2 - 2i...[3)4 

r=~22+(-2...[3)2 =4 

-2...[3 -fj 

tan6=--= 3 so 8=300°. 

2 

(2 -2i...[3)4 

= [4(cos 300° + i sin 300°)]4 

= 4 4 (cos1200° + isin 1200°) 

{ 1 i...[3) 

=25 -2'+2 

= -128+ 128;...[3 

10. 

(-;--;J 

..fi 

..{i 

x=T' Y=T 

r=~~+~ = 1 

tan 8 = -1 so 8 = -45° 

(-;--; I)' 

= [1(cos(-45°)+ isin(-45°))]8 

= 1 8 (cos(- 360°)+ i sin(-360°)) 

= 1(1 +0 ·i) 

=1+0ior 1 

11. (-2-2i)5 

r = ~(_2)2 +(_2)2 

=../8 = 2..{i 

-2 

tan8=-=1 so 8=225°. 

-2 

(-2-2i)5 

= (2..{i)5(cos225° + i sin 225°)5 

= 32.[32(cos1125°+ i sin 1125°) 

= 128..{i(cos45° + i sin 45°) 

=1284 ~+ ~i) 

= 128+128i 

12. (-I+i)7 

x=-I, y= 1 

r=~(-1)2+12 =...fi 

1 

tan 8 =-=-1 so 8= 135°. 

-1 

(_I+i)7 

= [..{i(cos135°+ isin 135°)f 

= 8...fi(cos 945° + i sin 945°) 

= s..J2(cos225° +;sin 225°) 

=8...fi 

(-2-'2 

..{i . ...fi) 

=-8-8i 

334



13. (cos 0° + i sin 0°) 

= l(cosO° + isin 0°) 

r= 1,0=0°, n=3 

r 1 / 3 =1 1 / 3 =1 

0+36O° ·k 

a=-....:..:..-~ 

n 

y 

I 

t----:¥=--------1- x 

- I 

-I 

15. 8 cis 60° 

r =8, 0 =60°, 

r 1l3 = 8 113 = 2 

n =3 

a = 600+36O·k 

3 ; k = 0, I, 2 

So the cube roots are 

(cos 0° + i sin 0°), 

(cos 120° + isin 120°), 

(cos 240° + i sin 240°). 

y 

a = 20°, 140°, 260° 


2 cis 20°. 

2 cis 140°, 

2 cis 260°. 

y 

~-+-¥~-+-_ x 

--;f---'+--*,=,..- x - 2 2 

-I 

-I 

- 2 

14. Find the cube roots of 

16. 27 cis 300° 

(cos 90° + i sin 90°). 

r=27, 0=300°, n=3 

r= 1,0=90°, n=3 r 1 / 3 =27 1 / 3 =3 

r 1/3= 1 113 =1 

300° + 360° . k 

a= 3 ; k =0, 1, -2 

0+360° ·k 

a=--"":":""":":" a = 100°, 220°, 340° 

n 

The cube roots are 

90 0+36O°·k 3 cis 100°, 

= 3 k =0, I, 2 

3 cis 220°, 

Ifk=O, a= 90°+360° ·0 =30°. 

3 cis 340°. 

y 

3 

90°+ 360°·1 

If k =1, a = 3 = 150° . 

Ifk=2, a= 90°+360° ·2 =270° 

3 . --::t-+-+~d--+-+-"'- X 

- 3 3 


(cos 30° + i sin 30°), 

(cos 150° + i sin 150°), - 3 

(cos 270° + i sin 270°) . 

335



17. -8; = 8(cos 270° + i sin 270°) 

r = 8, 8 = 270°, n = 3 

l /3 

=8 1/ 3 r =2 

270° + 360° . k 

a = 3 ; k = 0, 1, 2 

a = 90°, 210°, 330 0 


2(cos 90° + i sin 90°), 

2(cos 210° + i sin 210 0 ) , 

2(cos 330° + i sin 330 0 ) . 

y 

2 

~-+-_:-----l--+:--x 

-2 2 

-2 

18. 27; = 27(cos 90° +; sin 90 0 ) 

r=27, 8=90 0 , n=3 

ll3 

=27 1/ 3 r =3 

90 0+36O°·k 

a= 3 ;k=0,I,2 

a = 30 0 , 150 0 , 270° 


3(cos 30° + i sin 30 0 ) , 

3(cos 150 0 + i sin 150 0 ) , 

3(cos 270 0 + i sin 270 0 ) . 

y 

3 

y 

20. 27 =27(cos 0 0 + i sin 0°) 

r = 27, 8 = 0 0 , n =3 

r l/ 3 =271/3 =3 

00+36O°·k 

a= k=O, 1, 2 

3 

a = 0°, 120°, 240° 


3(cos 0 0 + i sin 0 0 ) , 

3(cos 120 0 + i sin 120°), 

3(cos 240 0 + i sin 240 0 ) . 

y 

3 

+-+-+--J::o-+....~- x 

-3 3 

21. I+;..fj 

-3 

r=~12+(..fj)2 =2 

+-+-+~~-+-1_ x 

tan 8 = ~ so 8=60 0 • 

1 

-3 

19. -64 = 64(cos 180 0 + i sin 180 0 ) 

r=64, 8=180°, n=3 

ll3 

=64 1/ 3 r =4 

180° + 360 0 • k 

a = 3 ; k =0, 1, 2 

a = 60 0 , 180 0 , 300 0 


4(cos 60 0 + i sin 60°), 

4(cos 180 0 + i sin 180°), 

4(oos 300 0 + i sin 300°). 

1+;..fj = 2(cos 60° + i sin 60 0 ) 

ll3 1/ 3 

r =2 ='Jfi 

60 0+360°·k 

a= k =0, 1, 2 

3 

a = 20°, 140 0 , 260 0 


V2(cos200+ isin200) 

V2(cos 140° + i sin 140°) 

V2



2 

-2 2 

y 


:v4(cos500+ isin500), 

:v4(cos 170° + isin 170°), 

:v4(cos290° + i sin 290°). 

y 

2 

- 2 

22. 2 - 2i../3 

r = ~r-22-+-(--2-../3-3)-2 = 4 

tan 8 = - 2../3 =--J3 

2 

8 = 300° (quadrant IV) 

ll 3 

r 

=:v4 

300° + 360° . k 

a = ; k =0, I, 2 

3 

a = 100°, 220°, 340° 


:v4(cos 100° + isin 100°), 

:v4(cos 220° + i sin 220°), 

:v4(cos 340° + i sin 340°). 

y 

2 

- 2 2 

-+-~~-4-+-I-+- .r 

24. ../3 - i 

-2 

r = ~r-(../3-3-:")2-+-(_-1)-=-2 = 2 

tan 8 = - ..[3 , 8 = 330° 

3 

..[3 - i = 2(cos 330° + isin 330°) 

l / 3 

r 

=Vi 

a = 330° + 360° .k; k = 0, I, 2 

3 

a = 110°, 230°, 350° 


Vi(cos 110° + isin 110°), 

Vi(cos230° + isin 230°), 

Vi(350° + i sin 350°). 

y 

2 

-2 

23. -2..[3 +2i 

r = ~r-(--2..[3--3--)2-+-2-:"'2 = 4 

tan 8 = 2 M so 8 = 150° . 

-2"\13 

-2..[3 + 2i = 4(cos 150° + isin 150°) 

ll 3 = 4 

1/ 3 

r = :v4 

150° + 360° . k 

a = k =0, I, 2 

3 

a = 50°, 170°, 290° 

-2 

25. Find all the second (or square) roots of 1. 

1= l(cosOO+ isin 0°) 

r=I, 8=0°, n=2 

r 1l2 =11/2 = 1 

00+36O°·k 

a = 

k =0, J 

2 

a =0°, 180° 

The second roots of 1 are 

(cos 0° + i sin 0°), 

(cos 180° + i sin 180°). 

337



y 28. Find all the eighth roots of 1. 

I = cos 0° + i sin 0° 

r = I, () = 0, n = 8 

00+360° ·k 

a=---- 

8 

k=O, 1,2,3,4,5,6,7 

a =0°, 45°, 90°, 135°, 180°, 225° 

270°, 315° 

The eighth roots of I are 

26. Find all the fourth roots of I. (cos 0° + i sin 0°), 

1 =cos 0° + i sin 0° (cos 45° + i sin 45°), 

r=I, (}=Oo, n=4 (cos 90° + i sin 90°), 

r l / 4 = 1 114 (cos 135° + i sin 135°), 

= 1 

(cos 180°+ i sin 180°), 

00+360° ·k 

a = 4 ; k =0, 1, 2, 3 

(cos 225° + i sin 225°), 

(cos 270° + i sin 270°), 

a =00, 90°, 180°, 270° (cos 315° + i sin 315°). 

The fourth roots of 1 are 

y 

1 

(cos 0° + i sin 0°), 

(cos 90° + i sin 90°) , 

(cos 180° + i sin 180°), 

(cos 270° + i sin 270°). 

y 

1 

*"__ ~ __~~ x 

-I o 

27. Find alI the sixth roots of 1. 

1 =1(cos 0° + i sin 0°) 

r = 1, () = 0°, n = 6 

-1 

r l/6 = 1 1/6 = 1 

00+360°·k 

a=--- 

6 

k = 0,1,2,3,4,5 

a =0°, 60°, 120°, 180°, 240°, 300° 

The sixth roots of 1 are 

(cos 0° + i sin 0°), 

(cos 60° + i sin 60°), 

-1 

29. Find all the second (square) roots of i. 

i = cos 90° + i sin 90° 

r = 1, () = 90°, n = ~ 

r 1l2 =1 11 2 =1 

(cos 120° + i sin 120°), 

(cos 180° + i sin 180°), 

(cos 240° + i sin 240°), 

-I 

(cos 300° + i sin 300 °). 

y 

90 0+360°·k 

a= 

k =0, 1 

2 

a= 45°, 225° 

The second roots of i are 

(cos 45° + i sin 45°), 

(cos 225° + i sin 225°). 

y 

-+--~


30. Find all the fourth roots of i. 

i =cos 90° + i sin 90° 

r =1. (J=90°. n =4 

90°+360° ·k 

a = ; k =0, I, 2, 3 

4 

a =22.5°, 112.5°, 202.5°, 292 .5° 

The fourth roots of i are 

(cos 22.5° + i sin 22.5°), 

(cos 112.5° + i sin 112.5°), 

(cos 202.5° + i sin 202.5°), 

(cos 292.5°.+ i sin 292.5°). 

, y 

1 

8.4 Powers aod Roots of Complex Numbers 339 

33. x 3 +i=O 

x 3 =-i 

=O-i 

= 1(cos 270° + i sin 270°) 

11/3 =1 

270° + 360° . k 

a= 3 ;k=O,I,2 

a = 90°, 210°, 330° 

x =(cos 90° .+ i sin 90°) , 

(cos 210° + i sin 210°) , 

(cos 330° + i sin 330°) 

34. x 4 +i =0 

4 . 

X =-1 

=O-i 

+------:::*::......---t.. x 

=1(cos 270° .+i sin 270°) 

31. x 3 -1=0 

- I 

x 3 =1 

= 1+0i 

=l(cosoo + isinOO) 

The modulus of the solutions is 

1 1 / 3 = 1. 

The arguments of the solutions are 

00+360° ·k 

a = ' ; k =0, I, 2. 

3 

Ifk =0, 0° + 360° . 0 = 00. 

3 

Ifk =I, 0° + 360° . 1 = 1200. 

3 

1 1 / 4 =1 

270° + 360° . k 

a = " k =0, 1, 2, 3 

·4 

x =(cos 67.5° + i sin 67.5°), 

(cos 157.5° + i sin 157.5°), 

(cos 247 .5° + i sin 247.5°), 

(cos 337 .5° + i sin 337 .5°) 

35. x 3 -8=0 

x 3 =8 

=8+0i 

=8(cosO° + isinOO) 

81/3 =2 

00+360°·k 

a = 3 ; k =0, I, 2 

a = 0°, 120°, 240° 

x = 2(cosO° + isinOO), 

2(cos 120° + i sin 120°), 

2(cos 240° + i sin 240°) 

Ifk = 2, 0°+360° ·2 = 2400. 

3 

x = (cos 0° + isinOO), 

(cos 120° + i sin 120°), 

36. 

x 3 +27 =0 

(cos 240° + isin 240°) x 3 =-27 

32. x 3 +1 =0 

=-27+0i 

=27(cos 180° + i sin 180°) 

x 3 =.:...1 

=-1+0i 

27 1 / 3 =3 

a = 180 0+360° ·k . k =0 1 2 

=l(cos 180° + i sin 180°) 3 ' " 

The modulus of the solutions is a = 60°,180°,300° 

1 1 / 3 =1. x = 3(cos60° + isin600) 

The arguments of the solutions are 3(cos 1800 + i sin 180°) 

a =1800+ 36QO · k ; k = 0, I, 2. 3(cos 300° .+i sin 300°) 

3 

a =60°, 180°, 300° 

x =(cos 60° + i sin 60°), 

(cos 180° + i sin 180°), 

(cos 300° + i sin 300°) 

339



37. x 4 +1 =0 

x 4 =-1 

=-1+0; 

= l(cos 180° + i sin 180°) 

4 

1 1 / =1 

0+360° 

a= 180 ·k . k =0 I 3 

4 ' " 2, 

a = 45°, 135°, 225°, 315° 

x = (cos 45° +;sin 45°), 

(cos 135° + i sin 135°), 

(cos 225° + i sin 225°), 

(cos315° + ;sin 315°) 

38. x 4 +16 =0 

4 

x =-16 

=-16+0i 

= 16(cos 180° + isin 180°) 

161/4 =2 

180° + 360°· k 

a= 4 ;k=0,1,2,3 

a = 45°, 135°, 225°, 315° 

x = 2(cos 45° + i sin 45°), 

2(cos 135° + i sin 135°), 

2(cos 225° + i sin 225°), 

2(cos 315° + i sin 315°) 

39. x 4 _ ; = 0 

x 4 =i 

=O+i 

= I(cos 90° + i sin 90°) 

4 

1 1 / =1 

90°+ 360° · k 

a= 4 ;k=0,1,2,3 

a = 22.5°, 112.5°, 202.5°, 292.5° 

x = (cos 22.5° +;sin 22.5°), 

(cos 112.5° + i sin 112.5°), 

(cos 202 .5° + i sin 202 .5°), 

(cos 292.5° + i sin 292.5°) 

40. x 5 -i= 0 

x 5 =i 

." =O+i 

= I(cos 90° + i sin 90°) 

1 1 / 5 =I 

90 0+36O°·k 

a = k =0, I, 2, 3, 4 

5 

a = 18°, 90°, 162°, 234°, 306° 

x = (cosI8° + i sin 18°), 

(cos 90° + i sin 90°), 

(cos 162° + i sin 162°), 

(cos 234° +;sin 234°), 

(cos 306° + isin306°) 

41. x 3 - (4 + 4i.J3) =0 

x 3 =4+ 4;.J3 

r = ~(4)2 + (4.J3)2 = 8 

4.J3 

tan8=-- 8=60° 

4 ' 

x 3 = 8(cos 60° +;sin 60°) 

8 1 / 3 = 2 

60°+ 360° · k 

a = 3 ; k =0, 1, 2 

a = 20°, 140°, 260° 

x = 2(cos 20° + i sin 20°), 

2(cos 140° + isin 140°), 

2(cos 260° + i sin 260°) 

42. x 4 - (8+8i.J3) =0 

4 

x =8+8;.J3 

r = ~r-:82'---+-(8-.J3-3)-2= ../256 = 16 

tan 8 = 8~ = .J3;' 8 = 60° 

4 

x = 16(cos600+;sin60 0) 

16 114 =2 

60°+ 360°· k . 

a = 4 ; k =0, 1, 2, 3 

a = 15°, 105°, 195°, 285° 

x = 2(cosI5° + ksin 15°), 

2(cos 105° + i sin 105°), 

2(cosI95° + isin 195°), 

2(cos 285° + i sin 285°) 

43. x 3 -1 =0 

(x -1)(x 2 + x + I) = 0 

x-I =0 

x= I 

or 

x 2 +x+1 =0 


a =1, b = I, and c = 1. 

340



-1±~12 -4·1 ·1 

x - --"---- 

2·1 

x= 

-1±H 

2 

-1±i../3 

x--- 

2 

x=_.!.±../3 i 

2 2 

1 ../3 . 1 ../3 . 

x=1 --+-1 ----I 

• 2 2' 2 2 

We see that the solutions are the same as 

Exercise 31. 

For Exercise 31. the solutions are 

(cos 0° + i sin 0°) 

(cos 120 0 + i sin 120°) 

(cos 240° + i sin 240°). 

In standard form. these are 

1, _.!. + ../3; _.!. _../3 ;. 

2 2' 2 2 

44. x 3 + 27 = 0 

(x+3)(x 2 - 3x+9) =0 

x+3=0 

x=-3 

or 

x 2 -3x+9 =0 


a =I, b=-3, and c =9. 

x 

-(-3)±~(-3)2-4·1·9 

2·1 

3±..J-27 

x=--'- 

2 

3±(3../3)i 

x= 

2 

x=~±3../3 i 

2 2 

x =-3 ~+ 3../3; ~_ 3../3; 

'2 2' 2 2 

For Exercise 36, the solutions are 

3(cos6O°+ ;sin600). 

(cos180° + i sin 180°), 

(cos300° + isin3000). 

In standard form, these are 

~ + 3../3 i _ 3 ~ _ 3../3t. 

2 2 ' , 2 2 

45. (cos6+;sin6)2 

=1 2(cos26+ isin 26) 

=cos 26 +; sin 26 

46. (cos6+;sin6)2 

=cos 2 6 + 2isin6cos6+ i 2 sin 2 6 

=cos 2 6+ 2isin6cos6-sin 2 6 

=(cos 2 6 -sin 2 6) + i(2sin 6 cos 6) 

47. Two complex numbers a + bi and c + di are equal 

only if a =c and b =d. Therefore, 

cos 2 6-sin 2 6 =cos26 

and 

2sin6cos6 =sin 26. 

48. cos 26 + i sin 26 

=(cos 2 6 - sin 2 6) + i(2 cos 6sin6) 

Since a + bi =c + di only if a =c and b = d. 

i sin 26 =i(2 cos 6 sin 6) and 

sin 26 =2cos6sin6 

49. (a) Ifz = 0 + Oi,we must make the following 

calculations: 

0 2 + 0 =0; 0 2 + 0 =O. 

The calculations repeat as 0, 0, 0, . .., and 

will never exceed a modulus of 2. The point 

(0, 0) is part of the Mandelbrot set. The pixel 

at the origin should be turned on. 

(b) Ifz=l-li. 

(1- ;)2 +(1- ;) =1-3i. 

The modulus of 1 .:.. 3i is .,flO > 2. 

Therefore, 1 - 1i is not part of the 

Mandelbrot set. and the pixel at (1. -1) 

should be left off. 

(c) Ifz = -.5i, 

(-.5;)2 - .5; =- .25 - .5;; 

(-.25)- .5i)2 + (-.25 -.5i) 

=-.4375 - .25i; 

(-.4375 - .25i)2 + (-.4375 - .25i) 

= -.308593- .03125;; 

(-.308593 - .03125i)2 

+(-.308593 - .03125i) 

=-.214339 - .0119629i; 

(-.214339 - .0119629i)2 

+(-.214339 - .0119629;) 

=-.16854 - .00683466i. 

This sequence appears to be approaching the 

origin. and no number has a modulus greater 

than 2. Thus, -.5; is part of the Mandelbrot 

set, and the pixel at (0, -.5) should be turned 

on. 

341



2z 3 + 1 . 

50. (a). Let f(z) =--2- and ZI =I. 

3z 

Then, 

2;3 + 1 

z2= f(zl) =3j2 

1 2 . 

=--+-,. 

3 3 

Similarly, 

3 

2z 2 + 1 

z3 = f(Z2) = 2 

3z 2 

'" -.58222 + .92444;, 

2z 3 + 1 

z4 = f(Z3) = --2 

3z 3 

'" -.50879 + .868165;. 

The values of z seem to approach 

1 .../3. 

w2 =--+-1. 

2 2 

Color the pixel at (0, 1) blue. 

2z 3 +1 . 

(b) Letf(z) =--2- and Zl = 2 + I. 

3z 

Then, 

z =f(z ) = 2(2 + ;)3 + 1 

2 I 3(2 + 02 

'" 1.37333 + .61333;. 

2z 2 

3 

+ 1 

z3=f(Z2)= 2 

3z 2 

'" 1.01389 + .29916li 

2z 3 

3 

+ 1 

Z4 =f(Z3) = 2 

3z 3 

'" .926439 +.0375086; 

2zl+1 

Zs = f(Z4) = 2 

3z 4 

'" 1.00409 - .00633912;. 

The values of z seem to approach WI =1. 

Color the pixel at (2, 1) red. 

2z 3 + 1 . 

(c) Let f(z) =--2- and zl =-I-I. 

3z 

Then, 

2(-1- ;)3 + 1 

z - f(z ) 

2- 1- 3(-1-0 2 

2 5. 

=----1. 

3 6 

2z 2 

3 

+ 1 

z3 =f(Z2) = 3 2 

(;2 

'" .508691 - .841099; 

2zl+1 

Z4 =f(Z3) = 2 

3z 3 

'" -.499330- .866269; 

The values of z seem to approach 

1 .../3 . 

W3 =----1. 

2 2 

Color the pixel at (-I, -1) yellow . 

51. Using the trace function, we find that the other 

four fifth roots of 1 are: .3090 1699 + .95105652;, 

- .809017 + .58778525;, 

-809017 - .5877853;, .30901699 - .9510565; 

52. 

-1.5 

-l.SI--+---t--iE-- 1.5 

T=O 

11=1 y=O 

1.5 

Using the trace function, we find that three of the 

tenth roots of 1 are: I, .8090.1699 + .58778525;, 

.30901699 + .95105652i 

53. 2 + 2.,[3i is one cube root. 

r=~22 +(2.../3)2 =.J4+12 =4 

2.../3 .../3 

tan8 =--= 3' 8=60° 

2 ' 

This root is 

4 cis 60° = 4 cis (60° . 1). 

Since the graphs of the other roots must be 

equally spaced around a circle and the graphs of 

these roots are all on a circle that has center at the 

origin and radius 4, the other roots are 

4 cis (60° + 120°) = 4 cis 180° 

and 

4 cis (60° + 2· 120°) = 4 cis 300°. 

342



4 cis 180° = 4(cos 180° + i sin 180°) 

=4(-I+i ·0) =-4 

4 cis 300° = 4(cos 300° + i sin 300°) 

=2-2i..[3 

The roots are 2 - 2i..[3, - 4 and 2 + 2i..[3 . 

54. x 3 + 4 - 5; =0 

x 3 =-4+5; 

r = ~(-4)2 +5 2 =..J41 

tan (} = -~, (} is in quadrant II. 

4 

(} = 128.65981° 

x 3 = ..J41(cosI28.65981° 

+ i sin 128.65981°) 

(..J41)1I3 = 1.8569376 

a=(}+36O°.k· k=0 1 2 

3 ' " 

Irk = 0, a = 128.65981° + O· 36QO 

3 

=42.886603°. 

Irk = I, a = 128.65981° + 1·360° 

3 

= 162.8866°. 

If k = 2, a = 128.65981° + 2.360° 

3 

= 282.8866°. 

x = 1.8569376 

·(cos 42.886603° +; sin 42.886603°), 

1.8569376 

· (cos 162.8866° +; sin 162.8866°), 

1.8569376 

· (cos 282.8866° +;sin 282.8866°) 

In standard form, 

x =1.3606 + 1.2637;, 

-1.7747 + .5464;, 

.4141-1.8102;. 

55. x 5 + 2 +3; =0 

x 5 =-2 -3; 

r= ../4+9 =../13 

r 1/ n =(../13)115 = 13 1/ 10 = 1.292 

-3 

tan(}=-=1.5 

-2 

(}is in quadrant III. 

(} = 236.30993° 

236.30993 ° + 360° . k 

a=------ 

5 

k=0,1,2,3,4 

a =47.26°, 119.26°, 191.62°, 

263.26°, 335.26° 

x'" 1.292(cos47.26°+;sin47.26°) 

'" .87708 + .94922i 

1.292(cosI19.26° + isin 119.26°) 

'" -.63173 + 1.1275i 

1.292(cos 191.26° + ;sin 191.26°) 

'" -1.2675 - .25240; 

1.292(cos 263.26° + i sin 263.26°) 

'" -.15164 -1.28347i 

1.292(cos 335.26° +; sin 335.26°) 

'" 1.1738 - .54083i 

56. The number 1 has 64 complex 64th roots . Two of 

them are real, 1 and -1, and 62 ofthem are not 

real. 

57. The statement, "Every real number must have two 

real square roots," is false. Consider, for example, 

the real number -4. Its two square roots are 2; 

and -2; which are not real. 

58. The statement, "Some real numbers have three 

real cube roots,"is false. Every real number has 

only one cube root that is also a real number. 

Then there are two cube roots, which are 

conjugates that are not real. 

Section 8.5 


1. y=x-3 

-x+ y =-3 

x-y=3 

2. y=~ 

i =4-x 2 

Exercises 

x 2 +i =4 

1. (a) II (since r » 0 and 90° < (}< 135°) 

(b) I (since r » 0 and 0° < (J < 90°) 

(c) IV (since r> 0 and _90° < (} < 0°) 

(d) III (since r > 0 and 180° < (J < 270°) 

2. (a) positive x-axis 

(b) negative x-axis 

343



(c) negative y-axis 

(d) positive y-axis (since 450 - 360 =90°) 

7. Two other pairs of polar coordinates for (5, -600) 

are (5, 300°) and (-5 , 120°). 

For Exercises 3-12, answers may vary. 

3. Two other pairs of polar coordinates for (1, 45°) 

are (1, 405°) and (-I, 225°). 

8. Two other pairs of polar coordinates for (2, - 45°) 

are (2, 315°) and (-2, 135°). 

90° 

270° 

9. Two other pairs of polar coordinates for (-3, 

_2 10°) are (-3, 150°) and (3, -30°). 

90° 

Two other pairs of polar coordinates for 

(- 2, 135°) are (-2, 495°) and (2, 315°). 

90° 

270· 

6. Two other pairs of polar coordinates for (-4,27°) 

are (-4, 387°) and (4, 207°). 

90° 

10. Two other pairs of polar coordinates for (-1, 

-120°) are (-I, 240°) and (1, 60°). 

90° , 

./)(;~r:=lt(\ , • 

18W~jW 

270° 

11. Two other pairs of polar coordinates for (3, 300°) 

are (3, 660°) and (-3, 120°). 

344



12. 

For Exercises 13-21, answers may vary. 

13. r = ~(_1)2 + 1 2 =.J1+l =..fi 

8 = arctan( ~1) = 135° 

(since x < 0 and y > 0) 

So one possibility is (..fi, 135°) 

Alternatively if r = -.{i, 8 = 135+ 180 = 315° 

A second possibility is (-.{i, 315°) 

Two pairs: 

(..fi, 135°), (-.{i, 315°) 

y 

3 

2 

.1 

-3 -2 -1 0 1 2 3 

-1 

-2 

-3 

14. r = ~12 + 1 2 =..fi 

8 = arcta{D= 45° 

(sincex>O and y > 0) 

One possibility is: (..fi, 45°) 

Alternatively, if r = -.{i, 8 = 45 + 180 = 225° 

A second possibility is: (-.{i, 225°). 

Two pairs: 

(..fi, 45°), (-Ii, 225°) 

y 

3 

2 

• 

+-+-+-:+---t--t-~ x 

-3 -2 -1 0 1 2 3 

-I 

-2 

-3 

15. r = ~02 + 3 2 =..[9 = 3 

The point (0. 3) is on the positive y-axis. Thus, 

8=90°. 

One possibility is (3. 90°). 

Alternatively, if r= -3, 8 = 90+ 180 = 270° 

A second possibility is: (-3,270°) 

Two pairs : 

(3,90°), (-3, 270°) 

y 

3 

2 

+-+-+-.l--+-+-~ .x 

-3 -2 _I 0 I 2 3 

-I 

-2 

-3 

16. r=~02+(-3)2 =..[9=3 

The point (0, -3) is on the negative y-axis. Thus, 

8=270°. 

One possibility is: (3, 270°) 

Alternatively, if r = -3, 8 = 270 -180 = 90° 

Two pairs : 

(3,270°), (-3, 90°) 

y 

3 

2 

+-+--t---:;d-+-+-~x 

-3 -2 -1 0 123 

-I 

-2 

-3 

17. r=~(..fi)2 + (..fi)2 =·./2+2 =../4 =2 

8 = arctan(~) = arctan (1)= 45° 

(since x > 0 and y > 0) 


Alternatively, if r =-2, 

8 = 45°+180= 225° 


Two pairs: 

(2,45°), (-2, 225°) 

y 

3 

2 

• 

+-+-+-nt--+--+-,. .x 

-3 -2 _I 0 1 2 3 

-I 

-2 

-3 

345



18. r=~(--J2)2 + (-J2)2 =../2+2 =../4 =2 

f) = arc 3 11 ( ~J= arctan (-1) = 135° 

(since x < 0 and y > 0) 


Alternatively, if r = -2, f) = 135+ 180= 315° 

A second possibility is: (-2, 315°) 

Two pairs: 

(2, 135°), (-2, 315°) 

y 

3 

2 

• 

-3 -2 -1 0 1 2 3 

-1 

-2 

-3 

19. r= (~r +(%y =~~+~ =~ 

3 

2 

-3 -2 -18 0 1 2 3 

-I 

-2 

-3 

21. r =~32 + 0 2 = ~ =3 

f) =arctan(j)= 0° 

y 

(since tan f) =0 and x > 0). 


Alternatively, if r= -3, f) = 0+180= 180° 

A second possibility is (-3,180°) 

Two pairs: 

(3,0°), (-3, 180°) 

y 

3 

2 

f) = arcta{%...lJ) =arctan (~) = 60° 


One possibility is: (.,fj, 60°) 

Alternatively, if r = -./3, f) = 60+ 180 = 240°. 

Two pairs: 

(..[3, 6QO), (-Jj, 240°). 

y 

3 

2 

• 

+-+-+-~-+-+-....... x 

- 3 -2 -1 0 1 2 3 

-1 

-2 

-3 

20. r= (~r+(-fY =~~+~=1 

f) = arctan( ~1 -( ~)) = arctan( ~) = 210° 

(since x < 0 and y < 0) 


Alternatively, if r = -I, f) = 210 -180 = 30° 

A second possibility is: (-1, 30°) 

Two pairs: 

(1,210°), (-1 ,30°) 

0 

-3 -2 -1 1 2 3 

-1 

22. (a) C 

(b) D 

(c) A 

(d) B 

-2 

-3 

x 

346



23. r = 2 + 2 cos 0 (cardioid) 

0 0° 30° 60° 90° 120° 150° 

cos 0 1 .9 .5 0 -.5 -.9 

r 4 3.8 3 2 1 .2 

0 180° 210° 240° 270° 300° 330° 

cos 0 -1 - .9 -.5 0 .5 .9 

r 0 .3 1 2 3 3.7 

270· 

r=2+2cos9 

24. r =2(4 + 3cos 0) (cardioid) 

=8+6cosO 

0 0° 30° 60° 90° 120° 150° 

cosO 1 .9 .5 0 -.5 - .9 

r 14 13.2 11 8 5 2.8 

0 180° 210° 240° 270° 300° 330° 

cos 0 -1 -.9 -.5 0 .5 .9 

r 2 2.8 5 2 11 13.2 

90· 

ISO· -+-+-++-+++-+tHlb. 

270· 

r= 2(4+ 3 cos 9) 

25. r =3 + cos 0 (limacon) 

0 0° 30° 60° 90° 120° 150° 

r 4 3.9 3.5 3 2.5 2.1 

0 180° 210° 240° 270° 300° 330° 

r 2 2.1 2.5 3 3.5 3.9 

347



90° 

26. r = 2 - cos 6 (limacon) 

--1 

0° 30° 60° 90° 135° 

1.1 1.5 2 2.7 

6 180° 225° 270° 315° 

r 3 2.7 2 1.3 

R90° 

'Y 

\.......+->~. 

~ ,..--r·y -, 

i l / \ i



Pattern 3,0, -3, 0, 3 continues for every 18°, 

90° 72D 

't9~ 

180· _.;.-. -+-j.- .....•...-+_.,-. O· 

,~~)i 

2~Oo 288 0 

r=3cos59 

29. r 2 =4 cos 29 (lemniscate) 

r =±2.Jcos29 

Graph only exists for [0°, 45°J, 

[135°, 225°J, and [315°, 360 0J 

because cos 29 must be positive. 

r ±2 ±1.4 0 0 ±1.4 

9 180 

r ±2 ±1.4 0 0 ±1.4 

90· 

30. r 2 = 4sin29 (lemniscate) 

r =±2.Jsin 29 

Graph only exists for [0°, 90 0J, [180°, 270 0J because sin 9 must be positive. 

r 0 ±1.86 ±2 ±1.86 0 o ±2 0 

90· 

270· 

2 r = 4 sin 29 

349



31. r = 4(1- cos 9) (cardioid) 

9 0° 30° 60° 90° 120° 150° 

r 0 .5 2 4 6 7.5 

f) 180° 210° 240° 270° 300° 330° 

r 8 7.5 6 4 2 .5 

90 0 

32. r = 3(2- cos f)) (cardioid) 

=6-3cos9 

r 3 3.9 6 8.1 9 6 3 

180" 

270 0 

r = 3(2 - cosO) 

33. r = 2 sin f) tan f) (cissoid) 

r is undefined at 9 = 90° and B = 270°. 

r 0 .6 1.4 3 -3 -1.4 -.6 0 

Notice that for [180°, 360°), the graph retraces the path traced for [0°,180°). 

90 0 

270 0 

r = 2 sin 0 tan 0 

350



cos20 (ci 'd ·th I ) 

34. r =-- CISSOI WI a oop 

cosO 

r is undefined at 0 = 90° and 0 = 270° 

r = 0 at 45°, 135°,225°, and 315°. 

0 0° 45° 60° 70° 80° 

r 0 - 1 -2.2 -5.4 

0 90° 100° 110° 135° 180° 

r 5.4 2.2 0 -1 

Notice that for [180°, 360°), the graph retraces the 

path traced for 

[0°, 180°). 

35. (a) (r,-0) 

36. -0 

(b) (r,1t - 0) or (-r, -0) 

(c) (r, 1t+ 0) or (-r, 0) 

37. 1t-0 

38. -r;-O 

39. -r 

40. 1t+ 0 

41. the polar axis 

42. the line 0 = ~ 

2 

43. r = 4 cos 20, 0° s 0 < 360° 

Since the largest value of cos 20 is 1 (the range 

of the cosine function is [-I, 1]), the largest value 

of r is 4 . 1 =4. 

4cos20=0 

cos20=0 

Since 0°::; 0 < 360°,0°::; 20 < 720°. 

The cosine is 0 at 90°, 270°, 450°, 630°. Thus, 

20 = 90° or 20 = 270° 

o= 45° or 0 = 135° 

20 = 450° or 20 = 630° 

0=225° or 0 = 315°. 

44. r = 5 sin 30, 0° ::; 0 < 180° 

Since -1 ::; sin 0::; I, the largest value for sin 30 

is 1.The largest value r can be is 

5 sin 30 = 5· 1 = 5. 

0= 5 sin 30 

0= sin 30 

Since 0° ::; 0 < 180°,0° ::; 30 < 540°. 

Therefore, 

30 = 0° or 30 = 180° or 30 = 360° 

0=0° or 0 = 60° or 0 = 120°. 

46. r = 4 sin 0, r = 1 + 2 sin 0 , 

0::; 0 < 21t 

4sinO = 1+ 2sinO 

2sinO = 1 

. n 1 

sm(7 = 

2 

1t 51t 

0=- or 

6 6 

The points of intersection are 

( 4 sin ~' : ) = (2, ~) and 

4 sin 51t 51t) = (2 51t). 

( 6' 6 ' 6 

47. r = 2 + sin 0, r = 2 + cos 0, 

0::; 0 < 21t 

2 + sin 0 = 2 + cos 0 

sinO=cosO 

1t 51t 

0=- or 

4 4 

. 1t -Ii 4+-li 

r =2 + sm - =2 +- =-- 

4 2 2 

. 51t -Ii 4 - -Ii 

r = 2 +sm- = 2 --=-- 

4 2 2 


(4+ :} (45;). 

2-1i, 2-1i, 

351



48. r = sin 20. r = .ficos o. 0 ~ 0 < 1t 52. r=2cos 0 

Multiply both sides by r. 

sin 20 = -ficos 0 

r 2 =2rcosO 

2sin OcosO = -ficosO 

Since r 2 = x 2 + y2 and x =rcosO. 

2sinOcosO - -ficosO = 0 

x 

cos 0(2 sin 0 -fi) 2 +i =2x. 

= 0 

Complete the square on x to get the equation of a 

cos 0 = 0 or 2 sin 0 - -fi = 0 

circle. 

2sinO =-fi x 2 -2x+i =0 

sinO = -fi 

2 

1t 

1t 

31t 

0=- or 0=- or 

244 y 


( s in 2 . ~ 2 ' 1t)=(0 2 ' ~I 2) 

( 

sin 2 .! ~) = (1 !) 

4' 4 • 4 

and (sin 2 . 31t 

4' 

51. r= 2 sin 0 

Multiply both sides by r . 

31t) = (-1 31t). 

4 • 4 

r 2 =2rsinO 

Since r 2 = x 2 + y2 and y = rsin O. 53. 

x 2+i =2y. 

Complete the square on y. 

x 2 +i - 2y + 1= 1 

x 2 + (y _1)2 =1 

The graph is a circle with center at (0. 1) and 

radius 1. 

y 

(x _1)2 +i = 1 

The graph is a circle with center at O. 0) and 

radius 1. 

-+-+--+-....,.f--+--I--+-. x 

o 

r=2cosO 

(x-I)2+ r=1 

2 

r=-- 

I-cosO 

Multiply both sides by 1 - cos O. 

r-rcosO =2 

~x2+i -x=2 . 

~x2 +i =2+x 

x 2 +i =(2 + x)2 

x 2 + i =4+4x+;2 

2 

i =4(I+x) 

The graph is a parabola with vertex at (-I, 0) and 

axis y = O. 

y 

r=2 sinO 

x 2 + (y_1)2= I 

-Hf-+-f-:+-++++-+-t-l-. x 

-1 

r-_2_ 

- I -cos9 

y2=4(x+l) 

352


3 

3 

54. r=--56. r=---- 

I-sinO 

4cosO-sinO 

r-rsinO =3 

4rcosO- rsinO = 3 

r= rsinO+3 

4x- y=3 

~x2+l =y+3 

The graph is a line with intercepts (0, -3) and 

x 2+l (3/4,0). 

=l +6y+9 

y 

x 2 =6y+9 

X~=6(Y+~) 

The graph is a parabola with axis x = 0 and vertex 

(0, -312). 

y 


_ 3 

r- 4cos8-sin8 

4x-y=3 

55. 

r+2cosO = -2sinO 

x 2 + i 

x 2 +i+2y+2x =0 

r 2 =-2rsinO-2rcosO 

=-2y-2x 

(x+l)2 +(Y+ 1)2 =2 

The graph is a circle with center (-1. -1) and 

radius .fi. 

y 

+-++-T-+--+-+-" x 

r + 2 cos 9 = -2 sin 9 

(x+I)2+(y+I)2=2 

57. r=2secO 

2 

r=- 

cosO 

rcosO = 2 

x=2 

The graph is a vertical line. 

y 

-1--4--++-+-+--i--- x 

o 

2 

1 

r=2sec9 

x=2 

58. r=-5cscO 

5 

r=-- 

sinO 

rsinO=-5 

y=-5 

The graph is a horizontal line. 

y 

1 

o 

~-5+---~ 

r=-5 esc 9 

y=-5 

353



59. r(cos 8 + sin 8) = 2 

rcos8+ rsin8 = 2 

x+y=2 

The graph is the line with intercepts (0, 2) and 

(2,0). 

y 

-+-+-+-:+-+""k--t-.. x 

65. y= 2 

rsin8 = 2 or 

2 

r=--=2csc8 

sin8 

66. x=4 

rcos8 = 4 or 

4 

r=--=4sec8 

cos8 

67. Graph r = 8, a spiral of Archimedes. 

, (cos 6 + sin 6) = 2 

x+y=2 

60. r(2 cos 8 + sin 8) = 2 

2rcos8+ rsin8= 2 

2x+y=2 

The graph is the line with intercepts (0, 2) and 

(1,0). 

y 

8 -6.3 -4.7 -3.1 -1.6 

(radians) 

r -6.3 -4.7 -3.1 -1.6 

o 1.6 3.1 4.7 6.3 

(radians) 

r o 1.6 3.1 4.7 6.3 

1< 

2 

,(2 cos 9 + sin 9) = 2 

2x+ y=2 

61. x+y=4 

rcos8 + rsin8 = 4 

r(cos8 + sin 8) = 4 

4 

r=--- 

cos8+sin8 

68. 

62. 2x-y=5 

2(rcos8) - rsin8 = 5 

r(2cos8 - sin 8) = 5 

5 

r=---- 

2cos8-sin8 

63. x 2 + y2 =16 

r 2 =16 

r=4 

Degree mode 

64. x 2 +i =9 

r 2 =9 

r=3 

354



69. In rectangular coordinates, the line passes through 

(1,0) and (0, 2). So 

2-0 2 

m =--=- =-2 and 

0-1 -1 

(y - 0) = -2(x -1) 

2x+y=2 

y = -2x+ 2 

Converting to polar form r = c. 0 

acosO+bsm 

2 

we have: r =---- 

2cosO + sin 0 

70. (a) Plot the following polar equations on the 

same polar axis: 

.39(1-.206 2) 

Mercury: r = ; 

1+ .206 cos0 

.78(1- .007 2 ). 

Venus: r 

1+ .007 cosO' 

2 

Earth: r = 1(1- .017 ) ; 

1+.017cosO 

2) 

Mars: r = 1.52(1- .093 . 

1+ .093cosO 

Use a viewing window of [-2.4,2.4] by 

[-1.6, 1.6] with Xscl =.2, Yscl =.2. 

1.6 

(c) We must determine if the orbit of Pluto is 

always outside the orbits of the other planets. 

Since Neptune is closest to Pluto, plot the 

orbits of Neptune and Pluto on the same 

PQlar axes. 

30.1(1 - .009 2 ) 

r = 1 + .009cos 0 

60 

2 

30.1(1-.009 ) 

Neptune: r = 1+.009 cos 0 

..:.... 

39_.4~ ( 1__ - . 2_ 4_~ 9 2 ) 

PIuto: r = 1+ .249cosO 

The graph shows that their orbits are very 

close near the polar axis. 

30.1 (1 - .009 2 ) 

r= 1 + .009 cos 8 

(b) Plot the following polar equations on the 

same polar axis: 

2 

Earth: r = 1(1- .017 ) ; 

1+.017cosO 

2 

. 5.2(1-.048 ) 

Jupiter: r = ; 

1+.048cosO 

19.2(1-.047 2) 

Uranus: r= ; 

1+ .047 cos 0 

2 

39.54(1-.249 ) 

PIuto: r =--..:..----~ 

1+ .249 cos 0 

Use a viewing window of [~, 60] by [-40, 

40] with Xscl =10, Yscl = 10. 

From the graph, Earth is closest to the sun of 

these four planets. 

Use ZOOM to determine that the orbit of 

Pluto does indeed pass inside the orbit of 

Neptune. 

Therefore, there are times when Neptune, not 

Pluto, is the farthest planet from the sun. 

(However, Pluto's average distance from the 

sun is considerably greater than Neptune's 

average distance.) 

355



Section 8.6 

1. C; at t = 2, x = 3(2) + 6 = 12 

y = -2(2)+ 4 =0 

2. D; att =:' x=co{:)=~ 

Y =Sin(:) =~ 

6. 

x=t +2, y=t 2,fortin[-I,I] 

t x=t+2 y = t 2 

-1 -1+2=1 (-1)2=1 

o 0+2 =2 0 2 =0 

1 1+2=3 1 2=1 

y 

3. A; at r = 5, x =5 

Y =5 2 =25 

4. B; at t =2, x =2 2 + 3 =7 

y=2 2-2=2 

5. x=2t,y=t+l,fortin[-2,3] 

x=2t 

y=t+l 

-2 2(-2)=-4 -2+1 =-1 

-1 2(-1) =-2 -1 + 1=0 

o 2(0) =0 0+1=1 

1 2(1)= 2 1+1=2 

2 2(2) = 4 2+1=3 

3 2(3) = 6 3+1=4 

x = 2t y 

Y= t+ I 

for t in [-2, 3] 

2 3 

x=t+2 

y=t 2 

for t in [-I. I] 

x - 2 = t, therefore 

y =(x - 2)2 or y =x 2 - 4x + 4. 

Since t is in [-1, 1], x is in [-1 + 2, 1 + 2] or 

[1,3]. 

7. x =..[i, y = 3t - 4, for t in [0,4]. 

x=..[i 

y=3t-4 

o .JO = 0 3(0) - 4 = -4 

1 .J!=1 3(1)-4=-1 

2 ..fi = 1.4 3(2) - 4 = 2 

3 ~ =1.7 3(3)-4=5 

4 ..[4=2 3(4)-4=8 

y 

8 (2. 8) 

x=-{i 

y= 3t-4 

for t in [0. 4] 

x=2t,y=t+ 1 

S . x x 

mce - =t, Y =- + 1. 

2 2 

Since t is in [-2,3],x is in [2(-2),2(3)] or [-4, 6]. 

(0. -4) 

o 2 

x=..[i y=3t-4 

Since x 2 =t, y=3x 2 -4. 

Since t is in [0, 4],x is in hlO, ..[4] or [0, 2]. 

356



8. x = t 2, Y = .,fi, for t in [0,4] 10. x=2t-l, y=t 2+2 ,fortin(-oo,oo) 

t x =t 2 Y =.,fi 

x=2t-l y = t 2 +2 

0 0 2 =0 .JO =0 

-3 2(-3)-1=-7 (-3) 2 +2=11 

1 1 2 = 1 ..JI = 1 

-2 2(-2)-1 =-5 (_2)2 +2 =6 

2 2 2 =4 -fi = 1.414 

-1 2(-1)-1=-3 (_1)2 +2 =3 

3 3 2 =9 ..fj = 1.732 

4 4 2 = 16 ..[4 =2 

y 

2 

0 

1 

2 

3 

2(0)-1 =-1 

2(1)-1 = 1 

2(2)-1 = 3 

2(3)-1 = 5 

0 2 +2=2 

1 2+2=3 

2 2+2=6 

3 2 +2 =11 

y 

9. 

++-t+t++lf+t++lf+t-H.- x 

5 16 

x= 1 2 

y={t 

for 1 in[0, 4] 

Since y =.,fi, y2 = t, and x = t 2 = (y2)2 =l. 

Since t is in [0, 4], x is in [0 2,42], or [0, 16]. 

x=t 3+1, y=t 3-1 ,fortin(-oo,00) 

t x =t 3 +1 

-2 (-2)3+ 1=-7 

-1 (_1)3 + 1= 0 

o 0 3 + 1=1 

1 1 3+1=2 

2 2 3 +1 =9 

3 3 3 +1 =28 

y 

(-2)3_ 1=-9 

(_1)3-1 =-2 

0 3 -1 =-1 

1 3 -1 =0 

2 3 -1 =7 

3 3 -1 =26 

+-+-+-+-++d-++-+-+.- o 

x 

x=21-1 

y=~+2 

for 1in(--, ...) 

x+l=2t 

x+l 

--=t 

2 

Since y = t 2 + 2, 

X +1)2 1 2 

y= -2 +2='4(x+l) +2. 

( 

Since t is in (-00, 00) and x = 2t 1, x is in 

(-eo, 00). 

11. x=2sint, y=2cost, fortin[0,2n] 

t x=2sint y=2cost 

o 2sinO = 0 2cos6 = 2 

f 2sinf = 1 2cosf =.J3 

t 2sint =.J2 2cost =.J2 

Since x = t 3 + 1, i 2sini =..fj 2cosi = 1 

x-I =t 3 . 

i 

Since y = t 3 -1, 

y =(x -1) - 1=x 2. 

Since t is in (-eo, 00), x is in (-eo, 00). 

2 

2sini = 2 2cosi = 0 

y 

x=2sinl 

y=2cosl 

for 1 in [0, 21r] 

Since x = 2sint and y = 2cost, 

~ = sint and L = cost. 

2 2 

357



Since sin 2 I + cos 2 I = 1, 

y 

x 2 

i 

-+-=1 

4 4 

x 2 + y2 =4. 

Since I is in [0, 2n] , x is in [-2.2] because the 

graph is a circle. centered at the origin. with 

radius 2. 

12. x =../5 sinr, y =../3 COS/. for I in [0. 21t] 

x . y 

../5 =smr• ../3 = cosr 

Since sin 2 I + cos 2 1= 1. 

Since 

'3...... 

...... 

...... ...... 

x= 3 tao r 

y=2secr 

for r io (-~, ~) 

~ = tan I and 1.= secI 

3 2' 

x 2 

and 1+tan 2 I = (ff = sec 2 I, 

i 

1+(~r =(ff 

5 3 

-+-=1. 

Since-d s sinr s I, -.J5 :s;../5sin/:S;../5. 

Therefore. x is in [-J5, ../5]. The graph is an 

ellipse. centered at the origin. with vertices 

(../5. 0), (-J5, 0), (0• ../3). (0. - ../3). 

13. x = 3tanI, 

y 

(lS",o) 

x='"ISsior 

y='Y3cosr 

for r io [0, 211'J 

y = 2 sec I. for I 

in (- ~, ~) 

I x = 3tanl y = 2secl 

-of 3tan(-t) = -3../3 2sec(-t) = 4 

-f 3tan(-t)=-13 2sec(-f)=¥ 

o 3tanO=0 2secO=2 

~ 

t 3tanf=../3 2sect= 4f 

t 3tant=3.J3 2sect=4 

x2 y2 

1+-= 

9 4 

i ={1+ x:) 

r7 

y=2 V l + 9 · 

Since this graph is the top half of a hyperbola, x is 

in (-00, 00). 

14. x=cot/. y=csct. for lin (0, It) 

Since 1+ cot 2 I = csc 2 I, 

l+x2 =y2 

~1+x2 =y. 

Since I is in (0. n) and the value of the cotangent 

of a value close to 0 is very large and the value of 

the cotangent of a value close to n is very small. x 

would be (--. 00). The graph is the top half of a 

hyperbola with vertex (0. 1). 

y 

7 

'/ 

/ 

/ 

/ ',I 

x=cotr 

y=cscr 

for r in (0, 1t) 

358



y=! x 

for x in (0, I] 

n. I) 

-t--+-:+-t-ll-t-t-t-- x 

o 

-H-t--t-ht--I-H-t--I-x 

o 

16. x = tant,y = cott, for t in (0, ~) 

1 

15. x=sint, y=csct=-.18. x = -Ii, y = t 2 -1. for t in [0,00) . 

smr 

2 

Since x = -Ii, x = t, Therefore, 

Therefore, 

1 

y = t 

2 -1 = (x 

2)2 -1 = x 

4 

-1. 

y = - . 

x 

Sincetis in [0, 00),x is in [../0, ~)or[O, 00). 

Since t is in (0, n), x is in (sin 0, sin 1&) or (0, 1). 

y 

y 

19. Sincex=2+sint andy=I+cost, 

1 1 

x-2 = sint andy-I = cost. 

y=cott=--=tant 

x 

2 

Sincesin 2 t + cos t = I, 

Since t is in (0, ;). x is in (0, 00). 

y 

y:x 4 - t 

for x in [0,00) 

(x - 2)2 +(y _1)2 =1. 

Since this is a circle centered at (2, 1) with radius 

I, and t is in [0, 2n]. x is in [1,3]. 

y 

2 

y= 1. 

x 

for x in (0, 00) 

17. Sincex=tand y=~t2+2, y=~x2+2. 

Since t is in (- 00)and x = t, x is in (-,00). 

Y 

+t-l+tffiP.lEt+t-t-H-t+. x 

Y : ~X2 + 2 for x in (_00, 00) 

+-t-H-f3'~I-+-+-- x 

o 

3 

(x _ 2)2 + (y _ t )2 : I 

for x in [1,3] 

20. x=I+2sint, y=2+3cost, 

for t in [0, 21&] 

x-I = 2sint 

x-I . 

--=smt 

2 

y-2=3cost 

y-2 

--=cost 

3 

2 

Since sin 2 t + cos t = I, 

(X~Ir +(Y;2r =1 

(x-I)2 (y_2)2 1 

~----:;,-+ 

4 9 -. 

Since - 1$ sin t $ I, 

-2 $2sint $2 

-1$I+2sint$3. 

359



Therefore, x is in [-1, 3]. The graph is an ellipse 

with center (1, 2) and vertices (3,2), (-1, 2), 

(1,5), (1, -1). 

y 

23. x = sin t, y =cos t 

This graph is a circle centered at the origin with 

radius 1. 

y 

x =sin / 

y e cosr 

-t--+---::-I-+--t--- x 

21. 

(x_1)2 ~-1 

4 + 9 - 

for x in [-1, 3] 

Since x =t + 2 and y =_1_, y =1.. 

t+2 x 

Since t ¢ -2, x 'I:- -2 + 2, x ¢ O. 

Therefore, x is in (-00, 0) V (0, 00). 

y 

24. 

~4-4t2 

X= t, Y = -'--- 

2 

This graph is the semicircle formed by the top 

half of the graph of #23 . 

y 

22. 

-2 

2 

2 

-2 y = 1 x 

for xin 

(-00,0) u (0, 00) 

2 

x=t-3, y=--, t'l:-3 

t-3 

S . 2 2 

mce y=--, y=-. 

t-3 x 

Since t'l:- 3, x ¢ 3 - 3 =O. 

Therefore, x is in (-00, 0) V (0, 00). 

y 

25. 

--.---::-I--+--- x 

-I 

x = / 

Y =-14 -41 2 

2 

x =t + 2, y = t 4 

This graph is a line through (6, 0) and (0, -6). 

y 

:E::t-++-tAf-+-t-++4" x 

y = 1 

x 

for x in 

(--,0) u (0, 00) 

26. 

x =t 2 +2, Y =t 2 - 4 

This graph is a line segment which has (2, -4) as 

its endpoint and goes up to the right. It contains 

the point (6, 0). 

y 

6 

HH--t-H-+::ofL-i-t--t- o 

x 

360



27. y 

2 

x = t -sin t 

y = 1 - cos t for t in [0, 4n) 

+----1f--~-l---+. x 

n 2n 31r 4n 

28. y x = 2t - 2 sin t • Y= 2 - 2 cos t 

for t in [0 , 81r] 

4 

For Exercises 29-32, recall that the motion of a projectile (neglecting air resistance) can be modeled by: 

x =(vO cos6)t, y =(vOsin6)t -16t 2 for t in [0, k). 

29. (a) x =(vo cosO)t y =(vo sin 6)t -16t 2 

=(48 cos 60 0 )t =(48sin 60 0 )t -16t 2 

..fj 2 

=48(~} =48--t-16t 

2 

=24t 

= -16t 2 +2Wt 

31. (a) x =(vo cosO)t y =(vosin6)t-16t 2 + 2 

=(88 cos 20 0 )t = (88sin200)t -16t 2 + 2 

(b) t=-- x ___ 

88 cos 20° 

)-l j )2 +2 

Y=88sin20 0 ( x x 

88 cos 20° '\88cos 20° 

= (tan200)x- 1 2 x 2 +2 

484cos 20° 

361



(c) 0=-161 2 +(88sin200)1+2 

-88 sin 20° ± ~r-(8-8-si-n-2-00-)2-_-4(---16-)-(2-) 

1 =-------:..:.....---....:....-....:....---=-~ 

(-16)(2) 

(using the quadratic form ula) 

-30.098 ± .../905.8759 + 128 

= 

-32 

=:: -0.064sec, 1.95sec 

Discard 1=-0.064 sec since it is an unacceptable answer. 

At 1 = 1.95 sec 

x = (88 cos 20°)(1.95) 

=:: 161ft 

The softball traveled 1.95 sec and 161 feet. 

32. (a) x = (vo cos 0)1 y = (vo sin 0)1 -161 2 + 2.5 

= (136 cos 29°)1 = (136 sin 29°)1-161 2 + 2.5 

= -161 2 + (136sin 29°)1+ 2.5 

(b) 

x 

1 =--- 

136 cos 29° 

Y=(136sin29 oj x )-li 

'\136cos29° \ 

= 1 2 x 2 + (tan 29°)x + 2.5 

1156cos 29° 

x )2 +2.5 

136cos 29° 

(c) 0 =-161 2 + (136sin 29°)1 + 2.5 

-136sin29o±~r-(1-3-6-si-n-29-0-)2-_-4-(---16-)(-2-.5-) 

1 =-----.!...:...---~-..:....--..:....:~.:.... 

2(-16) 

(using the quadratic formula) 

-65.934 ± ../4347.3066 + 160 

= 

-32 

=::-.04, 4.2 

(Discard 1 = -.04 since it gives an unacceptable answer). 

At 1= 4.2 sec, x =(136 cos 29°)(4.2) =:: 495 ft. 

The baseball traveled 4.2 sec and 495 feet. 

33. (a) 

x = 82.692650631 

Y = -161 2 + 30.09m2611 

, 

" 

: .. ; ". 

.",1; , 200 

(b) 

x = 82.692650631 = vo(COSO)1 

82.69265063 = vo cosO 

Y = -161 2 + 30.097772611 = vo(sinO)1-161 2 

30.09777261 = vo sin 0 

30.09777261 = vosinO 

82.69265063 vo cosO 

0.3697 = tan 0 

0=20.0° 

(c) 

30.09777261 = vosin20.0° 

vo = 88.0 ftlsec 

x =88(cos200)1 

Y = -161 2 + 88(sin 20°)1 

362


34. (a) 

x = 56.56530965t 

y = -16t 2 + 67.41191099t 

l 

2 

x 

37. ---=1 

2 

a b 2 


To find a parametric representation, let 

x=asecO. 

Then 

(a sec 0)2 i_ 

a2 b2- 1 

(b) x =56.565309651=vo(COSO)1 

56.56530965 = vocosO 

(e) 

y =-161 2 +67.411910991 

= -16/ 2 + vo(sinOi 

67.41191099 = Vo sinO 

67.41191099 = Vo sinO 

56.56530965 Vo cos 0 

1.1918 = tan 0 

0=50.0 0 

67.41191099 = vosin 500 

Vo = 88.0 ftIsec 

x =88(cos500)1 

Y = -161 2 + 88(sin 50 0)1 

35. The equation of a line with slope m through 

(Xl. Yl) is 

Y- YI =m(x-xl)' 

To find two parametric representations, let x = I. 

Then 

Y-Yl =m(/-xl) 

Y =m(/-xI)+ YI' 

2• 

Or, let x = 1 

Then 

2 

Y - Yl =m(/ - xl) 

2-XI)+Yl' 

y=m(/ 

36. Y=a(x _h)2 +k 

To find one parametric representation. let 

x e t, 

Then 

Y =a(/-h)2 +k. 

For another representation, let 

x=1 + h. 

Then 

Y =a(/+h _h)2 + k 

2 

=al +k. 

sec 2 0-i = 1 

b 2 y 

2 

20-1 

sec = 

b 2 

l 

2 

tan 20=L 

b 2 

b 2tan20=y2 

btanO= y. 

2 

x 

38. -+-=1 

2 

a b 2 

Let x = a sin I. 

Then 

a 2 sin21 y 

2 

2 

a b 2 

----;;--+- =1 

2 

sin 2 / + L =1 

b 2 

2 

L=I-sin 2 1 

b 2 

l = b 2 (l - sin 2 I) 

i =b 2 COS 2 1 

Y =b cost. 

41. The second set of equations x = cos I, Y =-sin I, I 

in [0, 2n] trace the circle out clockwise. A table of 

values confirms this. 

I x =cosr y= -sinl 

o cosO = 1 -sinO=O 

t cost=1/ -sint=-! 

t 

cost=1f - sint=-4 

t cost=! -sini=-~ 

t cos!=O -sin! =-1 

42. If x = j(I) is replaced by x = c +j(/), the graph will 

be translated c units horizontally. IfY = g(/) is 

replaced by Y = d + g(/), the graph is translated 

vertically d units. 

363



Chapter RReview Exercises 11. (2 + 6i)2 = 4 + 24i + 36i 2 

1. H = i.J9 = 3i 

2. -./-12 =iM=2i-J3 

= 4+24i +36(-1) 

=-32+24i 

12. (6 - 3i)2 = 36 - 36i + 9i 2 

3. x 2 =-81 = 27 -36i 

x = ±-./-81 13. 

x =±i.J8l 

(1-i)3 =(1-i)2(1-i) 

x = ±i(9) 

x=±9i 

=(1-2i+i 2)(I-i) 

The solutions are 9i and -9i. = -2i(1- i) 

4. x(2x + 3) =-4 

=-2i+2i 2 

=-2i-2 

2x 2 +3x+4 =0 

or-2 - 2i 

Use the quadratic formula with a = 2, b = 3, and 

c=4. 

14. (2 + i)3 = (2 + i)2(2 + i) 

-3±~32-4·2·4 

x= = (4 + 4i + i 2)(2 + i) 

2 ·2 

=(3 + 4i)(2 + i) 

-3±-./9-32 

x = -----:.- 

4 

= 6 + 3i + 8i + 4i 2 

-3±..J=23 

=2+ 11i 

x= 

4 

6 + 2i 6 + 2i 3 + i 

15. --=-_. 

-3± i..fi3 

x=-----' 3-i 3-i 3+i 

4 

18+12i+2i 2 

· 3..fi3. d 3 ..fi3 . 

=-----,.- 

The soIutions are --+--1 an -----1- 9-i 2 

4 4 4 4 

18+12i+2(-I) 

= 

5. (l -I) - (3 + 4i) + 2i 9-(-1) 

=(1-3)+(-1-4+2)i 

16+ 12i 

=-2- 3i = 

10 

6. (2 - 51)+ (9 - 101) - 3 8+6i 

=- 

=(2 + 9 - 3) + (-5 - 1O)i 5 

=8 -15i 8 6. 

=-+-1 

5 5 

7. (6-5/)+(2+7i)-(3-21) 

=(6 + 2 - 3) + (-5 + 7 + 2)i 

2 - 5i (2 - 5i)(I- i) 

=5 +4i 16. --=~-;....;.....---'- 

1+ i (1+ i)(I- i) 

8. (4 - 2i) - (6 + 5i) - (3 - i) 

2 - 2i - 5i + 5i 2 

=(4-6-3)+(-2-5+ l)i = 

= -5 - 6i 

1-i 2 

-3-7i 

9. (3+ 5i)(8 - i) 

=- 

2 

=24- 3i+ 40i - 5i 2 3 7. 

= 24 + 37i - 5(-1) 

= 29+ 37i 

10. (4 -i)(5+2i) =20+8i -5i -2i 2 

=22+3i 

=----1 

2 2 

364



2 +; 2 +; 1+5; 

17. --=-_.- 

1-5i 1-5i 1+5; 

18. 

2+ IIi +5i 2 

1-25;2 

2 + 11i+ 5(-1) 

= 

1-25(-1) 

-3 + IIi 

= 

26 

3 11 . 

=--+-1 

26 26 

3+2i (3+ 2i)(-i) 

(i)(-i) 

19. i 53 = i 52 . i 

-3i-2i 2 

= 

-I·2 

=2-3i 

=(;4pi 

=1 13; 

= I ·i 

=i 

or 0 +; 

·-41 ·-44·3 

20. 1 = 1 . 1 

= (;4)-1 I(-i) 

= I-II(_i) 

=-i or 0-; 

21. 1.i . i 2 . i 3 = 1. i 6 = i 6 = ;4 .;2 

= (1)(-1) =-1 

22. 1·;·i 2 ·...· i loo 

Notice that i 2 = -1 and i 3 = -i. 

Also notice that 

i 4 = ;8 = ;12 = I, 

;5 = ;9 = i l 3 = i, 

.6 .10 .14 1 

1=1 =1 =- , 

and; 7 = ill = ;15 = _i. 

(1 ..2 .3) ('4 ·5 ·6 .7 ) 

.1. 1 . 1 . 1 . 1 .1 .1 

. (i8 .;9 .;10 . ;11) ... ..;100 

= (1. i -(-1)(-;)) 

.(1. i- (-1) ' (-i)) 

. (1 . i .(-1) .(-i)) ..... iI00 

= (-1) · (-1) ' (-1) ' .... ;100 

There will be twenty-five -Is. 

=-I -i loo 

=(-1)'(1) 

=-1 

23. [5(cos 90° + i sin 90°)] 

{6(cos 180° + ;sin 1800)] 

= 5 . 6[oos(90° + 180°) 

+; sin(90° + 180°)) 

= 30(cos2700 + ;sin2700) 

= 30(0-i) 

= 0 - 30i or - 30i 

24. [3cis135°][2cisI05°] 

= 3· 2cis(135° + 105°) 

= 6cis240° 

= 6(cos 240° + isin 240°) 

=6(-~- ~ ;) 

=-3-M; 

25. 

2(cos 60° + i sin 60°) 

8(cos 300° + sin 300°) 

= ...!.[cos(-2400)+ isin(-2400)] 

4 

1 .J3 . 

=--+-, 

8 8 

365



26. 

4cis 270° 33. 

2cis90° 

y 

4 . -4 +2i 

=- cIs(270°- 90°) 2 

2 

+++++F!!H-H-H-f.- x 

= 2cis180° -4 

= 2(cos 180° + isin 180°) 

= 2(-1 + Oi) 

= -2+0i or -2 

27. (.J3 + i)3 

r = ~r-(.J3"""":3:-:)2"-+-17"2 = 2 

34. y 

1 .J3 

tan 0 = .J3 = 3""' 0 = 30° 

3 

( .J3 + i)3 = [2(cos 30° + i sin 30°)]3 

= 2 3(cos90° + isin 90°) 

=0 +8i or 8i 

35. (7+3i)+(-2+i) 

28. (2-2i)5 =(7-2)+(3+1)i 

=5+4i 

r =.J8 = 2..[i 

y 

-2 

tanO=-=-1 

2 

5 +4i 

Since 2 - 2i is in quadrant IV, 

0=315°. 

(2-2i)5 

= [2..[i(cos315° + isin 315°)]5 

= (2..[i)5[cos(5. 315°) + i sin(5· 315°)] 

= 128..[i(cosI575° + isin 1575°) 

= 12g..fi(cos 135° + i sin 135°) 

++++++-dG--H-H~x 

36. -2 + 2i 

r =~r-(_-2-)2=--+-27"2 

1284-'7 = +i '7) =../8 =2..[i 

=-128+128i 

2 

tanO=-=-1 

-2 

29. (cosl000+isinl000)6 ois in quadrant ITso 0 = 135°. 

= cos 600° + i sin 600° -2+2r 

1 .J3. = 2..[i(cos 135° + i sin 135°) 

=----1 

2 2 

37. 3(cos900+isin900) 

30. The vector representing a real number will lie on = 3(0+ i) 

the x-axis in the complex plane. 

= 3i or 0+ 3i 

32. 

y 

38. 2(cos 225° + i sin 225°) 

5i 

=2(-'7 _i~) 

++++++++++++--+-..- x 

o 

= --Ii -i..[i 

366



39. -4 +4i.J3 

r=..J16+48 =8 

tan8 = 4.J3 =-J3 

-4 

8 is in quadrant II. 

8 = 120 0 

-4+4i.J3 

= 8(cos120 0 + i sin 120 0) 

40. 1-i 

r = ~r-12-+-(_-1)-2 =.,fi 

-1 

tan8=-=-1 

1 

8 is in quadrant IV so 8 = 315 0. 

1-i = .,fi(cos315° + isin 315 0) 

41. 4 cis 240 0 

= 4(cos240° + i sin 240 0) 

={-±-i~) 

:: -2-2i.J3 

42. -4i 

r=4 

8=2700 

-4i = 4(cos270 0 + i sin 270 0) 

43. Since the modulus of z is 2, the graph would bea 

circle, centered at the origin, with radius 2. 

44. z=x+yi 

Since the imaginary part ofz is the negativeof the 

real part of z, we are saying 

y=-x. 

This is a straight line. 

45. Convert -2 + 2i to polar form. 

r=~(_2)2+22 =.J4+4 =../8 

8 = arctan(:J = arctan(-I) = 135 0 

(since x < 0 andy> 0) 

then -2 + 2i = ../8(cos135°+ isin 135 0) and the 

5th roots of ../8(cos135°+isin135°) are: 

l~(cosa + isina) 

8+360 0k 

where a = for k = 0, 1, 2, 3, 4 

n 

k = 0: a = 135±360(0) = 135 = 270 

5 5 

k = 1: a = 135+ 360(1) = 495 = 990 

5 5 

k = 2: a = 135+ 360(2) = 855 = 171° 

5 5 

k = 3: a = 135+ 360(3) = 1215 = 2430 

5 5 

k = 4: a = 135+360(4) = 1575 = 3150 

5 5 

So the fifth roots of -2 + 2i are: 

l~(cos 27 0 + isin 27 0) 

l~(cos 99 0 + i sin99 0) 

l!Y8(cos 171 0+ isin 171 0) 

l~(cos 243 0 + isin 243 0) 

l~(cos 315 0 + isin315°)' 

46. Convert 1 - i to polar form 

r=~12+(-1)2 =.J1+l =../2 

8 = arctan( ~ 1 ) = arctan(-1) = 315 0 

(since x > 0 andy < 0) 

then 1-i = .,fi(cos315° + isin315°) and the cube 

roots of .,fi(cos315°+isin315°) are: 

~(cosa+isina) where 

8+36O° ·k 

a= fork=O, 1,2 

n 

k=O: a> 315+(360)(0) = 315 =1050 

3 3 

k = 1: a = 315 + (360)(1) = 675 = 2250 

3 3 

k = 2: a = 315 + (360)(2) = 1035 = 3450 

3 3 

So the cube roots of 1 - i are: 

~(cos 105 0 + isin 105 0) 

~(cos225° + isin 225 0) 

~(cos 345 0 + isin 345 0) 

367



47. The real number - 32 has one real fifth root. The 

one real fifth root is -2, and all other fifth roots 

are not real. 

48. The number -64 has no real sixth roots because a 

real number raised to the sixth power will never 

be negative. 

49. x 3 + 125 = 0 

3 

x = -125 

-125 = -125 +Oi 

-125 = 125(cosI80° + isin 180°) 

1251/3 = 5 

180° + 360° . k 

a k =0, 1, 2 

3 

a = 60°, 180°, 300° 

x = 5(cos6O° + isin600), 

5(cos 180° + isin 180°), 

5(cos 300° + i sin 300°) 

50. x 4 + 16 = 0 

x 4 =-16 

=16(cos 180° + isin 180°) 

r =16, r ll4 =2 

_180 0+360° ·k. k=O 1 2 3 

a- 4 ' ' " 

a = 45°, 135°, 225°, 315° 

x = 2(cos 45° + i sin 45°), 

2(cos135° + isin 135°), 

2(cos 225° + i sin 225°), 

2(cos315° + isin 315°) 

51. x 2 +i =0 

x 2 =-i 

=O-i 

=1(cos 270° + i sin 270°) 

~=1 

a = 270° + 360° . k; k = 0, 

2 

a = 135°, 315° 

x = (cos 135° + i sin 135°), 

(cos315° + i sin 315°) 

52. r =~(_1)2 + (.J3)2 = -J1+3 =.J4 =2 

53. x =rcose y =rsine 

= 5cos315° =5sin315° 

5../2 ../2 

=-- =5- 

2 2 

5../2 

=-- 2 

the rectangular coordinates are: 

5../2 _5../2) 

( 2' 2 

54. the angle must be quadrantal. 

55. Since r is constant and e ~ e~ 2n, it will be a 

circle. 

56. r=4 cos e 

e 

r 4 3.5 2.8 2 0 

e 120° 135° 150° 180° 

r -2 -2.8 -3.5 -4 

Graph is retraced in the interval (180°, 360°) 

90" 

270" 

r=4cos8 

57. r = -1 + cos e 

e 

r o -.7 -.3 -.5 0 

r -1.5 -1.7 -1.9 -2 -1 -.3 

90" 

e=arctan(- ~ J=arctan(-J3) = 120° 

(sin x < 0 and y > 0) 

the polar coordinates are: (2, 120°) 

270" 

r=-I+cos8 

368



58. r= 2 sin 4£} 

£} 0° 7.5° 15° 22.5° 0° 37.5° 45° 

r 0 

-13 

2 

-13 

1 0 

£} 52.5° 60° 67S 75° 82.5° 90° 

r -1 

--13 

-2 

-J3 -1 0 

The graph continues to form eight petals for the 

interval [0°,360°). 

90° 

i 

64. r =sin £)+ cos£} 

r 2 =rsin £} + rcos£} 

x 

2+l 

=x+y 

x 

2+l-x-y=0 

or 

65. r=2 

~x2+l =2 

x 2 +l =4 

r=2sin 48 

59. Suppose (r, £}) is on a graph. -£} reflects this 

point with respect to the x-axis, and -r reflects the 

resulting point with respect to the origin. The net 

result is that the original point is reflected with 

respect to the y-axis (B). 

60. If(r, £}) lies on a graph, (-r, £}) would reflect that 

point into the quadrant diagonal to the original 

quadrant. Therefore, there is symmetry about the 

origin (A). 

61. If(r, £}) lies on a graph, (r, -£}) would reflect that 

point across the x-axis, Therefore, there is 

symmetry about the x-axis (C). 

62. If(r, £}) lies on a graph, (r, 1t - £}) would reflect 

that point into the other quadrant on the same side 

of the x-axis as the original quadrant. Therefore, 

there is symmetry about the y-axis (B). 

3 

2 

63. r=--r=- 

J + cos£} 

cos£} 

r(1 + cose) =3 

I 

r=2·- 

r+ reas£} =3 

cos£} 

r =2 sec£}o 

~x2+l +x=3 

x 2 + l =(3-x)2 

70. This is a circle centered at the origin with radius 

x 2 + l =9 - 6x + x 2 

2. Its equation is 

x 2 +l =4. 

l =9-6x 

l+6x-9=0 

or l =-6x+9 

66. y=x 

rsin£} = rcos£} 

sin£}= cos£} 

or 

tan £} =1 

67. 

y=x 

2 

rsin £} =(rcos£})2 

sinO 

r =--2- =tan £)sec £} 

cos £} 

tan £} 

or r= - 

cos£} 

68. This is the line y = 2. Since y = r sin £}, 

rsin£} =2 

1 

r=2·- 

sin£} 

r = 2 csc£}. 

69. The line shown is a vertical line, x == 2. Since 

x = 2 and x =r cos£}, 

rcos£} = 2 

Since r = ~r-x2--+y-2, 

r 2 =x2 + y2 

r 2 =4 

r=2. 

369



71. x = t + cost, Y = sint for t in [0,27t] 

t x::::: t + cos t y::::: sinr 

o O+cosO:::::I sin 0 =0 

7t 7t 7t 7t+ 3.J3 · 7t 1 

-+cos-=-- sm-= 

6 666 

6 2 

7t 7t 7t 27t+ 3 

-+cos-=- sm-= 

· 7t .J3 

3 3 3 6 3 2 

7t 7t 7t 7t 

-+cos-=- sm-= · 7t 1 

2 2 2 2 2 

37t 37t 37t 37t- 2...{i · 37t Ii 

-+cos-=---- sm-= 

4 4 4 4 4 2 

7t 7t+cos7t=7t-I sin 7t= 0 

y 

I 

o 

(~ ,-1) 

x=t+cost 

y= sin t 

for t in 10,2n) 

72. x::::: 3t + 2, Y =t - 1, for t in [-5,5] 

Solve for tin terms of y. 

y+I:::::t 

Substitute y + 1 for t in the equation for x. 

x =3(y+I)+2 

x=3y+3+2 

x=3y+5 

x-3y=5 

Since t is in [-5,5], x is in [3(-5) + 2,3(5) + 2] or 

[-13, 17]. 

73. x =.Jt::J" y =.Ji, for t in [1, 00) 

Solve for t in terms of x. 

x 2 =t-I 

x 2 +1 =t 

Substitute x 2 + 1 for t in the equation for y. 

y :::::~x2 +1 

Since t is in [1, 00), x is in [.JI=l, 00) or [0, 00). 

74. x =t 2 +5, y =~, for t in (-00, 00) 

t +1 

Then x - 5 = t 2 . Substitute x - 5 for t 2 in the 

equation for y. 

1 

y=- 

t 

2 

+1 

1 

y=-- 

x-5+I 

1 

y=- 

x-4 

2 2 

Since x = t + 5 and t ~ O. x ~ 0 + 5 =5. 

Therefore, x is in [5,00). 

75. x = 5 tan t, y ::::: 3 sec t, for t in ( - ~ , ~) 

Then 

~ = tant and L = sect. 

5 3 

2 2 

Since 1+ tan t = sec t 

I+(~r =(fr 

x2 

y2 

1+-= 

25 9 

{I+~:)=i 

A=y 

3~I+x2 

25 

=y. 

S· mce t .. ( tr tr) id 5 . 

IS 10 - '2''2 ' an x = tan t IS 

undefined at - tr and tr, x is in (-00, 00). 

2 2 

76. x =cos 2t, Y = sin t for tin (-tr, n) 

cos 2t = cos 2 t - sin 2 t (double angle formula) 

Pythagorean Theorem: 

cos 2 t + sin 2 t =1 

. cos 2 t -sin 2 t+2sin 2 t =1 

x+2i =1 

2/ =I-x 

2 1 

y =-(I-x) 

2 

Since t is in e-x. n), and cos 2t is in [-1, 1], x is in 

[-1,1]. 

370


77. The vertex occurs halfway between the zeros of the given equation. If y = O. then 

16 2 

0= (tanO)x - 2 2 x 

vo cos 0 

-16x 2 + v0 2(cos2 O)(tan 0) x = 0 

j -16x+ Vo2 cos 2 0 sin 0) = 0 

A\ cosO 

x=O. -16x+vo 2cosOsinO=0 

16x = Vo 2 cosOsinO 

vo 2 cosOsinO 

x =--"---- 

16 

v02 sin20) 

or x=~--- 

( 32 

the midpoint between these values is: 

vQ2sinOcosO - 

0 2· ( 2· ) 

= 

16 vo smOcosO or vo sm20 

2 32 64 

v02 sinOcosO 

At x =-"---- 

32 • 

= (tano{V 02 sinOcosO)_( 16 Xv 2 sinOCoso)2 

y 32 vo2 cos 2 0 0 32 

_ V0 2 sinOcosOsin 0 16vo 4 sin 2 Ocos 2 0 

- 32 cosO (32)2v02 cos 2 0 

= v 2 sin 2 0 v 2 sin 2 0 _0"'--__ 0 

32 64 

= 2vo2 sin 2 0 - Vo 2 sin 2 0 

64 

v 2 sin 2 0 

= 0 

64 

The vertex is: 

2 2 2 2 

V 02 sinOcosO V0 sin 0) or (V 02 sin 20 V0 sin 0) 

( 32 '64 64' 64 

78. Since x = a + rcos t, y =b + r sin t, tin [0. 21l'l. is a circle with center (a. b) and radius r 

a circle with center (3. 4) and radius 


r= ~(3-0)2 +(4-0)2 =.,fi5 =5 

would be 

x =3+Scost 

Y =4 +Ssint 

tin [0. 21t]. Note that the radius is the distance between (3,4) and (0.0) since (3.4) is the center and the 

circle must contain the origin. 

80. (a) x = (vo cosO)t y = (vO sin O)t-16t 2 + 3.2 

= (118cos27°)t y =(118sin 27°)t -16t 2 + 3.2 

371



(b) 

x 

t=-- 

118cos27° 

y=118sin270 ' x -16( x )2 +3.2 

118cos27° 118cos27° 

4 2 

= (tan 27°)x - 2 x + 3.2 

3481cos 27° 

(c) 0=(118sin27°)t -16t 2 +3 .2 

-118sin 27° ± ~r-(l-18-s-in-2-7-0)-2---4-(--1-6)-(3-.2-) 

t =------'--------.....;...;----' 

2(-16) 

-53.571±.J2869.839 + 204.8 

= 

-32 

== -.059, 3.4 sec. 

(Discard t =-0.59 sec since it is an unacceptable answer.) 

At t = 3.4 sec, 

x = (118cos27°)(3.4) 

== 357 ft 

The baseball traveled for 3.4 sec and 357 feet. 

Chapter 8 Test Exercises (d) ~= 2-4i. 5-i 

z 5+i 5-i 

1. (8) W+Z =(2-4i)+(5+0 

=(2+5)+(-4+1)i 

1O-2i -20i+4;2 

=7-3i - 25 - 5; + 5i - i 2 

y 

10+ 4(-1) - 22; 

= 

25-(-1) 

6-22i . 

=- 

26 

3 11 . 

=---1 

7 -3; 

13 13 

2. (a) il5 =;4 .;4 .;4 . i3 

(b) w-z=(2-4i)-(5+;) = 1.1.1 .i 3 

=(2-5)+(-4-1)i =i 3 

=-3-5i 

=-i 

(c) WZ :::: (2 - 4i)(5 +;) 

(b) (l+i)2 =O+;)(I+i) 

:::: 10+ 2i - 20; - 4i 2 = l+i+;+i 2 

= 10-4(-1)-18; 

=2;+1+(-1) 

=14-18; 

=2i 

372



3. 2x 2 -x+4 = 0 

Using the quadratic formula : 

1±~(-1)2 -4(2)(4) 

x =--'-----'---- 

2(2) 

1±.Jl=32 

= 

4 

1±.J=3l 

= 

4 

1±i.J3i 

= 

4 

x ={1+ i.J3i 

4 ' 

4. (a) r = ~02 + 3 2 =.J9 = 3 

The point (0, 3) is on the positive y-axis. 

Thus, () = 90°. 

3i = 3(cos 90° + i sin 90°) 

(b) r=~12+22 =-./5 

() = arctan(T) = arctan(2) = 63.43° 


1+ 2i = .J5(cos63.43° + isin63.43°) 

(c) r=~(-1)2 + (-J3)2 =.../1+3 =-14 =2 

() = arctan( -!!) = arctan(.J3) = 240° 

(since (x < 0 and y < 0) 

-1- .J3i =2(cos 240° + i sin 240°) 

S. (a) a =rcos() b =rsin() 

= 3cos30° =3sin30° 

=\±) 

3 

= 2 

3(cos30° + isin300) = 3.J3 +~i 

2 2 

(b) a = rcos() 

b = rsin() 

= 4 cos 40° =4sin40° 

=3.06 = 2.57 

4 cis 4° = 3.06 + 2.57i 

(c) a = rcos() b = rsin() 

= 3cos90° = 3sin90° 

=0 =3 

3(cos 90° + i sin 90°) = 3i 

6. (a) wz= 8 .2(cos(40° + 10°)+ i sin(40° + 10°)] 

= 16(cos500+isin500) 

w 8 

(b) - = -[cos(40° _10°) + isin(40° _10°)] 

z 2 

= 4(cos 30° + i sin 30°) 

a = r cos () b = r sin () 

=4cos30° 

=4sin30° 

.J3 = 4 . ..!. 

=4·- 2 

2 

=2 

=2.J3 

a+bi = 2.J3 +2i 

(c) 

z3 = [2(cos1Oo+ isin100)]3 

= 2 3(cos3 · 10° + isin3 · 10°) 

= 8(cos30° + isin300) 

a = 8cos 30° b = 8sin30° 

=8 . .J3 =8 . ..!. 

2 

2 

=4.J3 

=4 

a + bi = 4.J3 + 4i 

7. Put -16i into polar form 

r=~02+(-16)2 =../256=16 

() =arctan( -~ 6 ) = 270° 

(since tan «() is undefined and y < 0) 

then -16i =16(cos 270° +i sin 270°). 

The fourth roots of -16i have the fonn 

q{;(cosa+isina) where 

a =() + 360 0k, k =0, I, 2, 3 

n 

k = 0: a = 270° + 360(0) = 67.5 

4 

k = 1: a = 270 ± 360(1) = 157.5 

4 

k = 2: a = 270 + 360(2) = 247.5 

4 

k = 3: a = 270 + 360(3) = 337.5 

4 

The fourth roots of-16i are: 

2(cos67.5° + sin67.5) 

2(cos 157.5° + i sin 157.5°) 

2(cos247.5° + isin 247S) 

2(cos337.5° + isin 337.5°) 

373



For Exercise 8, answers may vary. 

8. (a) r =~02 + 52 =.J25 =5 

The point (0, 5) is on the positive y-axis. 

Thus, 9 =90". 


Alternatively, if,=-5, 

9 = 90 + 180 =270° 


Two pairs : 

(5,90°), (-5, 270°). 

(b) '=~(_2)2+(_2)2=.,J4+4=..[8=2.,fi 

9 = arctan( =~) = arctan(l) = 225° 

(since x < 0 and y < 0) 

One possibility is: (2.,fi, 225°) 

Alternatively, if r = -2.,fi 

9 = 225-180 = 45° 

A second possibility if: (-2.,fi, 45°) 

Two pairs: 

(2.,fi, 225°), (-2.,fi, 45°) 

9. (a) x ='cosO y = ,sinO 

= 3cos315° = 3sin315° 

.,fi 

=3· 

2 ={- ~) 

3.,fi 

=-- -3.J2 

2 =- 

2 

The rectangular coordinates are: 

3.,fi - 3.,fi). 

( 2' 2 

11. r = 3cos39 

90· 

r=l-cosO 

r 3 0 -2.1 -3 0 3 

o 135° 150 0 180 0 

r 2.1 0 -3 

Graph is retraced in the interval (180 0 , 360°). 

90· 

)/-±--,; 

-/.. \~ rA.'I'').,· 

I 7'. -< K'\ \ 

18O·t ._-...- -' 3~ 0" 

-\- . Y/. 

/ 

I 

270· 

r=3OO530 

12. This is the line where a = -1, b = 2, and c = 4 

The rectangular form is: 

-x+2y=4, so 

x-2y=-4 

y 

(b) x = ,cos9 

y = ,sinO 

= -4 cos 90° = -4sin 90° 

=0 =-4 

The rectangular coordinates are: (0, -4). 

10. ,=I-cosO 

r o .1 .3 .5 1.7 

r 2 1.7 1 .3 o 

374



13. x = 4t 3, y = t 2 for tin [-3,4] 

x y 

-3 -15 9 

-1 -7 

0 -3 0 

1 1 

2 5 4 

4 13 16 

y 

(13, 16) 

y 

~~ 

x=2cos 21 

y =2 sin 21 

for 1 in [0, 21f] 

15. z2 - 1= (-1 + ;)2 - 1 

2 

=1-;-;+;2 -1 

=-2;-1 

= -1-2; 

r= ~(_1)2 +(_2)2 =../1 +4 =.J5 > 2 

Since r » 2, z is not in the Julia set. 

-16 (-3,0) 

x 

2 

x=41-3 

Y=1 2 

for 1in [-3.4] 

14. x = cos 2t, y = sin 2t for tin [0, 21t] 

t x Y 

0 0 

f 4 4 

i 

0 

t -4 4 

t 

-1 0 

-Sf -4 -4 

¥ 

0 -1 

1t 1 0 

¥ 

0 

'* 

-1 0 

1Jr.. 0 -1 

4 

21t 1 0 

375


376 Chapter 9 Exponential and Logarithmic Functions 

CHAPTER 9 EXPONENTIAL AND 

LOGARITHMIC FUNCTIONS 7. g(X)=(~r 

Section 9.1 1) -2 2=16 

g(-2)= ( '4 =4 


1 2 1 3 1 4 

1. 1+1+-+--+-- 8. g(X)=(±r 

2 ·1 3·2 ·1 4 ·3 ·2·1 

15 1) -3 

+ ,z 2.717. g(-3)= ( '4 =4 3 =64 

5 ·4 ·3·2 ·1 

The calculator gives 2.718. 

9. /(x) = 3 x 

2. 1+ (-.05) + (-.05)2 + (-.05)3 + (_.05)4 /(1.5) = 31.5 ,z 5.196152423 

2 ·1 3 ·2·1 4 ·3 ·2 ·1 

5 

+ (-.05) ,z .9512. 

5 ·4·3 ·2 ·1 10. g(X)=(~r 

The calculator gives..9512. 

g(1.5) =(±r =i 

3. 6·5·4 ·3 ·2 ·1 

11. g(x) =(±r 

Exercises 

1. /(x) =3 x g(2.34) =(±f.34 

/(2)= 3 2 =9 

,z .0390103297 

2. /(x) = 3 x 12. /(x) =3 x 

/(3)= 3 3 =27 f(1.68) =31.68 

,z 6.332332457 

3. /(x) = 3 x 13. Yes;f(x) =aX is a one-to-one function. 

/(-2) =r 2 =.!. 

9 Therefore, an inverse function exists for f 

4. /(x) =3 x 14. Since f(x) =aX has an inverse, the graph of 

/(-3)=r 3=-.!.. 

5. g(x) =(~r 

g(2)=(.!.J2 =-.!.. 

4 16 

/-I(x) will be the reflection of/across the line 

27 y=x. 

y 

6. g(X)=(±r 

g(3)=(.!.)3 =_1 

15. Since /(x) =aX has an inverse, we find it as 

. 4 64 follows. 

y=a 

x 

x=a Y 

376



16. Ifa =10, the equation for I-I(x) will be given by 

x = IO", 

17. Ifa =e, the equation for I-I(x) will be given by 

x=e Y • 

18. If the point (P, q) is on the graph of f, then the 

point (q, p) is in the graph of 1-1. 

19. I(x) = 3 x 

Make a table of values. 

x -2 -1 0 2 

1 

y 3 9 

9 3 

Plot these points and draw a smooth curve 

through them. This is an increasing function. The 

domain is (- 00,00) and the range is (0,00). The 

x-axis is a horizontal asymptote. 

20. I(x) = 4 x 

y 


2 

x -2 -1 0 1 2 

I 1 4 16 

Y 16 4 


through them. This is an increasing function. The 

domain is (- 00,00) and the range is (0, 00). The 


y 

x -2 -1 0 2 

1 1 

y 9 3 

3 9 


through them. This is a decreasing function . The 

domain is (- 00,00) and the range is (0,00). The 


y 

_1 0 1 

3 !(x) = (Ir 

I 3 

22. I(x) =(±r 


x -1 0 1 2 

I I 

Y 4 1 - 

4 16 


through them. This is a decreasing function. The 

domain is (- 00,00) and the range is (0, 00). The 


y 

23. I(X)=(%r 

4 

!(x) =(~r 

The domain is (- 00, 00) and the range is (0, 00). 


x -2 -1 0 2 

y 

4 2 3 9 

9 3 2 4 

21. I(X)=Gr 


377



y 

y 

!(X) = (~r 

5 

o 

3 

24. f(X)=(%r 

The domain is (- 00,00) and the range is (0, 00). 


x -2 -1 o 2 

y 

9 3 2 4 

1 

4 2 3 9 

y 

27. f(x)=e- x 

Use a calculator to find approximate values of y . 

x -2 -1 o 1 2 

y 7.39 2.72 1 .368 .135 

The graph of f(x) =e- x is the reflection of the 

graph f(x) =eX about the y-axis. 

y 

+-+---1--::1--+--+---'1- x 

f(x) =e- X 

x 

25. f(x) =e 

The domain is (- 00,00)and the range is (0, 00). 

Use a calculator to find approximate values of y. 

x 1 0 -1 

y 2 .36 

y 

28. f(x) = lO-x 


x -1 0 

y 10 1 

3 

1 

10 

The graph of f(x) =lO-x is the reflection of the 

graph of f(x) =lOX about the y-axis. 

y 

10 

26. f(x) =lOx 

The domain is (- 00,00) and the range is (0, 00). 


x 0 -1 

y 

10 

1 

10 

!(x) = 10..... 

1 

_2_1 0 

I 2 

29. f(x) =2 1xl 


x -3 -2 -1 0 2 3 

y 8 4 2 1 2 4 8 

x 

378



Notice that for x < 0, Ix 1= -x, so the graph is the 32. f(x) =:' 2 x - 4 

This graph is obtained by translating the graph of 

same as that of f(x) =2- x . For x ~ 0, Ix 1= x, so 

f(x) =2 x down 4 units. 

the graph is the same as that of f(x) =2 x . Since 

y 

1- x I=Ix ~ the graph is symmetric with respect 

to the y-axis. 

y 

y=-4 

1 33. f(x) =2 x + 1 

o 

3 


f(x)=PI 

f(x) =2 x to the left 1 unit. 

30. f(x)=2-!xl y 


x ~ -I 0 1 2 

y 

1 1 1 1 1 

4 2 2 4 

The domain is (- 00,00) and the range is (0, 1]. 

The graph is symmetric with respect to the y-axis. 

14~+-I-d--+-+-+-+-+~ x 

y o 

4 

34. f(x) =2 x - 4 


f(x) =2 x to the right 4 units. 

y 

+-+-+--:d--+---+~- x 

-2 2 

8 

31. f(x)=2 x + l 


f(x) =2 x up 1 unit. 2 

y 

o 

4 

8 

f(x) = 2" + 1 

y=1 

H-+-H-"t-+-+-+-+-+- x 

-2 0 2 

379



I)X-4 

35. f(X)=(~r -2 38. f(x)= ("3 


f(x) =(~r down 2 units. 

y 


f(x) = (~r 4 units to the right. 

y 

9 

!(x)=(tY-2 

+++++-+-H\tR--H--x 

4 

39. a =2.3 

36. f(X)=(~r +4 Since graph A is increasing, a > 1. Since graph A 


f(x) =(~r up 4 units. 

!(x)=HY+4 

y 

is the middle of the three increasing graphs, the 

value of a must be the middle of the three values 

of a greater than I. 

40. a = 1.8 

Since graph B is increasing, a > 1. Since graph B 

increases at the slowest rate of the three 

increasing graphs, the value of a must be the 

smallest of the three values of a greater than 1. 

41. a =.75 

y:4 Since graph C is decreasing, 0 < a < 1. Since 

graph C decreases at the slowest rate of the three 

+t-+++-H~++++-x 

o 

decreasing graphs, the value of a must be the 

3 

closest to 1 of the three values of less than 1. 

1)X+2 42. a =.4 

37. f(x)= ( "3 

Since graph D is decreasing, 0 < a < 1. Since 

The graph is obtained by translating the graph of 

graph D is the middle of the three decreasing 

f(x) =(~r 2 units to the left. 

graphs, the value of a must be the middle of the 

three values of a less than 1. 

y 

43. a =.31 

Since graph E is decreasing, 0 < a < 1. Since 

graph E decreases at the fastest rate of the three 

decreasing graphs, the value of a must be the 

closest to 0 of the three values of a less than 1. 

44. a =3.2 

+t-+++-HH-ff"l'"'l'-X 

Since graph F is increasing, a > 1. Since graph F 

2 increases at the fastest rate of the three increasing 

graphs, the value of a must be the largest of the 

three values of a greater than 1. 

380



48. I(x) =x 2 . T X 

This function may be graphed in the standard 

viewing window . but a better picture of the graph 

is obtained by using the window 

[-2.8] by [-2. 5]. 

~.....:.._-- 

I(x) = x2 • 2- .1 

-10 10 

46. 

x -x 

f(x) =e +e 

2 

Graph this function in the standard viewing 

window. 

e.r + e-.r 

I(x) =----r 

10 

49. 4 x =2 

Write both sides as powers of 2. 

(2 2 )x =2 1 

2 2x =2 1 

2x=1 Property (b) 

1 

x= 2 

47. 

-10 

f(x) =x·2 X 

Although this function can be graphed in the 

standard viewing window. a better picture of the 

graph is obtained by using window 

[-5.5] by [-2. 5]. 

I(x) = x· 2.1 

5 

10 

SO. 

51. 

125' =5 


(53), =51 

53' = 51 

3r= 1 Property (b) 

1 

r= 3 

(~r =4 

(2-1)k =2 2 .!. =2- 1 

2 

2- k =2 

-k=2 

k=-2 

Property (b) 

52. 

(~r =~ 

(~r =(%r 

(~r =(~r2 

x=-2 

Property (b) 

381



53. 2 3 - y =8 4 =8 213 4 =(_8)213 

2 3 - y =2 3 

3-y=3 

-y=O 

y=O 

54. 52p+1 =25 

5 2p + 1 = 52 

2p+l =2 

2p=1 

1 

p= 2 

Property (b) 

Property (b) 

= (WY = (~t=8Y 

=2 2 =(_2)2 

=4 =4 

Both proposed solutions check. 

58. z5/2 =32 

Raise both sides to the ~ power. 

5 

2/5 

( z5/2 

) 

=(32)215 

z=(32)2/5 

= (32 215)2 

=2 2 

=4 

59. 27 4z =9 z +1 

(3 3 )4Z =(3 2 )z+1 

312z =3 2z + 2 

12z =2z+2 

lOz=2 

1 

z= 5 

56. ..!- -_ k- 4 

81 

81- 1 =k- 4 

([(±3)]4fl =r 4 60. 32 1 =16 1 - 1 

(±3r 4 =k- 4 

[(±3)r 4-1I4 =(k- 4 ) 1/ 4 

±3=k 


(2 5 )1 =(2 4 )]- 1 

2 51 =2 4 - 41 

5t =4 - 4t 

9t=4 

4 

t= 9 

Property (b) 

.... 

57. 4 =,213 

Raise both sides of the equation to the 

3/ 2 

since 

( 

,213 

) 

=,I =r. 

43i2 =, 

(213)3/2 

(±~)3 =,1 

~ power, 

2 

(±2)3 =, 

±8=, 

Since raising both sides of an equation to the 

same power may result in false "solutions," it is 

necessary to check all proposed solutions in the 

original equation. 

61. (~rX =(~r+1 

(w=[mT' 

(~rX =(~rX+2 

-x=2x+2 

-3x=2 

2 

x=- 3 

382



62. (~Y-I =G~Y+I 

(~r~[mT+I) 

(~r =[(~rrl) 

k-I=4k-4 

5k=-3 

k=-~ 

5 

63. Use the compound interest formula to find the 

future amount A if the present amount 

P = 8906.54. r= .05. m = 2. and t = 9 

A=~I+ ~)nn 

I )9(2) 

= (8906.54,-1+ .~5 

= (8906.54)(1.025)18 

= (8906.54)(1.5596) 

= 13,891.16 

The future value is $13.891.16. 

23 

64. P=56.780. t =-, m =4. r=.053 

4 

A=~I+ ~)un 

053)(2t~4) 

=56.78 1+-' 

~ 4 

= 56.780(1 .01325)23 

=76.855.95 

The future value is $76.855.95. 

65. Use the compound interest formula to find P if 

11 

A =25.000. t =-. m = 4. and r= .06. 

4 

A=~l+ ~)nn 

.06)(1114)



(1.4833)1/20 = (1.4833)"05 

'" 1.01991. 

1.01991=1+.!:. 

4 

.01991=.!:. 

4 

.07964 = r 

The interest rate , to the nearest tenth, is 8.0% . 

69. (a) 1200 

(b) 

(c) 

-1000 11,000 

° 

From the graph above, we can see that the 

data are not linear but exponentially 

decreasing. 

71• 

(c) T(5) = 1.03(5) =5.15 

When R = 5 w/m 2 the global temperature 

increase is 5.15 DF . 

y = 448kolS6x 

(a) 1990 is 10 years after 1980, so use t = 10. 

A(lO) = 448kO IS6(10) 

= 4481e,IS6 

'" 4481(1.168826) 

= 5238 

The function gives a population of about 

5238 million , which differs from the actual 

value by 82 million. 

(b) For 1995, use t = 1995 - 1980 =15. 

A(15) =448kO IS6(IS) 

=448le"234 

'" 4481(1.2636445) 

=5662 


5662 million. 

(c) For 2005, use t = 25. 

A(25) = 448kO IS6(2S) 

(d) 

-1000 

° 

P(x) = 1013e-·000134lx 

P(l5OO) = 1013e-0001341(ISOO) 

=828 

P(ll,ooo) = 1013e-·0001341(11000) 

=232 

When the altitude is 1500 m, the function P 

gives a pressure of 828 mb, which is less 

than the actual value of 846 mb. When the 

altitude is 11,000 m, the function P gives a 

pressure of 232 mb, which is more than the 

actual value of 227 mb. 

70. (a) T is a linear function, not an exponential 

function. 

(b) The graph of T(R) passes through the points 

(0, 0) and (20, 20.6). The slope between 

th . . 20.6-0 103 

ese points IS m = =.. 

20-0 

Since the slope is 1.03 and the y-intercept is 

0, the equation is T(R) =1.03R. 

= 4481e.39 

'" 4481~1.416991) 

=6618 


6618 million. 

72. (a) T = 50,000{1 + .00)R. 

After 4 years, n =4. 

(b) 

T= 50,000(1+ .06)4 

=63,000 

Total population after 4 years is about 

63,000. 

T = 30,000{1 + .12)R 

After 3 years, n = 3. 

T =30,000(1 + .12)3 

=: 42,000 

There would be about 42,000 deer after 

3 years. 

384



(c) T = 45,000(1 + .08)n 

After 5 years, n = 5. 

T = 45,000(1 + .08)5 

=66,000 

We can expect about 

66,000 - 45,000 = 21,000 additional deer 

after 5 years. 

(d) 

73. p(t) = 250 -120(2.8)-·51 

(a) P(2) =250 -120(2.8)-·5(2) 

=250 -120(2.8)-1 

=207 

After 2 months, a person will type about 

207 symbols per minute, 

(b) P(4) = 250-120(2.8)-,5(4) 

= 250 -120(2.8)-2 

=235 

After 4 months , a person will type about 

235 symbols per minute. 

(c) P(1O) = 250 -120(2.8)-·5(10) 

= 250 -120(2.8)-5 

=249 

After 10 months, a person will type about 

249 symbols per minute. 

74. C(x) =130,700e·027x 

(a) 1990 is 0 years after 1990, so use x =O. 

027(0) 

C(O) = 130,700e · 

=130,700(1) 

= 130,700 

The median cost in 1990 was $130,700. 

(b) 1993 is 3 years after 1990, so use x =3. 

C(3) =130,700e,027(3) 

=130,700e·081 

= 130,700(1.08437) 

= 141,700 . 


(c) 1998 is 8 years after 1990, so use x =8. 

C(8) =130,700e·027(8 ) 

=130,700e·216 

= 130,700(1.241102) 

= 162,200 


Section 9.2 


1. loglO 458.3 = 2.661149857 

+ loglO 294.6 = 2.469232743 

"" 5.130382600 

105.130382600 "" 135,015.18 

2. Answers will vary. 

Exercises 

1. E; log216 = 4 because2 4 = 16. 

2. A; logj l =0 because 3 0 =1. 

3. G; loglO.1=-1 because In"" =.1. 

4. B;log2 .J2 =.!.. because2 112 =.J2. 

2 

5. C; log1010 5 = 5 because 10 5 = 10 5. 

6. H; lOge(e; ) = -2 because e-2 = e12 ' 

cr3 

7. F;logI/28=-3 because"2 =8. 

8. D; log5(-I) is not a real number. 

9. 3 4 = 81 is equivalent to log3 81 = 4. 

10. 2 5 = 32 is equivalent to log2 32 = 5. 

11. (~r3 = 2; is equivalentto IOg2/3(2;) = -3. 

12. 10- 4 = .0001 is equivalent to loglO .~1 = -4. 

385


386 

Chapter 9 Exponential and Logarithmic Functions 

13. 

14. 

15. 

16. 

18. 

19. 

20. 

log6 36 :;::; 2 is equivalent to 6 2 :;::; 36. 

logs 5 :;::; 1 is equivalent to 51 :;::; 5. 

10g.J381 :;::; 8 is equivalent to (.J3t :::: 81. 

I (1) . . I 4-3 1 

og4 64 :::: -3 IS equiva ent to :;::; 64 . 

loga 1= 0 for all real numbers a, because aO = 1, 

(a *-0) for all real numbers a. 

X::::IOgs(_I) 

625 

5x ::::_1_ 

625 

5x __1 

- 54 

5 x = 5-4 

x=-4 

x= IOg3(8\) 

24. 

25. 

26. 

x:;::; 8 10ggll 

x:;::; 11 

log, 25 :;::;-2 

x- 2 :;::; 25 

x- 2 :::: 52 

(x- 2 )- 1I2 :::: (52)-112 

x=5- 1 

1 

x:::: 

5 

IOgxL~)=-2 

-2 1 x :;::; 

16 

-2 1 

x = 

2 4 

x- 2 = 2- 4 

(x-2)-1/2 = (r4)- 1I2 

x=2 2 

x:;::;4 

21. 

22. 

3 x =..!. 

81 

3x =_1 

3 4 

3 x = 3- 4 

x=-4 

x = 10gJO .001 

lOX = .001 

lOx:::: 10-3 

x=-3 

X::::IOg 6(_I ) 

216 

6x =_1_ 

216 

6x =_1 

6 3 

6 x = 6- 3 

x=-3 

27. 

28. 

29. 

log4x = 3 

4 3 =x 

64=x 

log2x =-1 

2- 1 =x 

1 

-=x 

2 

x=log 4 Vl6 

4x=~ 

4 x = (16)113 

4 x =(4 2 ) 113 

4 x = 4 2 / 3 

2 

x= 3 

23. 

x =2log2 9 

x=9 

386



30. x =logs V25 

5 x =tfi5 

5 x =(25)114 

5 x =(5 2 )1/4 

5 x =5 2/4 

= 

5 x 5 112 

1 

x= 

2 

32. y (2.2319281) represents the exponent to which 2 

must be raised in order to obtain x (5). 

35. fex)=llog2eX+3)1 

First. translate the graph of fex) =log2 x to the 

left 3 units to obtain the graph of log2 (x + 3). 

(See Exercise 34.) For the portion of the graph 

where f(x) ~ O. that is. where x ~ -2. use the 

same graph as in 35. For the portion of the graph 

in 35 where fex) < 0, x < -2, reflect the graph 

about the x-axis. In this way. each negative value 

of f(x) on the graph in 35 is replaced by its 

opposite. which is positive. The graph has a 

vertical asymptote at x = -3. 

y 

33. f(x) =(log2 x) +3 


f(x) =]og2 x up 3 units. 

y 

6 

4 

2 

+-+AIt+-+-++-+-++-+.. x 

o 

2 4 6 8 

36. fex)=(logll2x)-2 

This is the graph of fex) =10gl12 x translated 

down 2 units. 

y 

34. f(x) = log2(X +3) 


f(x) =]og2 x to the left 3 units. The graph has a 

vertical asymptote at x =-3. 

y 

+-+-+M-::+-+-+-+-H" x 

!(x) = log2(x+ 3) 

+lH\++++-H-+-++- x 

-4 

f(x) = (log 1/2 x) - 2 

37. f(x) =10gll2(x - 2) 

This is the graph of fex) =10gl12x translated to 

the right 2 units. The graph has a vertical 

asymptote at x = 2. 

y 

!(x) =log 1/2 (x - 2) 

2 

+ll-=t-t+~--H-+t+- x 

o 

-2 

387



38. f(x) =Ilog1l2(x - 2) I 41. f(x) =logIl2(1-x) 

This is the same graph as 37, with the part below 


the x-axis reflected about the x-axis. 

, 

1 

x -7 -3 -I 0 

3 7 

2 4 8 

4 I-x 8 4 2 

2 

1 1 1 

248 

-Hf:t-H-H+t-Hf-t-- x y -3 -2 -1 0 1 2 3 

o I 3 6 9 

-2 Ix=2 The graph has a vertical asymptote at x =1. 

(x) = /Iog 1/2(x - 2)1 

39. f(x) =log3 x 

I 

Write y =log3 x in exponential form as x =3 Y to 

find ordered pairs that satisfy the equation. It is 

easier to choose values for y and find the 

corresponding values of x. Make a table of values. 

y 

Ix= I 

-Hf-t-H--+-+-hI~H--- x 

f(x) =log IflO - x) 

x 1 1 1 3 9 

42. f(x) = log1l3(3 - x) 

9 3 


y -2 -1 0 1 2 

The graph can also be found by reflecting the x -6 0 2 2! 

graph of f(x) =3 x about the line y = x. The graph 1 1 

3-x 9 3 

has the y-axis as a vertical asymptote. 3 9 

J 

y -2 -1 0 1 2 

f(x) = log3 x 

x =3 is a vertical asymptote. 

y 

2 

-H-±li~H-++-H-+-~ x 

3 9 

-3 2 I 

-2 I 

I 

40. f(x) =10gIO x I 

I 

Write y = 10gIO x in exponential form as x = lOy 

f(x) = log 113(3 - x) 

to find ordered pairs that satisfy the equation. It is 

easier to choose values for y and find the 

43. f(x) =log3(x -1) 

corresponding values of x. Make a table of values. To graph the function, translate the graph of 

f(x) =log3 x (from Exercise 39) 1 unit to the 

~ .001 .0 1 .1 1 10 100 1000 

right. The vertical asymptote will be x =1. 

~ -3 -2 -1 0 1 2 3 y 

I 

x =0 is a vertical asymptote. 

JX=I 

y 

1 

2 

2 

3p _ 

- I 

0- , 2 S 10 x 

-3 1 

+-t-;d-,I9--+-1I-++-i-+-+-i_x I f(x) =log3(x- I) 

5 10 

-1 f(x) = log JO X 

44. f(x) =log2(X 2) 

-2 


388



1 

x .354 - .707 1.414 2 2.828 

2 

I 1 1 

xl- - - - 2 4 8 

8 4 2 

y -3 -2 -1 0 1 2 3 

x =0 is an asymptote. The graph is symmetric 

with respect to the y-axis, Sketch the graph. 

y 

so. f(x) = logz( -x) 

The graph of this function is the reflection of the 

graph of f(x) = logz x about the y-axis. Note that 

the x-intercept is -I since 

logz[-(-l)] = logj l = O. The correct graph is A. 

51. f(x) =x loglo x 

Using a graphing calculator, enter x log x for Y l 

The graph is as shown. 

f(x) = x logJOx 

+-+--+-;;t--+-t~- x 

45. f(x) = logz x 

The x-intercept is 1, and the graph increases. The 

correct graph is E. 

46. f(x) =logz(2x) 

The graph will be similar to that of f(x) =logz x 

(graph E) but will increase more rapidly. Note 

that while the x-intercept for f(x) =logz 2x will 

-1 

52. f(x) = x Z loglO x 

Graph f(x) = x Z log x in the window 

[-1,5] by r[-.;:..,,5,'-'-Z..... ).__""'""""\ 

f(x) = x 21ogJOx 

be ±' since logz ~~) = logz 1= O. The correct 

graph isD. 

47. f(X)=IOgz(~)=IOgZX-l =-Iogzx 

The graph will be similar to that of f(x) = logz x, 

but will be reflected across the x-axis. The 

x-intercept is 1. The correct graph is B. 

48. f(x) =10gz(~) 

1 

=Iogz-x 

2 

The graph will be similar to that of f(x) = logz x 

but will increase less rapidly. The z-intercept is 2 

since 10gz(~) =logz 1=O. The correct graph 

isC. 

49. f(x) =logz(x -1) 

The graph will be similar to that of f(x) =logi x, 

but will be shifted 1 unit to the right. The 

x-intercept is 2 since logz(2 -1) = logzl = O. 

The correct graph is F. 

-5 

53. Ifx and y are positive numbers, 

x 

log, - = loga x - loga y. 

y 

54. Since IOgz(1")= logz x - logz 4 by the quotient 

rule, the graph of y = IOgz(1") can be obtained 

by shifting the graph of y = logz x down 

logi 4 =2 units. 

55. Graph f(x) = 10gz(~) and g(x) = logz x on the 

same axes. The graph offis 2 units below the 

graph of g. This supports the answer in 

Exercise 54. 

389



y 

60. log2 ( -5- 2../3J 

=log2 2../3 -log2 5 

Logarithm ofa quotient 

=log2 2 + log2 ../3 -log2 5 

Logarithm of a product 

1 

=1+-log2 3 -log2 5 

56. Ifx =4,Iog 2 

2 ~ =0; since log2 x =2 and 

4 Logarithm of a power 

log2 4 =2, log2 x -log2 4 =O. 

By the quotient rule, 61. log4(2x+ 5y) 

IOg2(~) =log2 x -log2 4. 

Both sides should equal O. Since 2 - 2 = 0, they 

do. 

Since this is a sum, none of the logarithm 

properties apply, so this expression cannot be 

simplified. 

62. log6(7m + 3q) 

Since this is a sum, none of the logarithm 

57. IOg2(6: ) properties apply, so this expression cannot be 

simplified. 

=log2 6x -log2 y 

Logarithm of a quotient 

=log2 6+log2 x-log2 y 


63. 10gmH 

, [5r 3 )112 

=logm -2 

z 

58. IOg3(~) 1 5r 3 

=log3 4p-Iog3 

='2 logm q 

-;s 


Logarithm of a power 

=log-,4 + log3 P - log3 q 

=.!..Oogm 5r 3 -logm zS) 

Logarithm of a product 2 


=~ (logm5 + logm r 3 -.1og m zS) 

59. lOgs( 5~ J 


=logs(5.J7)-log S 3 1 

=-(logm 5+3 logm r-510gm z) 

Logarithm of a quotient 2 

=logs 5 + logs .fi-logs 3 



64. 

=1+ logs .J7- logs 3 10gbb =1 

=1+ logs 71/2 -logs 3 .J7 =7 112 

1 

= 1+ -logs 7 -logs 3 

2 


390



65. loga x + logo Y - log, m 

=log, xy - logo m 


= 10ga(:) Logarithm of a quotient 

66. (log, k -10gb a) =logj, a 

k 

=10gb- - 10gb a Logarithm of a quotient 

m 

=logb(':) 


2) 

67. 210gma - 310gm(b 

2 

=logm a 2 -log m(b )3 


=logm a 2 -Iogmb 6 

68. 

=10gm(::) Logarithm of a quotient 

72. log10 12=loglO(3· 2 2 ) 

= 10g1O 3+2 . log10 2 

=.4771+ 2(.3010) 

= 1.0791 

9 

73. 10g1O - =loglo 9 -loglO 4 

4 

=10g1O 3 

2 -log lO 2 2 

= 210g 1O 3-210g 10 2 

::: 2(.4771) - 2(.3010) 

=.9542 - .6020 

=.3522 

74. 10gIOG~) 

=10g1O 20 -loglO 27 

=log10 2 ·10 -log10 3 3 

= loglo 2 + log1010 - 3 log10 3 

....3010+ 1-3(.4771) 

=-.1303 

75. (a) 

69. 210ga(z -1) + 10ga(3z + 2) 

=loga(z _1)2 + loga(3z + 2) 


=loga[(z -1)2(3z + 2)] 

Logarithhm of a product 

1 

70. logb(2y + 5) - -logb(Y + 3) 

2 

2y+5 

=10gb 1/2 

(y+3) 

2 Y + 5 ) =10gb ~ 

( vy+3 . 

71. 10glO 6 =loglO(2 · 3) 

=log10 2 + loglo 3 

=.3010 + .477 1 

=.7781 

(b) The interest rates increase with time, but not 

at a constant rate. Therefore, a linear function 

would not model the data well. The interest 

rates gradually level off. This resembles a 

shifted logarithmic function. An (increasing) 

exponential function does not level off, but 

rather continues to increase at an even faster 

rate. 

76. (a) From the graph.Iogj .3 =-1 .1 

(b) From the graph.IogjB =-.2 

77. /(3)= 2 

/(x) =logax 

2 =loga 3 

a 2 =3 

(a 2 )112 =3 112 

a=.,J3 

391



(a) I(x) = log, x 3. I(x) = 5 x 

I(x) = log.J3 x 

y=5 x 

x = 5 Y Exchange x and y to find the inverse. 

~i) = IOg.J3(i) 

or logsx = y 

y = log.J39 I If I(x) = 5 x , then I-I(x) = logs x. 

(~r=i 

(3 112)y =_1 

3 2 

3y12 =r 2 

L=-2 

2 logarithm. 

y=-4 

(b) I(x) = loga x 

Inl2 

6. log3 12=- 

In3 

I(x) = log.J3 x 

1(27) = log.J3 27 

y = log.J3 27 

(-\J3Y = 27 

4. 4l0g4Jl=1l 

The exponent to which 4 must be raised to obtain 

11 is log4 II. 

5. A base e logarithm is called a natural logarithm, 

while a base 10 logarithm is called a common 

7. logj O is undefined because there is no power of 

2 that yields a result of O. 

8. x = log212 

2 x =12 

(3 112)y = 3 3 Since2 3 = 8 and 2 4 = 16. log2 12 must lie 

3y/2 = 33 between 3 and 4. 

I=3 

2 

9. log 36 = 1.5563 

y=6 10. log 72 = 1.8573 

78. (a) lOOlog3 = 102log 3 11. log .042 = -1.3768 

= IOlog32 

= IO Iog9 

12. log .319 = -.4962 

=9 13. log(2 x 10 4 ) = 4.3010 

(b) loglo .01 3 = loglO(IO- 2)3 

14. log(2 x 10- 4) = -3.6990 

= loglOIO- 6 15. In 36 = 3.5835 

=-6 

Section 9.3 

Ell:ercises 

1. For I(x) = a", where a > 0, the function is 

increasing over its entire domain. 

2. For g(x) = log, x, where a> I, the function is 

increasing over its entire domain . 

16. In 72 = 4.2767 

17. In .042 =3.1701 

18. In .319 =-1.1426 

4 

19. In(2 x e ) = 4.6931 

20. In(2 x e- 4 ) = -3.3069 

392


9.3 Evaluating Logarithms and the Change-of-Base Theorem 393 

21. Grapefruit, 6.3 x 10- 4 26. Wine, 3.4 

pH = -log[H 30+] 

3.4 = -log[H 30+] 

= -log(6.3 x 10- 4 ) -3.4 = log[H 3O+] 

= -(log6.3 + loglO- 4 ) 

[H 30+] = 10- 3 .4 

= -(.7793-4) [H 30+] "" 4.0 X 10- 4 

= -.7993+4 

pH=3.2 27. Beer, 4.8 

The answer is rounded to the nearest tenth pH = -log[H 30+] 

because it is customary to round pH values to the 

4.8 = -log[H 

nearest tenth. The pH of grapefruit is 3.2. 

30+] 

-4.8 = log[H 3O+] 

22. Crackers, 3.9 x 10-9 [H 30+] 

= 10-4·8 

pH = -log[H 3 0+] 

[H 30+] = 1.6 x 10- 5 

pH = ~log(3.9 x 10- 9 ) 

= -(log 3.9 + log 10- 9 ) 28. Drinking water, 6.5 

= -(.59106-9) 6.5 = -log[H30+] 

=-(-8.409) -6.5 = log[H 3O+] 

pH=8.4 

[H 30+] 

= 10-6.5 

The answer is rounded to the nearest tenth 

because it is customary to round pH values to the [H 30+] = 3.2 X 10- 7 

nearest tenth. The pH of crackers is 8.4. 

29. Wetland, 2.49 x 10- 5 

23. Limes, 1.6x 10- 2 pH = -log(2.49 x 10- 5 ) 

pH = -log[H 30+] = -(log2.49+ loglO- 5 ) 

= -log(1.6 x 10- 2 ) = -log2.49- (-5) 

= -(log1.6+ log10- 2 ) =-log2.49+5 

=-(.2041-2) ""4.6 

= -(-1.7959) Since the pH is between 4.0 and 6.0, it is a 

pH = 1.8 

poor fen. 

The pH of limes is 1.8. 

30. Wetland, 2.49 x 10- 2 

24. Sodium hydroxide (lye), 3.2 x 10- 4 pH = -log(2.49 x 10- 2 ) 

pH = -log[H30+] = -(1og2.49 + log10- 2 ) 

pH = -log(3.2 x 10- 14 ) 

= -log2.49-(-2) 

= -(log3.2 + log10- 14 ) 

= -log2.49+2 

= -(.5052 -14) 

= 1.6 

Since the pH is 3.0 or less, it is a bog. 

= -(-13.49) 

pH=13.5 

31. Wetland, 2.49 x 10- 

The pH of sodium hydroxide is 13.5 

7 

pH = -log(2.49 x 10- 7 ) 

... 25. Soda pop, 2.7 = -(log 2.49 + log 10- 7 ) 

pH = -log[H 30+] = -log2.49 - (-7) 

2.7 = -log[H 30+] = -log 2.49 + 7 

-2.7 = log[H =6.6 

3O+] 

Since the pH is greater than 6.0, it is a rich fen. 

[H30+] = 2.0x 10- 3 32. (a) log 398.4 = 2.60031933 

(b) log 39.84 "" 1.60031933 

(c) log 3.984 "" .6003193298 

393



33. 

34. 

(d) The whole number parts will vary, but the 

decimal parts will be the same. 

In5 

log25= 

In2 

1.6094 

"'- 

.6931 

'" 2.3219 

35. logs .59 = log .59 

log8 

-.2291 

"'- 

.9031 

'" - .2537 

36. I ogs· 71 -- 

- log.71 

log8 

-.1487 

"'- 

.9031 

'" -.1647 

37. 

In12 

log.Ji312 = In.Ji3 

2.4849 

"'- 

1.2825 

'" 1.9376 

41. 21n3x = In(3x)2 

=ln3 2x 2 

= In9x 2 

It is the same as D. 

42. In4x - In2x 

4x 

=In 

2x 

4 

=In 2 

= In2 

It is the same as D. 

43. In(b4~) = In(b 4aIl 2 ) 

44. 

=lnb 4+lnaIl2 

= 4Inb+.!.lna . 

2 

1 

=4v+-u 

2 

a 3 

In2" = Ina 3 -In b 2 

b 

=3Ina-2Iob 

=3u-2v 

45. InB=In(::)'" 

=I{::;~) 

In5 =Ina 3/ 2 _lnb 5/ 2 

38. 10g,Ji9 5 = In.J19 

1.6094 

"'- 

1.4722 

'" 1.0932 

=~Ina-~Inb 

2 2 

3 5 

=-u--v 

2 2 

In9 

log29= 

In2 

2.1972 

"'- 

.6931 

'" 3.1699 

.. 

39. 

40. 

log5 

log325=- 

46. In(~. b 4 ) =In(a t / 3 . b 4 ) 

. log.32 

.6990 

= Ina t/3 + Inb 4 

"'- 

-.4949 = .!.Ina + 41n b 

'" -1.4125 3 

log8 

log918=- 

. log.91 

.9031 

47. g(x) =eX 

1 

=-u+4v 

3 

"'- 

-.0410 

",-22.0488 (a) By the theorem on inverses, 

g(lo 3) =e\n3 

=3. 

394


9.3 Evaluating Logarithms and the Change-or-Base Theorem 395 

(b) By the theorem on inverses, 

2 

g[ln (52)] = e'nS 

= 52 = 25 

(e) By the theorem on inverses, 

48. f(x) =3 X 

1n 

g[ln(;)] = e (1l e ) 

1 

=-. 

e 

(a) By the theorem on inverses, 

1og37 

f(log3 7) = 3 

=7. 

(b) JIlog 3(1n 

3)] = 3 Iog3(ln3) 

=ln3 

(e) JIlog 3(2ln 

3)] = 3 Iog3(2In3) 

49. f(x) = Inx 

=21n3 

or In9 


feeS) = Ine s 

=5. 

(b) By the theorem on inverses, 

f(e 1n3) = Ine 1n3 

=ln3. 

(e) By the theorem on inverses, 

f(e 2 In3) = Ine 21n3 

=21n3 

or In9. 

50. f(x) = log2 x 


f(2 3) = log2 2 3 

=3. 

(b) f(210822) = log2 2'082 2 

= log2 2 = I 

(e) f(2 210 822) = log2 2210822 

=210g22 

=2·1=2 

51. f(x) =In Ix I 

The domain is (-00,0) u (0,00). 

The range is (- 00,00). 

It is symmetric with respect to the y-axis. 

52. f(x) =log31 x I 

(a) The domain off is all real numbers except 0: 

(-00, O)u(O, 00). 

(b) The graph of f(x) is given below. 

[ex) = log31xl 

4 

-4 

(e) The graph appears to show a point with 

x-value zero, indicating a domain of 

(- 00,00), which is incorrect. 

53. The table for Y 1 = log3(4 - x) shows "ERROR" 

for x ~ 4. This is because the function is 

undefined for x ~ 4. The domain ofy = loga x is 

x > 0, which means that for Y 1 , the domain is 

4 - x > 0, or x < 4. 

54. y= log, x 

Since 1 = log., 4, then a l = 4 or a = 4 

Check: 2 = log, 16, then a 2 = 16 

or ..r;;z =.J16 or a = 4 

a=4 

55. Let r = the decibel rating of a sound . 

1 

r= 1O ·log lO 

1 0 

100 ·1 

(a) r=IO·log 0 10--"' 

1 0 

= 10 · log lO 100 

= 10·2 

=20 

1000· 1 0 

(b) r = 10 ·log 10 ---" 

1 0 

= 1O ·log lO 1000 

=10 ·3 

=30 

395


396 Chapter 9 Exponential and Logaritbmic Functions 

(c) r= 10· log 10 100,000./ 0 

1 0 

=10·5 

=50 

(a) r = 10glO 1000·1 

1 0 

= 10gl01000 

=3 

(d) r= 1O·10g lO 

1,000,000 ./ 0 

1 0 

=10·6 

=60 

(b) 

r= 10g\0 1,000,000 . I 

1 0 

= 10glO 1,000,000 

=6 

(e) 1=2/ 0 

21 

r = 10· log 10 0 

10 

r = 10 . log 10 2 

r=3.0103 

The described rating is increased by 3.0103. 

58. 

56. Let d represent the loudness in decibels. 

I 

d=10 ·1og lO 

1 0 

(a) d = 10· log 10 115./ 0 

1 0 

= 10 log 115 

d",21 

(c) 

100,000,000 ./ 

r= 10glO 

0 

1 0 

= 10glO 100,000,000 

=8 

r=IOgIOC:J 

6.7 =10glO( I: ) 

10 6 . 7 .s. 

1 0 

1 010 

6 . 

7 = I 

1=5,011,872/ 0 

. ... 

(b) d = 10· log 10 9,500,000./ 0 

1 0 

= IOlog lO 9,500,000 

d",70 

(c) 

(d) 

1,200,000,000./ 0 

d = 1O·10g 10 

1 0 

= 10 10g\0 1,200,000,000 

d",,91 

d = 1O·log lO 

895,000,000,000· 1 0 

1 0 

= 10 . loglO 895,000,000,000 

d= 120 

(e) d = 1O·10g 10 

109,000,000,000,000 ./ 0 

1 0 

= 1010g l0 109,000,000,000,000 

do:: 140 

57. Let r =the Richter scaJe rating of an earthquake. 

r= IOglO(~ J 

59. r= 10glO(~) 

60. 

8.1 = 10gl~( ~) 

10 8 .1 .:': 

1 0 

/ 0 10 8.1 = I 

1=125,892,5411 0 

108.1 

---;j'f = 10 8 . 1 - 6 . 7 = 101.4 = 25 

10' 

It was about 25 times greater. 

61. /(x) = -273 + 74 In x 

In the year 2000, x = 100. 

/(100) = -273 + 74 In 100 

=68 

Thus, the number of visitors in the year 2000 will 

be about 68 million. Beyond 1997, we must 

assume that the rate of increase continues to be 

logarithmic. 

396


9.3 Evaluating Logarithms and the Change-or-Base Theorem 397 

62. (a) /(I)=-608.5+149Inl (c) Substitute a = .88, n = 250. 

In the year 1999,1= 99 

/(99) = -608.5 + 1491n99 = 76 

Thus, the percent of freshmen entering 

college in 1999 who performed volunteer 

work during their last year of high school 

will be about 76%. 

63. s; = a In(1+;) 

= .36In(I+~) 

.36 

(8) S(I00) = .36 In( 1+ 100) 

.36 

= .36 In 278.77 

= (.36)(5.6304) 

=2 

(b) S(200) = .36In(1+ 200) 

.36 

= .361n(556.555) 

= .36(6.322) 

::::2 

(c) S(l50) = .36 In ( 1+ -150) 

.36 

= .36In417.666 

= .36(6.0347) 

::::2 

(d) S(IO)= .36 In(I+..!Q..) 

.36 

=.36 In 28.777 

= .36(3.3596) 

::::1 

S(250) = .88 In(1 + 250) 

.88 

=4.97 

The equation predicts there will be 5 species. 

65. The index of diversity H for 2 species is given by 

H = -[1\ log21\ + P2 log2 P 2 ] 

50 

1\ = 100 

=.5 

50 

P 2 = 100 

= .5 

Substituting into the formula gives 

H = -[.5log2 .5 + .5 1og 2 .5]. 

Since log2 .5 = log2 .!. = -1, we have 

2 

H = -{ .5(-1)+ .5(-1)] 

=-{-I) 

=1 

The index of diversity is 1. 

66. H = [1\ log21\ + P 2 log2 P 2 

+ P:3 log2 P:3 + P 4 log2 P 4 ] 

= -[.5211og 2 .521 + .324Iog 2.324 

+ .081log2 .081+ .0741og 2 .074] 

=1.59 

The index of diversity is 1.59. 

67. From Example 5, 

T(R) = 1.03R 

and R=kln( ~} 

By substitution. we have 

64. S(n)=aln(I+;) 

(8) Substitute a = .88, n = 50. 

S(50) = .88 In( 1+ 50) 

.88 

=3.57 


(b) Substitute a = .88, n =100. 

S(I00)=.881n(l+ 100) 

.88 

=4.17 


T(k) =1.03k In( ~ ) 

Since 10 s k ~ 16 and ( ~) = 2, the range for 

T = 1.03k In( ~) will be between 

T = 1.03(1O)ln2 = 7.1 

and T = 1.03(16) In 2 = 11.4. 

The predicted increased global temperature due to 

the greenhouse effect from a doubling of the 

carbon dioxide in the atmosphere is between 

7°F and 11°F. 

397



68. (a) C(x) =353(1.006}~-1990 

(b) 

T(x) =6.489In(.£) 

280 

=6.489In[353(1 .006)X-I990] 

280 

2100 

69. (a) The table of natural logarithms takes the 

following form. Let x =In D and y = In P for 

each planet. 

Planet InD InP 

Mercury - .94 -1.43 

Venus - .33 -.48 

Earth 0 0 

Mars .42 .64 

Jupiter 1.65 2.48 

1990 1:.: • •:....:....;.:;........:..~=--,:;,:..;~...J " 

2275 

300 

T(x) = 6.489In(2~ ).Where 

C(x) is defined as given earlier 

Saturn 2.26 3.38 

Uranus 2.95 4.43 

Neptune 3.40 5.10 

Plot this data in the window [-2,4] by 

[-2,6]. 

C is an exponential function and T is a linear 

function over the same time period. While 

the carbon dioxide levels in the atmosphere 

increase at an exponential rate, the average 

global temperature will rise at a linear rate. 

(c) The slope of the graph of T can be 

approximated using two points. Since 

T(2000) == 1.89 and T(2100) == 5.77 , the slope 

. _ 5.77 -1.89 _ 0388 

IS m- -.. 

2100-2000 

This means that temperature is expected to 

rise at an average rate of .04°P per year from 

1990 to 2275. 

(d) Graph y = T(x) and y = 10 on the same 

screen. Their graphs intersect when 

.... x == 2208.9. C(2208.9) :: 1308 ppm. which 

would be about 4.67 times above 

preindustrial carbon dioxide levels. 

-2 

Yes, the data points appear to be linear. 

(b) Choose two points from the table showing 

In D and In P and find the equation through 

them. If we use (0, 0), representing Earth and 

(3.40,5.10) representing Neptune, we obtain 

m= 5.10-0 =1.5. 

3.40-0 

Since the y-intercept is O. the equation is 

y =1.5x 

or In P =1.5 In D. 

Since the points lie approximately but not 

exactly on a line, a slightly different equation 

will be found if a different pair of points is 

used. 

6 

-21--~o':---"""'---"1 4 

-2 

398



Section 9.4 

Exercises 

(c) For Pluto, D = 39.S, so 

InP= l.5lnD 

= I.S In 39.5 

Then 

P = e1.5ln39.5 

1039.51.5 

=e 

=(39.5)1.5 

:::: 248.3. 

The linear equation predicts that the period 

of the planet Pluto is 248.3 years, which is 

very close to the true value of 248.5 years. 

1. Since x is the exponent to which 7 must be raised 

in order to obtain 19, the solution is 

log19 In19 

Iog7 19 or-- or --. 

log7 1n7 

2. Since x is the exponent to which 3 must be raised 


10g1O In10 

Iog3 10 or-- or - 

log3 In3 

3. Since x is the exponent to which .!.. must be 

2 

raised in order to obtain 12, the solution is . 

log12 

In12 

logl/2 12 or -(.1) or (.1) 

log 2 In I 

4. Since x is the exponent to which .!. must be raised 

3 


log 4 In4 

4 

logl/3 or log(t) or In(t)' 

6. 4 x = 12 

Take natural logarithms (base e) of both sides 

In4 x = In12 

x In4 = In 12 Logarithm of a power 

In12 

x=- 

In4 

2.4849 

x=-- 

1.3863 

==: 1.792S 

7. 6 1 - 2x =8 

Take base 10 (common) logarithms of both sides. 

(This exercise can also be done using natural 

logarithms .) 

log6 1 - 

2x = log8 

(l-2x)log6 = log8 

1-2x = log8 

log6 

log 8 

2x = 1_ 

log 6 

8. 3 2x - 5 =13 

x = .!.(1- IOg8) 

2 log6 

In3 2x - 5 = In13 

(2x - S) . In 3 = In13 

==: .!.(1- .9031) 

2 .7782 

==: -.0803 

2x-S = In 13 

In3 

2x =S+ In13 

In3 

x = .!..[s+ In13] 

2 In3 

x==: 3.6674 

5. 3 x =6 

Take base e (natural) logarithms of both sides. 

In3 x = In6 

x In3 = In6 Logarithm of a power 

In6 

x= 

In3 

1.7918 

=- 

1.0986 

:::: 1.6309 

9. 

2 x + 3 =SX 

log2 x + 3 = 10gSX 

(x+3)log2 = xlogS 

xlog2+310g2 =xlogS 

xlog2 -xlogS = -310g2 

x(log2 -logS) = -310g2 

x = _-_3_lo....:g::,.2_ 

log2-logS 

x ==:2.2694 

399


400 Chapter 9 Exponential and Logarithmic Equations 

10. 6 x + 3 =4 x 

log6x+3 = log4 X 

(x+3)log6 = xlog4 

x log6 + 310g6 = x log4 

xlog6-xlog4 = -3 log 6 

x(log6-log4) = -310g6 

-310g6 

x= log6-log4 

x =-13.2571 

x-l 

11. e =4 

Ine x-l = In4 

x-l= In4 

x = In4+1 

= 1.3863+ 1 

=2.3863 

2 x 

12. e - =12 

Ine 2 - x =lnl2 

(2 -x) ·Ine = In12 

2-x = In12 

x=2-ln2 

x =-.4849 

13. 2e Sx + 2 =8 

e Sx + 2 =4 

Ine Sx + 2 = In 4 

5x+2=ln4 

5x=ln4-2 

1 

x = -(ln4-2) 

5 

... .!. (1.3863 - 2) 

5 

= -.1227 

14. 10e 3x - 7 = 5 

In(1O 'e 3x-7 ) = In5 

InlO+ Ine 3x-7 = In5 

In1O+(3x -7)-Ine = In5 

3x - 7 =In5 - In 10 

3x = In5 -In 10 + 7 

1 

x =-[ln5 -In 10+ 7] 

3 

x =2.1023 

15. 2 x = -3 has no solution since 2 raised to any 

power is positive. 

8x 2x 20 

17. e . e = e 

lOx 20 

e =e 

lOx =20 

x=2 

18. e 

6x 

. eX =e 

2l 

7x 21 

e =e 

7x=21 

x=3 

19. loo(1.02)x/4 = 200 

(1.02)x/4 =2 

(~) log 1.02 = log2 

x log2 

-=--=- 

4 log 1.02 

x =4 log2 

log 1.02 

x'" 140.0112 

20. 500(1.05)x/4 =200 

(1.05)x/4 =~ 

5 

In(1.05)x/4 = In~ 

5 

x 2 

- In(I .05) =In 

4 5 

x 

In! 

21. logx + log(x - 21) =2 

log[x(x - 21)]=2 

-=- 

4 In1.05 

lln.4 ) 

x =\ln1.05 

x'" -75.1209 

10 2 =x(x - 21) 

loo=x 2-21x 

0=x 2 -2lx-loo 

0= (x - 25)(x+4) 

x - 25 = 0 or x + 4 = 0 

x=250rx=-4 

Since x =- 4 is not in the domain of log x or 

log (x - 21), it cannot be used. So x =25. 

16. 3 x =-6 

Since 3 raised to a power will never result in a 

negative value, this equation has no solution. 

400


22. log x + log(3x - 13 = 1 

log[x(3x -13)] = 1 


27. SX+2 =2 2x-1 

logSx+2 = log 2 2x - 1 

WI = x(3x -13) 

(x+2)logS =(2x -1)log2 

1O=3x 2 -13x 

xlogS+ 210gS = 2xlog2-log2 

O=3x 2 xlogS-2xlog2 = -2 logS -log2 

-13x-1O 

x(logS-210g2) = -2 logS -log2 

O=(3x+2)(x-S) 

x = -210gS-log2 

3x + 2 =0 or x - S=0 

logS -210g2 

2 

x=-- or x=S 

x ...-17.S314 

3 

Since x = -~ is not in the domain of 

28. 6 x 

3 =3 4x + 1 

3 

log6 

log x or log (3x - 13). it cannot be used. So x X 3 

- = log3 

= S. 

4x+1 

(x - 3)log6 = (4x + 1)log3 

23. In(S+ 4x) -10(3 + x) = In3 

xlog6-310g6 = 4xlog3+ log 3 

xlog6-4xlog3 = 310g6+ log3 

In S+4x =ln3 

x(log6-410g3) = 310g6+ log 3 

3+x 

S+4x =3 

x = 310g6+ log3 

3+x 

log6-410g3 

S+4x=3(3+x) 

x ...-2.4874 

S+4x =9+3x 

x=4 

29. In eX -lne 3 = Ine s 

x-3=S 

24. In(2x+S)+ Inx = In7 

x=8 

In[x(2x + S)] = In7 

x(2x+S) =7 

30. ln e" -2Ine= Ine 4 

2x 2 +Sx-7 =0 

eX 

(2x+ 7)(x -1) =0 

In-=lne 

2 

4 

e 

7 

x=-- orx=1 

4 

2 

:: =e 

We reject _2, since it is not in the domain of 

eX =e 6 

2 

either In x or In (2x + S). So x = 1. 

x=6 

31. 

25. log6 4x -log 6 (x - 3) = log6 12 

log2(log2 x) = 1 

4x 

log2(log2 x) = log2 2 since log22 = 1 

log6--= log6 12 

log2 x=2 

x-3 

x=4 

~=12 

x-3 

4x= 12(x-3) 

32. log x = .Jlog x 

4x=12x-36 

(logx)2 = (.Jlogx)2 

36=8x 

36 

(logx)2 = log x 

-=x 

8 

(logx)2 -Iogx = 0 

4.5=x 

logx(logx-l) = 0 

log x = 0 or log x -1 = 0 

26. log2 3x + log2 3 = log2(2x + IS) 

log x = 1 

log2[3x(3)] = log2(2x + IS) 

3x(3)=2x + 

10° =x WI =x 

IS 

l=x or 10=x 

9x=2x+lS 

7x= IS 

IS 

x= 7 

Since the work involves squaring both sides, both 

proposed solutions must be checked in the 

original equation. Both answers check. 

401



33. 10gx Z =(logx)Z 

210gx =(logx)Z 

Let w = logx. 

2w=w z 

o=w z -2w 

0=w(w-2) 

w=O or w=2 

Replace w with log x to find the values for x. 

o= log x or 2 =log x 

1=x or 100 =x 

34. logz ~2xZ = ~ 

2 

2 3 /2 =~2xZ 

(2 3IZ)Z =(~2XZ)Z 

2 3 =2x Z 

8=2x Z 

4=x Z 

±2=x 

Since the solution involves squaring both sides, 

both proposed solutions must be checked in the 

original equation. Both answers check. 

36. loga(4x -7) + 10ga(x Z + 4) = 0 

log, t is defined when t is in the interval (0, 00), 

but it is undefined when t is in the interval 

(- 00,0]. Sincex z + 4 is positive for all values of 

.r, loga(x Z + 4) will be defined for all values of x. 

However, loga(4x -7) will be undefined when 

4x - 7 is in the interval (- 00,0], which 

corresponds to 4x - 7 S 0 

4xS7 

7 

xS-. 

4 

Thus, the numbers in the interval ( -00, ~J could 

not be solutions of the given equation. 

37. j(x) =e x + 1 - 4 

x + 

I 

y=e - 4 

x =e Y +I - 4 Exchange x and y. 

y 

x+4 =e + 

1 

y 1 

In(x + 4) =In e + 

In(x+ 4) =y+ 1 

In(x + 4)-1 =y 

j-l(x) =In(x + 4)-1 

Domain: (-4,00) 

Range: (- 00,00) 

38. j(x) = 21n3x 

y = 21n3x 

x = 21n3y Exchange x and y. 

x 

-=ln3y 

2 

e X 

Z 1n 3y 

/ = e 

e X / Z = 3y 

e X / Z 

--=y 

3 

x/Z 

j-I(x)=_e_ 

3 

Domain: (- 00, 00) 

Range: (0, 00) 

39. eX +lnx=5 

Graph Y 1 = eX + In x and Yz = 5 on the same 

screen. 

Using the "intersect" option in the CALC menu, 

we find that the two graphs intersect at 

approximately (1.52, 5). The x-coordinate of this 

point is the solution of the equation. So x::: 1.52. 

40. e" -In(x + 1) =3 

Graph Y 1 = eX -In(x+ 1) and Yz =3 on the 

same screen. 

Use the "intersect" option in the CALC menu. 

The intersection points are at x « -.93 and 

x::: 1.35. 

41. 2e x +1=3e- x 

Graph Y 1 =2e x + 1 and Y z =3e- x on the same 

screen. The two curves intersect at the point (0, 

3). The x-coordinate of this point is the solution 

of the equation. So x =O. 

42. eX +6e- x =5 

Graph Y 1 =eX+ 6e- x and Yz = 5 on the same 

screen. 

The intersection points are at x ::: .69 and x::: 1.10. 

43. logx = x Z -8x+ 14 

Graph Y 1 =logx and Yz =x Z -8x+ 14. 

The intersection points are at x = 2.45 and 

X= 5.66. 

44.lnx=-'J.jx+3 

Graph Y 1 = In x and YZ =--l[;+3. 

The intersection point is at x ::: .23. 

402



45. Double the 1998 value is 2(97 .3) = 194.6. 

48. f(x) =-301ln(~) 

f(x) =23 · 1.2 x 207 

194.6 =23 ·1.2 x 

194.6 =1.2x 

23 

194.6 1 12x 

1ogn= og . 

194.6 

log-- =x log 1.2 

23 

10g.l!M& ______2""3_ =x 

log 1.2 

x == 11.7 

1990+ 11.7 = 2001.7 

During 2001, the worldwide shipments will be 

double their 1998 value. 

46. (a) At the finish line t = 9.86. 

(b) 

f(t) =11.65(1-e- tI 1. 27 ) 

f(9.86) =11.65(1-e-9·86/1.27) 

f(9.86) == 11.6451 

He was running 11.6451 meters per second 

as he crossed the finish line. 

f(t)=11.6~I-e-tI1.27) 

10 =11.6* _e- tI 1. 27 ) 

~ =l_e- tl l. 27 

11.65 

e- tl 1. 27 =1-~ 

11.65 

Ine- t l 1.27 =In(I-~) 

11.65 

- 1.~7 =In(1- 1:~5) 

I = -1.27In(I-~) 

11.65 

I"" 2.4823 

After 2.4823 seconds he was running at a 

rate of 10 meters per second. 

(a) The left side is a reflection of the right side 

across the vertical axis of the tower; the 

graph of f(-x) is exactly the reflection of the 

graph of f(x) across the y-axis. 

(b) x is half the length 

x = 15.7488 =7.8744 

2 

f(7 .8744) =-301ln 7.8744 

207 

y==984 

The height of the Eiffel Tower is 984 feet. 

(c) Let y =500 and solve for x. 

500=-301ln~ 

207 

500 x 

--=In 

-301 207 

eSOO/-301 =eln(xI207) 

eSOO/-301 =.z, 

207 

207eSOO/-301 =x 

x=39. 

The height is 500 feet, about 39 feet from the 

center. 

49. (a) In(1- P) = -.0034 .0053T 

Change this equation to exponential form 

1- P =e- .OO34-.00S3T 

(b) 

P(T) =1-e- .OO34-.0034T 

47. f(x) = 25 

1+ 1364.3exI9.316 

In 1997, x =97. 

f(97) _ 25 

- 1+1364.3e-97/9.316 

f(97) =24 

In 1997, about 24% of U.S. children lived in a 

home without a father. 

(c) 

From the graph one can see that initialIy 

there is a rapid reduction of carbon dioxide 

emissions. However, after a while there is 

little benefit in raising taxes further. 

P(T) =1-e-·OO34-.00S3T 

P(60) =1-e-·OO340.00S3(60) 

== .275 or 27.5% 

The reduction in carbon emissions from a tax 

of $60 per ton of carbon is 27.5%. 

403



(d) We must determine T when P = .5. 

pen = 1-e-.OO34-.0053T =.5 

52. A =~1+ ~r" 

.5 = 1-e-.OO34-.0053T To solve for t, substitute 

.5 = e-.OO34-.0053T A =2063.40, P =1786, r = .116, and m = 12. 

In.5 = -.0034 - .0053T rJ 6)(')(12) 

2063.40 =178\.1 + .~~ 

T= In.5+.0034 == 130.14 

-.0053 

The value T= $130.14 will give a 50% 

reduction in carbon emissions. 

2063.40 ( .116)12' 

---= 1+- 

1786 12 

In 2063.40 = 121 In(1 + .116) 

1786 12 

50. (a) Letting k =6.3 and ( ~ ) = 2 results in In~ 

----;'.......~~ =1 

12In(I+,*) 

R = k In( ~ ) = 6.3 In 2 1.25 == 1 

To the nearest hundredth, 1 =1.25 yr. 

w 

==4.4---y. 

m 53. 

(b) 

T(R) =1.03R 

T(4.4) = 1.03(4.4) == 4.5°F 

This is less than that predicted by Arrhenius 

in 1896; however, his values are still 

consistent with some current computer 

models. 

To solve to I, substitute 

A =30,000, P =27,000, r =.06, and m =4. 

tvvI 06)(')(4) 

30,000=27,\1+ 7 The interest rate is about 6.48%. 

1.1111,.. (1.015)4' 

In1.1111 = In(1.015 

4,) 54. 

In1.1111 = 41 In 1.015 

Inl.l111 

t 

41n1.015 

1.8", t 

To the nearest tenth of a year, Tom will be ready 

to buy a car in 1.8 yr. 

The interest rate is about 4.27%. 

404


55. (a) At the age of 60 your investment of $2000 

has earned interest for 35 years . 

Your investment has grown to 

A=~I+ .r 

.08)(1)(35) 

A =2 1+ 000( 

1 

=29,570.69 

The tax is 40% of this, or 

(.40)(29,570.69) = 11,828.28 

So you have $17,742.41 left. 

(b) A=~I+ ~rl andr=.048,P=1200, 

A = 1200(1 + .048)(1)(35) = $6191.93 

(c) With the IRA you earn 

17,742.41 - 6191.93 = $11,550.48 more. 

(d) If you pay taxes on the original $2000, then 

you have $1200 to invest. 

A= ~1+ ~)mt 

.08)(1)(35) 

A=12 00( 1+l 

1 

= 17,742.41 

This is the same as with the IRA. 


1. y= log.3x 

The point (1, 0) is on the graph of every function 

of the form y =loga x, so the correct choice must 

beeither B or C. 

Since the base is a = .3 and 

o< .3 < I, y =log.3 x is a decreasing function, 

and so the correct choice must be B. 


3. y = Inx 

or 

y = loge x 

The point (1, 0) is on the graph of every function 

of the form y = log., x, so the correct choice must 

beeither B or C. 

Since the base is a = e and e > I, y = In x is an 

increasing function, and so the correct choice 

must be C. 

4. y =(.3)x 

The point (0, 1) is on the graph since (.3)° =1, so 

the correct choice must beeither A or D. 

Since the base is .3 and 0 < .3 < I, y =(.3)x is a 

decreasing function, and so the correct choice 

must be D. 

5. 2 5 = 32 is written in logarithmic form as 

log2 32 =5. 

6. 100 1/2 =10 is written in logarithmic form as 

1 

logloo 10 =2' 

7. (%r =~ is written in logarithmic form as 

8. 

IOg3/4(~) =-1. 

y 

+--+-,..+--+-+-+-1- .x 

2. y=e x 

The point (0, 1) is on the graph sincee'' = I, so 

the correct choice must be either A or D. 

Since the base is e and e> l.y = eX is an 

increasing function and so the correct choice must 

be A. 

9. 

2 

y 

!(x) = 108) x 

+---!~-+-+---1_ x 

10 20 30 

-1 

10. log3 4 is the logarithm to the base 3 of 4. 

(log, 3 would be the logarithm to the base 

40f3.) 

405



11. The exact value oflog-, 9 is 2 since 3 2 =9. 

12. The exact value oflog-, 27 is 3 since 3 3 =27. 

13. log316 must lie between 2 and 3. Because the 

function defined by y =log3 x is increasing and 

9 < 16 < 27, we have logj 9 < log316 < log, 27. 

14. By the change-of-base theorem, 

_ log16 _ In16 

Iog3 16----- 

log3 In3 

== 2.523719014. 

This value is between 2 and 3, as predicted in 

Exercise 21. 

15. The exact value of logj (~) is -1 since 5- 1 =~. 

The exact value oflog, 1 is 0 since 5° =1. 

16. logs .68 must lie between -I and O.Since the 

function defined by y =logs x is increasing and 

I 

- = .2 < .68 < 1, we must have 

5 

logs .2 < logs .68 < logs 1. 

By the change-of-base theorem, 

_ log.68 _ In.68 

Iogs'68----- 

log5 In5 

...-.239625573. 

This value is between -1 and 0, as predicted 

above. 

3 

17. log927 = 

2 

9 312 =27 

18. log3.45 == .5378 

IO Ṣ378 == 3.45 

19. In 45 == 3.8067 

3 8067 e . == 45 

20. Letf(x) =log; x be the required function. Then 

f(81) =4 

loga 81 =4 

a 4 =81 

a 4 =3 4 

a=3. 

The base is 3. 

21. Letf(x) = aX be the required function. Then 

1 

f(-4) =16 

-4 1 

a = 

16 

a-4 =2-4 

a=2. 

The base is 2. 

22. IOg3( ~; ) 

=log3 mn - log3 5r 


=log-,m+ log3 n -log 3 5 -log 3 r 


23. logs(x2y4~m3p) 

=logs x 2 y4(m 3 p)I/S 

=logs x 2 + logs i + logs(m 3 p)IIS 


=2 logs x+ 4 logs y+ 5 1 (1ogs m 3 p) 


=210g sx + 4 logs y + 5 1 (1ogs m 3 + logs p) 


. I 

= 2log sx+4logs y+-(3Iog s m+ logs p) 

5 


24. log7 (7k +5r2 ) 

Since this is the logarithm ofa sum, this 

expression cannot be simplified. 

25. log 45.6 =1.6590 

26. log .0411 = -1.3862 

27. In 470 =6.1527 

28. In 144,000 =11.8776 

29. To find log , 769, use the change-of-base theorem. 

In 769 

Iog3 769 =- 

In3 

== 6.0486 

406



35. ex+1 = 10 

30. IOg2l3(%) 

X 

Ine + 1 = In 10 

_Iogi 

(x + 1)Ine = In10 

-log! 

x+l =In 10 

-.20412 

x = InlO-1 

= 

-.17609 

=2.3026-1 

= 1.1S92 

= 1.3026 

gX =32 

1 

31. 36. 10g64 x = 

(2 3)x = 2 5 3 

64 1 3 

/ = x Change to exponential form 

2 3x = 2 5 4=:x 

3x=S 

S 

37. In(6x) -In(x + 1)= In 4 

x= 3 

In(~)=ln4 

x+l 

-3 8 Logarithm of a quotient 

32. x = 

27 

~=4 

x+l 

(;f =(~f 

4x+4=6x 

1 2 

4=2x 

-=- 2=x 

x 3 

3 

x=- 

2 

38. logl6.JX + 1 =.!. 

4 

16 114 =.Jx+l 

33. 10 2x - 3 =17 

Take common logarithms of both sides. 

Change to exponential form 

log 102x-3 = log 17 (16 114)2 = (.Jx + 1)2 

(2x- 3) log 10 = log17 

Square both sides 

2x-3= log17 

16 1 / 2 =x+l 

2x= logl7+3 

4=x+l 

x = _lo...::g:...l_7_+..:..3 

3=x 

2 The proposed solution must be checked in the 

x =2.11S2 

original equation since both sides were squared. 

The proposed solution checks, so x = 3. 

34. 4 x + 3 =S2- x 

Take common logarithms of both sides . 

log4 x + 3 = logS2 -x 

(x + 3) log4 = (2 - x) log S 

xlog4+ 310g4 = 210gS -x log S 

x log4 + x log S = 2 log S - 310g4 

x(log4+ log S) = 210gS - 310g4 

x = 2 log S- 3 log 4 

log4+ log S 

x::: -.3138 

407



1 

39. Inx+3ln2 = In~ 43. (a) 6.6= 10glOx 

1 0 

Inx+ In2 3 = In~ Logarithm of a power .!... =10 6 . 

x 

1 0 

In(x. 2 3 ) = In ~ Logarithm of a product 1 

x 

- ::: 4,000,000 

1 

2 

0 

In8x = In I::: 4,000,000./ 0 

x 

The magnitude was about 4,000,000/ 0 , 

Thus, 

2 

8x= 1 

x 

(b) 6.5 =10glO 

1 0 

8x 2 =2 

5 

2 I L =10 6 . 

x =- 1 

4 

0 

1 

I - ::: 3,200,000 

x=±-. 1 

2 

0 

I::: 3,200,000./ 0 

- -I IS . not m . the d ornaṁ 0 

fl n x an d I n-, 2 so It . IS . 

The magnitude was about 3,200,0001 2 x 

0 , 

not a solution. So x = ~. 

(c) Consider the ratio of the magnitudes. 

4,000,000 10 = 40 = ~ = 1.25 

40. log x + log(x - 3) = I 3,200,000 1 0 32 4 

log x(x - 3) = 1 Logarithm of a product The earthquake with a measure of 6.6 was 

10 1 =x(x-3) 

about 1.25 times as great. 

lO=x 2 -3x 1 

44. (a) 8.3= loglo 

O=x 2 -3x-1O 1 0 

0= (x - 5)(x + 2) 

L =10 8 . 

x-5=O or x+2=O 3 

1 0 

x =5 or x =-2 

1=10 8 . 3 -2 is not in the domain of log x or log (x - 3), so 

./ 0 

it is not a solution. So x = 5. 1 = 200,000,000/ 0 


41. In[ln(e- x»)= In3 

In(-x) = In3 Ine- x =-x 

1 

(b) 7.1=loglo 

by theorem on inverses 

1 0 

-x=3 L =10 7 .1 

x=-3 

1 0 

1=10 7 . 1 . / 0 

42. 1 =13,000,000/ 0 

S = a In( I + ~) for n 


;= In(I+~) 

(c) 200,000,000./0 ::: 15.38 

13,000,000· 1 

e Sla -I =!!.. 

0 

a 

The 1906 earthquake had a magnitude more 

Sla 

a(e -I) =n 

than 15 times greater than the 1989 

earthquake. 

n=~eSla_l) 

6 

408



45. For 89 decibels, we have 

I 

89 = 10 log 

1 0 

I 

8.9=log-. 

1 0 

Change this equation to exponential form. 

.l: =10 8 . 9 

1 0 

1=10 8 . 9 / 0 

For 86 decibels, we have 

I 

86= 10log 

1 0 

L =10"8.6 

1 0 

1=10 8 . 6 / 0 

, 

To compare these intensities, find their ratio . 

10 8 . 9 t 10 8 . 9 

---;~O~ =__ "" 2 

10 8 . 6 / 10 8 . 6 

0 

From this calculation, we see that 89 decibels is 

about twice as loud as 86 decibels, for a 100% 

increase. 

46. A = 8780, P = 3500, t= 10, m = 1 

A =~1+ ~)nn 

rvI )10(1) 

8780 = 35'\.1+f 

2.50857 =(1+ r)1O 

(2.50857)1110 =1+ r 

(2.50857)1110 -1 = r 

.09631 =r 

The annual interest rate, to the nearest tenth, is 

9.6%. 

47. Substitute P =48,000, A =58,344, r= .05, and 

m =2 into the formula. 

A=~I+ ~)'m 

58, 344 =48, ~1+ .~5 f)(2) 

58,344 = 48,000(1.025)21 

1.2155 =(1.025)21 

In1.2155 =In(1.025) 21 

In1.2155 = 2t In1.025 

In1.2155 

---=t 

21n 1.025 

4.0""t 

$48,000 will increase to $58,344 in about 4.0 yr. 

48. A =~1+ ~)'m 

First, substitute P = 10,000, r= .12, t =12, and 


.12)12(1) 

A=10, I 

m{ 

I 

=10,000(1.12)12 

""10,000(3.8960) 

=38,959.76 

After the first 12 yr, there would be about 

$38,959.76 in the account. To finish off the 

21-year period, substitute P =38,959.76, r= .10, 

t =9, and m =2 into the original formula, 

A = 38, 959.7{1 + .~0)9(2) 

=38,959.76(1.05)18 

= 38,959.76(2.4066) 

""93,761.31 

At the end of the 21-year period, about 

$93,761.31 would be in the account. 

49. A =~1+ ~)tm 

First, substitute P =12,000, r =.05, t =8, and 


A =12,m{I+ .~5)8(1) 

=12,000(1.05)8 . 

= 12,000(1.4775) 

= 17,729.47 

After the first 8 yr, there would be $17,729.47 in 

the account. To finish off the 14-year period, 

substitute P =17,729.47, r= .06, t = 6, and m = 1 

into the original formula, 

A = 17,729.47(1 + .~)6(I) 

= 17,729.47(1.06)6 

= 17,729.47(1.4185) 

= 25,149.59 

At the end of the 14-year period, $25,149.59 

would be in the account. 

50. A =Pen 

To find t, substitute a =2, P = 1, and r =.1. 

2 =I·e·11 

Take natural logarithms of both sides. 

In2 =In e o)1 

In2 =.It 

In2 

-=t 

.1 

6.93""t 

It would take about 6.9 yr. 

409



51. f(x);;: 6.2(1O)-12(1.4)X 

Find x when f(x);;: 2000. 

2000;;: 6.2(10)-12(1.4)x 

322.6 x 10 12 ;;:(1.4)x 

log(322.6 x 10 12);;: x log 1.4 

99= x 

Software exports will double their 1997 value in 

1999. 

52. Let K be the total number of executives, 

managers, and administrators in communications. 

The number of black executives, managers, and 

administrators is .034K. If this number increases 

at a rate of 5% per year, we must determine when 

it will reach .12K. That is, we must determine 

when .034K(1.05)' ;;:.12K. 

Use logarithms to solve this equation for 1. 

(1.05)' ;;:-E.. 

.034 

1In1.05 =In(-E..) 

.034 

1=In(1&) = 25.8 

In 1.05 

Since 1991+ 25.8;;: 2020.8, this will occur during 

the year 2020 . 

53. (a) Compute the natural logarithm of P for each 

value of P given in the table in the textbook. 

x 

InP 

0 6.921 

1000 6.801 

2000 6.678 

3000 6.553 

4000 6.425 

5000 6.293 

6000 6.157 

7000 6.019 

8000 5.878 

9000 5.730 

10,000 5.580 

Plot this data on a graphing calculator using 

the window [-100, 11,000] by [5, 7] with 

Xscl ;;: 1000 and Ysci ;;: 1. 

54. 

(b) P= Ce kx 

Take natural logarithms on both sides. 

InP;;: InCe kx 

In P= InC+lne kx 

InP;;:kx+lnC 

Thus, y = In P = ax + b where a ;;: k and 

b ;;: In C, which is a linear function. 

(a) Plot the year on the x-axis and the number of 

processors on the y-axis. Let x =0 

correspond to the year 1971. 

10,000,000 

-1,000,000 

(b) The data are clearly not linear and do not 

level off like a logarithmic function. The data 

are increasing at a faster rate as x increases. 

Of the three choices, the exponential function 

(iii) will describe this data best . 

(c) 

Let x;;: 0 correspond to the year 1971. We 

can require that the exponential function pass 

through the points (0, 2300) and 

(26, 9,500,(00). 

We want to find a function of the form 

f(x);;: ae bx . 

First, find the value of a. 

f(O) =aeo =2300 

a;;: 2300 

Now find the value of b. 

f(26);;: 23OOe 26b ;;: 9,500,000 

26b 9,500,000 

e ;;: 

2300 

26b 

e = 4130 

26b =In 4130 

b == In4130 

26 

b = .3202 

410



Thus, we obtain the-exponential function 

I(x) =2300e,3202x. 

(If a different pair of points from the table is 

used, a slightly different function will be 

found.) 

(d) We can predict the number of transistors on a 

chip in the year 2002 by evaluating 

1(31) =2300e·3202(31) :: 47,000,000. 

55. (a) A(t) =t 2 - t + 350 

~-----~-~--... 

Fort = x,A(x) = x 2 - x +350 

56. Graph Y 1 =x 2 and Y2 =2 x on the same screen. 

Since the range of both functions is (0, 00), there 

is no need to include Quadrants ill and N in the 

viewing window. A good choice for the window 

is [-5, 5] by [0, 20]. 

The graph shows that the two curves intersect in 

three points. Two of these points are (2, 4) and 

(4, 16), as given in the exercise. To find the 

coordinates of the third point, use the "intersect" 

option in the CALC menu. The calculator 

displays the coordinates (-.7666647, .58777476). 

(a) Use the change-of-base theorem with base e 

to write the function as 

2 

I(x) = In(2x - x) 

In4 

o 

(b) A(t) =350 ]og(t +1) 

--....;;;..------., 

(b) Graph the function. 

In(Zx2 - x) 

1B4 

(e) 

(d) 

0ll:======== 

Fort = x,A(x) = l00(.95)~ 

(e) From the graph, the x-intercepts are . 

_.!.. and 1. 

2 

(d) From the graph, the vertical asymptotes are 

1 

x=Oandx=-. 

2 

(e) To make a y-lntercept, x = 0 must be in the 

domain, which is not the case here. 


1. (a) I and g are inverse functions. 

(b) The graph of g is the reflection of the graph 

ofI across the line y = x. 

(e) (9.2) 

o 

o 

Function (c) best describes A(t). 

411



2. (a) y = logl/3 x 

The point (1, 0) is on the graph of every 

function of the form y = loga x, so the 

correct choice must be either B or C. 

(b) 

(c) 

(d) 

i 

Since the base is a = 1. and 

3 

O l,y = eX is an 

increasing function, and so the correct choice 

must be A. 

Y = In x or y = loge x 

The point (1, 0) is on the graph of every 

function of the form y = log, x, so the 

correct choice must be B or C. 

Since the base is a = e and e> l,y = In x is 

an increasing function, and the correct choice 

must be C. 

y=Gr 

The point (0, 1) is on the graph since 

(~r = 1, so the correct choice must be 

either A or D. 

Since the base is.!.. and 0 < 1. < 1, y = (1.r 

3 3 3 

is a decreasing function, and so the correct 

choice must be D. 

( )2x-3 1 

3. = 16 x + 

(2-3)2x-3 = (24 ) x+ 1 

2-3(2x-3) = 24(x+1) 

2- 6x+9 = 2 4x+4 

-6x+9=4x+4 

-6x-4x=4-9 

-lOx =-5 

1 

x= 2 

4. (a) 4 3 / 2 = 8 is written in logarithmic form as 

3 

log4 8 = 

2 

(b) logg 4 = ~ is written in exponential form as 

3 

8 2/3 = 4 

S. f(X)=(~r andg(x)=logll2 x 

y 

A: 

g(x) =108 11 

2 X 

They are inverses of each other. 

6. [x2~) 

log7~ 

=log7x 2 +log7 ~ y-Iog 7z 

3 

1 

= 210g7x+-log7 y-310g 7 z 

4 

7. log 237.4 = 2.3755 

8. In.0467 = -3.0640 

In13 

9. log913 = --= 1.1674 

In9 

10. log(2.49 x 10- 3) = -2.6038 

11. log; 25 =2 

x 2 =25 

x 2 =5 2 

x=5 

Solution set: {5} 

12. log, 32 =x 

4 x =32 

2 2x =2 5 

2x=5 

-- 

5 

x= 2 

13. log2 x + log2(x + 2) = 3 

log2 x(x+2) = 3 

2 3 = x(x+2) 

8 = x 2 +2x 

0=x 2 +2x-8 

0= (x+4)(x-2) 

412



x+4=0 

or x-2=0 

x=-4 or x=2 

x =- 4 is not in the domain of log2 x or 

log2(x + 2), so it is not a solution. So x =2. 

14. SX+l =7 x 

InS x + l =In 7 x 

(x + 1)InS =x In 7 

x InS + InS =x In7 

InS = xln7 -xlnS 

InS =x(ln 7 -InS) 

InS 

---=x 

In7 -InS 

4.7833 = x 

15. In x - 41n 3 =In(';) 

In(; ) =In(';) 

x 5 

-= 

81 x 

x 2 =4OS 

x =± 20. 1246 

Since the domain of In x is (0, 00),only the 

positive square root of 40S can be a solution. 

So x = 20.1246. 

16. logs 27 is the exponent to which S must be raised 

in order to obtain 27. 

To approximate log5 27 on your calculator use the 

change-of-base formula; 

log 27 

Iog5 27 =- 

logS 

1.43136 

= .69897 

=2.048. 

17. v(1)=176(I-e-.J8t) 

Find the time 1 at which v(1)=147. 

147 =176(1-e-.J8t) 

147 =1_e-.J8r 

176 

-.J8r 176 147 

e =-- 

176 176 

-.J8t _ 29 

e - 

176 

Ine-·18t = In 29 

176 

- .lSt = In.I6477 

In.I6477 

1=-- 

-.18 

= -1.8 = 10 

-.18 

It will take the skydiver about 10 sec to attain the 

speed of 147 ft per sec (100 mph). 

18. (8) A=~I+ ~)hn 

(b) 

18,000 =s


414 Chapter9 Exponential and Logarithmic Equations 

20. Let x =0 correspond to the year 2000. The 

population ofNew York (in millions) can be 

approximated by y =18. 15e·OO21x • 

The population ofFIorida (in millions) can be 

approximated by y =15.23e·0131x. 

Graph the two functions on the same screen and 

use the "intersect" option to find the coordinates 

of the intersection. 

30 

X"" 15.9 corresponds to the end ofthe year 2015. 

Near the end of2015 the populations will be 

equal. 

414

Trigonometry Solutions Manual

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