Chapter 4 Trigonometric Functions

Chapter 4 

Trigonometric Functions 

Section 4.1 

Check Point Exercises 

1. The radian measure of a central angle is the length of 

the intercepted arc, s, divided by the circle’s radius, r. 

The length of the intercepted arc is 42 feet: s = 42 

feet. The circle’s radius is 12 feet: r = 12 feet. Now 

use the formula for radian measure to find the radian 

measure of θ . 

s 42 feet 

θ = = = 3.5 

r 12 feet 

Thus, the radian measure of θ is 3.5 

4. a. 

b. 

2. a. 

b. 

c. 

3. a. 

b. 

c. 

π radians 60π 

60°= 60 °⋅ = radians 

180° 

180 

π 

= radians 

3 


270°= 270 °⋅ = radians 

180° 

180 

3π 

= radians 

2 

π radians −300π 

− 300°= − 300 °⋅ = radians 

180° 

180 

5π 

=− radians 

3 

π π radians 180 

o 

radians = ⋅ 

4 

4 π radians 

180 

o 

= = 45 

o 

4 

4π 

4 π radians 180 

o 

− radians = − ⋅ 

3 3 π 

4180 ⋅ 

o 

=− =− 240 

o 

3 

180 

o 

6 radians = 6 radians ⋅ 

π radians 

6⋅180 o 

= ≈ 343.8 

o 

π 

c. 

d. 

5. a. For a 400º angle, subtract 360º to find a positive 

coterminal angle. 

400 o – 360 o = 40 o 

6. a. 

b. For a –135º angle, add 360º to find a positive 

coterminal angle. 

–135 o + 360 o = 225 o 

b. 

13 π 13 10 3 

− 2π 

= π − π = 

π 

5 5 5 5 

30 29 

− π + 2π 

= − π + π = 

π 

15 15 15 15 

489 

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.


7. a. 855°− 360°⋅ 2 = 855°− 720°= 135° 

b. 

c. 

17π 

17π 

−2π 

⋅ 2= − 4π 

3 3 

17π 12π 5π 

= − = 

3 3 3 

25π 

25π 

− + 2π 

⋅ 3=− + 6π 

6 6 

25π 36π 11π 

=− + = 

6 6 6 

8. The formula s = rθ can only be used when θ is 

expressed in radians. Thus, we begin by converting 

π radians 

45° to radians. Multiply by . 

180° 

π radians 45 

45°= 45 °⋅ = π radians 

180° 

180 

π 

= radians 

4 

Now we can use the formula s = rθ to find the 

length of the arc. The circle’s radius is 6 inches : r = 

6 inches. The measure of the central angle in radians 

π π 

is : θ = . The length of the arc intercepted by this 

4 4 

central angle is 

⎛π 

⎞ 6π 

s = rθ 

= (6 inches) ⎜ ⎟= inches ≈ 4.71 inches. 

⎝ 4⎠ 

4 

9. We are given ω , the angular speed. 

ω = 45 revolutions per minute 

We use the formula ν = rω 

to find v , the linear 

speed. Before applying the formula, we must express 

ω in radians per minute. 

45 revolutions 2 π radians 

ω = ⋅ 

1 minute 1 revolution 

90 π radians 

= 

1 minute 

The angular speed of the propeller is 90π radians per 

minute. The linear speed is 

90π 

135 π inches 

ν = rω 

= 1.5 inches ⋅ = 1 minute minute 

The linear speed is 135π inches per minute, which is 

approximately 424 inches per minute. 

Exercise Set 4.1 

1. obtuse 

2. obtuse 

3. acute 

4. acute 

5. straight 

6. right 

7. 

8. 

9. 

10. 

11. 

12. 

13. 

14. 

θ = s 40 inches 

4 radians 

r 

= 10 inches 

= 

θ = s 30 feet 

6 radians 

r 

= 5 feet 

= 

s 8 yards 4 

θ = = = radians 

r 6 yards 3 

θ = s 18 yards 

2.25 radians 

r 

= 8 yards 

= 

θ = s 400 centimeters 

4 radians 

r 

= 100 centimeters 

= 

θ = s 600 centimeters 

6 radians 

r 

= 100 centimeters 

= 

π radians 

45°= 45°⋅ 

180° 

45π 

= radians 

180 

π 

= radians 

4 

π radians 

18°= 18°⋅ 

180° 

18π 

= radians 

180 

π 

= radians 

10 

15. 

π radians 

135°= 135°⋅ 

180° 

135π 

= radians 

180 

3π 

= radians 

4 

490 


PreCalculus 4E Section 4.1 

16. 

17. 

18. 

19. 

20. 

21. 

22. 

π radians 

150°= 150°⋅ 

180° 

150π 

= radians 

180 

5π 

= radians 

6 

π radians 

300°= 300°⋅ 

180° 

300π 

= radians 

180 

5π 

= radians 

3 

π radians 

330°= 330°⋅ 

180° 

330π 

= radians 

180 

11π 

= radians 

6 

π radians 

− 225°= − 225°⋅ 

180° 

225π 

=− radians 

180 

5π 

=− radians 

4 

π radians 

− 270°= − 270°⋅ 

180° 

270π 

=− radians 

180 

3π 

=− radians 

2 


o 

radians = ⋅ 

2 2 π radians 

180 

o 

= 

2 

= 90 o 


o 

radians = ⋅ 


180 

o 

= = 20 

o 

9 

23. 

24. 

25. 

26. 

27. 

28. 

29. 

30. 

31. 

2π 


o 

radians = ⋅ 


2180 ⋅ 

o 

= 

3 

= 120 o 


0 

3⋅180 o 

⋅ = = 135 

o 


7π 


o 

radians = ⋅ 


7180 ⋅ 

o 

= 

6 

= 210 o 


o 

11⋅180 o 

⋅ = = 330 

o 


180 

o 

− 3 π radians = −3 π radians ⋅ 

π radians 

=−3180 

⋅ 

o 

=−540 

o 

180 

o 

−4 π radians ⋅ = −4 ⋅ 180 

o 

=− 720 

o 

π radians 

π radians 

18°= 18°⋅ 

180° 

18π 

= radians 

180 

≈ 0.31 radians 

π radians 

76°= 76°⋅ 

180° 

76π 

= radians 

180 


π radians 

− 40°=− 40°⋅ 

180° 

40π 

=− radians 

180 

≈−0.70 radians 

491 



32. 

33. 

34. 

35. 

36. 

π radians 

− 50°=− 50°⋅ 

180° 

50π 

=− radians 

180 

≈−0.87 radians 

π radians 

200°= 200°⋅ 

180° 

200π 

= radians 

180 


π radians 

250°= 250°⋅ 

180° 

250π 

= radians 

180 


180 

o 

2 radians = 2 radians ⋅ 

π radians 

2180 ⋅ 

o 

= 

π 

≈ 114.59 

o 

180 

o 

3⋅180 o 

3 radians ⋅ = ≈ 171.89 

o 

π radians π 

41. 

42. 

43. 

44. 

37. 

38. 

39. 

40. 


o 

radians = ⋅ 


180 

o 

= 

13 

≈ 13.85 o 

π 180 

o 

180 

o 

radians ⋅ = ≈ 10.59 

o 


180 

o 

− 4.8 radians =−4.8 radians ⋅ 

π radians 

−4.8⋅180 

o 

= 

π 

≈−275.02 

o 

180 

o 

−5.2⋅180 

o 

−5.2 radians⋅ = 

πradians 

π 

≈− 297.94 

o 

45. 

46. 

47. 

492 



48. 

55. 

49. 

56. 

50. 

57. 395°− 360°= 35° 

58. 415°− 360°= 55° 

59. − 150°+ 360°= 210° 

51. 

60. − 160°+ 360°= 200° 

61. − 765°+ 360°⋅ 3 =− 765°+ 1080°= 315° 

62. − 760°+ 360°⋅ 3 =− 760°+ 1080°= 320° 

52. 

53. 

54. 

63. 

64. 

65. 

66. 

67. 

68. 

69. 

19 π 19 12 7 

− 2π 

= π − π = 

π 

6 6 6 6 

17 π 17 10 7 

− 2π 

= π − π = 

π 

5 5 5 5 

23 π 23 23 20 3 

−2π 

⋅ 2= π − 4π 

= π − π = 

π 

5 5 5 5 5 

25 π 25 25 24 

−2π 

⋅ 2= π − 4π 

= π − π = 

π 

6 6 6 6 6 

100 99 

− π + 2π 

=− π + π = 

π 

50 50 50 50 

80 79 

− π + 2π 

= − π + π = 

π 

40 40 40 40 

31π 

31π 

− + 2π 

⋅ 3=− + 6π 

7 7 

31π 42π 11π 

=− + = 

7 7 7 

493 



38π 

38π 

70. − + 2π 

⋅ 3= − + 6π 

9 9 

38π 54π 16π 

=− + = 

9 9 9 

71. r = 12 inches, θ = 45° 

Begin by converting 45° to radians, in order to use 

the formula s = rθ . 

π radians π 

45°= 45°⋅ = radians 

180° 

4 

Now use the formula s = rθ . 

π 

s = rθ 

= 12⋅ = 3π 

inches ≈ 9.42 inches 

4 

72. r = 16 inches, θ = 60° 



π radians π 

60°= 60°⋅ = radians 

180° 

3 


π 16π 

s = rθ 

= 16 ⋅ = inches ≈ 16.76 inches 

3 3 

73. r = 8feet, θ = 225° 




225°= 225°⋅ = radians 

180° 

4 


5π 

s = rθ 

= 8⋅ = 10π 

feet ≈ 31.42 feet 

4 

74. r = 9 yards, θ = 315° 




315°= 315°⋅ = radians 

180° 

4 


7π 

63π 

s = rθ 

= 9 ⋅ = yards ≈ 49.48 yards 

4 4 

75. 6 revolutions per second 

6 revolutions 2 π radians 12 π radians 

= ⋅ = 

1 second 1 revolutions 1 seconds 

= 12 π radians per second 

77. 

78. 

79. 

80. 

81. 

4π 

2π 

− and 

3 3 

7π 

5π 

− and 

6 6 

3π 

5π 

− and 

4 4 

π 7π 

− and 

4 4 

π 3π 

− and 

2 2 

82. −π and π 

83. 

84. 

55 11π 

⋅ 2π 

= 

60 6 

35 7π 

⋅ 2π 

= 

60 6 

85. 3 minutes and 40 seconds equals 220 seconds. 

220 22π 

⋅ 2π 

= 

60 3 

86. 4 minutes and 25 seconds equals 265 seconds. 

265 53π 

⋅ 2π 

= 

60 6 

87. First, convert to degrees. 

1 1 360° 

revolution = revolution ⋅ 

6 6 1 revolution 

1 

= ⋅ 360 °= 60 ° 

6 

Now, convert 60° to radians. 


60°= 60°⋅ = radians 

180° 

180 

π 

= radians 

3 

Therefore, 1 6 revolution is equivalent to 60° or π 

3 

radians. 

76. 20 revolutions per second 

20 revolutions 2 π radians 40 π radians 

= ⋅ = 

1 second 1 revolution 1 second 

= 40 π radians per second 

494 



88. First, convert to degrees. 

1 1 360 

o 

revolutions = revolutions ⋅ 


1 

= ⋅ 360 

o 

= 120 

o 

3 

Now, convert 120° to radians. 


120°= 120°⋅ = radians 

180° 

180 

2π 

= radians 

3 

Therefore, 1 3 revolution is equivalent to 120° or 2 π 

3 

radians. 

89. The distance that the tip of the minute hand moves is 

given by its arc length, s. Since s = rθ , we begin by 

finding r and θ . We are given that r = 8 inches. The 

minute hand moves from 12 to 2 o'clock, or 1 6 of a 

complete revolution. The formula s = rθ can only be 

used when θ is expressed in radians. We must 

convert 1 revolution to radians. 

6 

1 1 2 π radians 



π 

= radians 

3 

The distance the tip of the minute hand moves is 

⎛π 

⎞ 8π 

s = rθ 

= (8 inches) ⎜ ⎟= 

inches 

⎝ 3 ⎠ 3 

≈ 8.38 inches. 

90. The distance that the tip of the minute hand moves is 

given by its arc length, s. Since s = rθ , we begin by 

finding r and θ . We are given that 

r = 6 inches. The minute hand moves from 12 to 4 

o’clock, or 1 of a complete revolution. The formula 

3 

s = rθ can only be used when θ is expressed in 

radians. We must convert 1 revolution to radians. 

3 

1 1 2 π radians 



2π 

= radians 

3 

The distance the tip of the minute hand moves is 

⎛2π 

⎞ 12π 

s = rθ 

= (6 inches) ⎜ ⎟= 

inches 

⎝ 3 ⎠ 3 

= 4π 

inches ≈12.57 inches. 

91. The length of each arc is given by s = rθ . We are 

given that r = 24 inches and 

θ = 90 ° . The formula s = rθ can only be used when 

θ is expressed in radians. 


90°= 90°⋅ = radians 

180° 

180 

π 

= radians 

2 

The length of each arc is 

⎛π 

⎞ 

s = rθ 

= (24 inches) ⎜ ⎟= 

12π 

inches 

⎝ 2 ⎠ 

≈ 37.70 inches. 

92. The distance that the wheel moves is given by 

s = rθ . We are given that r = 80 centimeters and θ 

= 60°. The formula s = rθ can only be used when θ 

is expressed in radians. 


60°= 60°⋅ = radians 

180° 

180 

π 

= radians 

3 

The length that the wheel moves is 

⎛π 

⎞ 80π 

s = rθ 

= (80 centimeters) ⎜ ⎟ = centimeters 

⎝ 3⎠ 

3 

≈ 83.78 centimeters. 

s 

93. Recall that θ = . We are given that 

r 

s = 8000 miles and r = 4000 miles. 

s 8000 miles 

θ = = = 2 radians 

r 4000 miles 

Now, convert 2 radians to degrees. 

180 

o 

2 radians = 2 radians ⋅ ≈ 114.59 

o 

π radians 

s 

94. Recall that θ = . We are given that 

r 

s = 10,000 miles and r = 4000 miles. 

s 10,000 miles 

θ = = = 2.5 radians 

r 4000 miles 

Now, convert 2.5 radians to degrees. 

180 

o 

2.5 radians⋅ ≈ 143.24 

o 

2 π radians 

495 



95. Recall that s = rθ . We are given that 

r = 4000 miles and θ = 30°. The formula s = rθ can 

only be used when θ is expressed in radians. 


30°= 30°⋅ = radians 

180° 

180 

π 

= radians 

6 

⎛π 

⎞ 

s= rθ 

=(4000 miles) ⎜ ⎟ ≈ 2094 miles 

⎝ 6 ⎠ 

To the nearest mile, the distance from A to B is 

2094 miles. 

96. Recall that s = rθ . We are given that 

r = 4000 miles and θ = 10°. We can only use the 

formula s = rθ when θ is expressed in radians. 


10°= 10°⋅ = radians 

180° 

180 

π 

= radians 

18 

⎛ π ⎞ 

s= rθ 

=(4000 miles) ⎜ ⎟ ≈ 698 miles 

⎝18 

⎠ 

To the nearest mile, the distance from A to B is 

698 miles. 

97. Linear speed is given by ν = rω 

. We are given that 

π 

ω = radians per hour and 

12 

r = 4000 miles. Therefore, 

⎛ π ⎞ 

ν = rω 

= (4000 miles) ⎜ ⎟ 

⎝ 12 ⎠ 

4000π 

= miles per hour 

12 

≈ 1047 miles per hour 

The linear speed is about 1047 miles per hour. 



r = 25 feet and the wheel rotates at 3 revolutions per 

minute. We need to convert 3 revolutions per minute 

to radians per minute. 

3 revolutions per minute 

2π 

radians 

= 3 revolutions per minute ⋅ 1 revolution 

= 6π 

radians per minute 

ν = rω = (25 feet)(6 π) ≈ 471 feet per minute 

The linear speed of the Ferris wheel is about 

471 feet per minute. 



r = 12 feet and the wheel rotates at 20 revolutions per 

minute. 

20 revolutions per minute 

2π 

radians 

= 20 revolutions per minute⋅ 

1 revolution 

= 40π 


ν = rω = (12 feet)(40 π) 

≈ 1508 feet per minute 

The linear speed of the wheel is about 

1508 feet per minute. 

100. Begin by converting 2.5 revolutions per minute to 

radians per minute. 

2.5 revolutions per minute 

2π 

radians 

= 2.5 revolutions per minute ⋅ 1 revolution 

= 5π 


The linear speed of the animals in the outer rows is 


The linear speed of the animals in the inner rows is 


The difference is 100π − 50π = 50π 

feet per minute 

or about 157.08 feet per minute. 

101. – 112. Answers may vary. 

113. 

114. 

115. 

116. 

496 



117. does not make sense; Explanations will vary. Sample 

explanation: Angles greater than π will exceed a 

straight angle. 


explanation: It is possible for π to be used in an 

angle measured using degrees. 

119. makes sense 


explanation: That will not be possible if the angle is a 

multiple of 2 π . 

121. A right angle measures 90° and 

π 

90 

≈ °= radians 1.57 radians. 

2 

3 

If θ = radians = 1.5 radians, θ is smaller than a 

2 

right angle. 

122. s = rθ 

Begin by changing θ = 20° to radians. 

π π 

20°= 20°⋅ = radians 

180° 

9 

π 

100= r 9 

900 

r = ≈ 286 miles 

π 

To the nearest mile, a radius of 286 miles should be 

used. 

123. s = rθ 

Begin by changing θ = 26° to radians. 

124. 

π 13π 

26°= 26°⋅ = radians 

180° 

90 

13π 

s=4000⋅ 

90 

≈ 1815 miles 

To the nearest mile, Miami, Florida is 1815 miles 

north of the equator. 

125. domain: { x −1≤ x ≤ 1} 

or [ − 1,1] 

range: { y −1≤ y ≤ 1} 

or [ − 1,1] 

126. 

1 3 

x =− ; y = 

2 2 

1 

− 

x 2 1 1 3 3 

= = − =− =− 

y 3 3 3 3 3 

2 

Section 4.2 


1. 

P ⎛ 

⎜ 

⎝ 

sin t 

3 1 

, 

2 2 

= y = 

⎞ 

⎟ 

⎠ 

1 

2 

cost 

= x = 

3 

2 

1 

2 

3 

2 

y 

tan t = = = 

x 

1 

csct 

= = 2 

y 

1 2 3 

sect 

= = 

x 3 

cot t = x = 3 

y 

3 

3 

2. The point P on the unit circle that corresponds to 

t = π has coordinates (–1, 0). Use x = –1 and y = 0 to 

find the values of the trigonometric functions. 

sinπ 

= y = 0 

cosπ 

= x = −1 

y 0 

tanπ 

= = = 0 

x −1 

1 1 

secπ 

= = = −1 

x −1 

x −1 

cotπ 

= = = undefined 

y 0 

1 1 

cscπ 

= = = undefined 

y 0 

497 



3. 

π ⎛ 1 1 ⎞ 

t = , P⎜ , ⎟ 

4 ⎝ 2 2 ⎠ 

π 1 

csc = = 

4 y 

2 

π 1 

sec = = 

4 x 

2 

π x 

cot = = 

4 y 

= 1 

1 

y 

1 

2 

⎛ π ⎞ ⎛π 

⎞ 

4. a. sec⎜− = sec = 2 

4 

⎟ ⎜ 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 

5. 

b. 

⎛ π ⎞ ⎛π 

⎞ 2 

sin ⎜− sin 

4 

⎟ =− ⎜ =− 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 2 

2 

sinθ 

tanθ 

= = 

3 

cosθ 

5 

3 

2 3 2 

= ⋅ = 

3 5 5 

2 5 2 5 

= ⋅ = 

5 5 5 

1 1 3 

cscθ 

= = = 

sinθ 

2 2 

3 

1 1 3 

secθ 

= = = 

cosθ 

5 5 

3 

3 5 3 5 

= ⋅ = 

5 5 5 

1 1 5 

cotθ 

= = = 

tanθ 

2 2 

5 

6. 

7. a. 

1 π 

sin t = , 0 ≤ t < 

2 2 

2 2 

sin t+ cos t = 1 

2 

⎛1 

⎞ + 

2 = 

⎜ cos 1 

2 

⎟ t 

⎝ ⎠ 

2 1 

cos t = 1− 

4 

3 3 

cost 

= = 

4 2 

π 

Because 0 ≤ t < , cos t is positive. 

2 

b. 

5π ⎛π ⎞ π 

cot = cot π cot 1 

4 

⎜ + = = 

4 

⎟ 

⎝ ⎠ 4 

⎛ 9π 

⎞ ⎛ 9π 

⎞ 

cos ⎜− cos 4π 

4 

⎟= ⎜− + 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 

7π 

= cos 4 

= 

π 

8. a. sin ≈ 0.7071 

4 

2 

2 

b. csc 1.5 ≈ 1.0025 


1. The point P on the unit circle has coordinates 

⎛ 15 8 ⎞ 

⎜− 

, 

17 17 

⎟ 

⎝ ⎠ . Use 15 8 

x =− and y = to find the 

17 17 

values of the trigonometric functions. 

8 

sin t = y = 

17 

15 

cost 

= x = − 

17 

8 

y 17 8 

tan t = = = − 

15 

x − 15 

17 

1 17 

csct 

= = 

y 8 

1 17 

sect 

= = − 

x 15 

x 15 

cot t = = − 

y 8 

498 



2. The point P on the unit circle has coordinates 

⎛ 5 12⎞ 

⎜− 

, − 

13 13 

⎟ 

⎝ ⎠ Use 5 12 

x =− and y =− to find the 

13 13 

values of the trigonometric functions. 

12 

sin t = y = − 

13 

5 

cost 

= x =− 

13 

12 

y − 

13 12 

tan t = = = 

5 

x − 5 

13 

1 13 

csct 

= = − 

y 12 

1 13 

sect 

= =− 

x 5 

x 5 

cot t = = 

y 12 


t = − π 

4 has coordinates ⎛ 2 2 ⎞ 

, − 

. Use x = 

⎜ 2 2 ⎟ 

⎝ ⎠ 

2 

and y =− to find the values of the trigonometric 

2 

functions. 

sin t = y = − 

2 

2 

cost 

= x = 

2 

2 

2 

y − 

2 

tan t = = 

x 2 

2 

=−1 

1 

csct 

= = − 

y 

2 

1 

sect 

= = 

x 

2 

x 

cot t = = −1 

y 

2 

2 


3π 

⎛ 2 2 ⎞ 

t = has coordinates 

− , . Use 

4 

⎜ 2 2 ⎟ 

⎝ ⎠ 

5. 

6. 

7. 

8. 

9. 

10. 

11. 

x =− 

2 2 

and y = to find the values of the 

2 2 

trigonometric functions. 

sin t = y = 

2 

2 

cost 

= x =− 

2 

2 

2 

y 2 

tan t = = 

x 2 

− 

2 

=−1 

1 

csct 

= = 

y 

2 

1 

sect 

= = − 

x 

2 

x 

cot t = = −1 

y 

π 1 

sin = 6 2 

π 3 

sin = 3 2 

5π 3 

cos =− 6 2 

2π 1 

cos =− 3 2 

0 

tan π= = 0 

−1 

0 

tan 0 = = 0 

1 

7π 1 

csc = =− 2 

6 − 

1 

2 

12. 

13. 

4π 

1 −2 3 

csc = = 3 − 3 

3 

2 

3 

2 

11π 1 2 3 

sec = = 6 3 

499 



14. 

5π 1 

sec = = 2 

3 

1 

2 

24. a. 

11π 

− 3 

tan = = − 

6 3 

1 

2 

3 

2 

15. 

16. 

17. 

18. 

19. a. 

3π sin =− 1 

2 

3π cos = 0 

2 

3π sec = undefined 

2 

3π tan = undefined 

2 

b. 

20. a. 

b. 

21. a. 

π 3 

cos = 6 2 

⎛ π ⎞ π 3 

cos⎜− cos 

6 

⎟ = = 

⎝ ⎠ 6 2 

π 1 

cos = 3 2 

⎛ π ⎞ π 1 

cos⎜− cos 

3 

⎟= = 

⎝ ⎠ 3 2 

5π 1 

sin = 6 2 

25. 

26. 

b. 

⎛ 11π 

⎞ 11π 

3 

tan ⎜− tan 

6 

⎟= − = 

⎝ ⎠ 6 3 

8 15 

sin t = , cost 

= 

17 17 

8 

17 8 

tan t = = 

15 

15 

17 

17 

csct 

= 

8 

17 

sect 

= 

15 

15 

cot t = 

8 

3 4 

sin t = , cos t = 

5 5 

3 

5 3 

tan t = = 

4 

4 

5 

5 

csct 

= 

3 

5 

sect 

= 

4 

4 

cot t = 

3 

b. 

22. a. 

b. 

23. a. 

⎛ 5π 

⎞ 5π 

1 


6 

⎟ =− = − 

⎝ ⎠ 6 2 

2π 3 

sin = 3 2 

⎛ 2π 

⎞ 2π 

3 


3 

⎟ =− =− 

⎝ ⎠ 3 2 

3 

2 

1 

2 

5π 

− 

tan = = − 3 

3 

27. 

1 2 2 

sin t = , cos t = 

3 3 

1 

3 2 

tan t = = 

2 2 4 

3 

csct 

= 3 

3 2 

sect 

= 

4 

cot t = 2 2 

b. 

⎛ 5π 

⎞ 5π 

tan ⎜− = − tan = 3 

3 

⎟ 

⎝ ⎠ 3 

500 



28. 

29. 

30. 

2 5 

sin t = , cost 

= 

3 3 

2 

3 2 5 

tan t = = 

5 

5 

3 

3 

csct 

= 

2 

3 5 

sect 

= 

5 

5 

cot t = 

2 

6 π 

sin t = , 0 ≤ t < 

7 2 

2 2 


2 

⎛6 

⎞ + 

2 

t = 

⎜ cos 1 

7 

⎟ 

⎝ ⎠ 

2 36 

cos t = 1− 

49 

13 13 

cost 

= = 

49 7 

π 

Because 0 ≤ t < , cost 

is positive. 

2 

7 π 

sin t = , 0 ≤ t < 

8 2 

2 2 


2 

⎛7 

⎞ + 

2 

t = 

⎜ 

8 

⎟ 

⎝ ⎠ 

cos 1 

2 49 

cos t = 1− 

64 

31. 

32. 

33. 

34. 

39 π 

sin t = , 0 ≤ t < 

8 2 

2 2 


2 

⎛ 39 ⎞ 

+ 

2 = 

cos t 1 

⎜ 

8 ⎟ 

⎝ ⎠ 

2 39 

cos t = 1− 

64 

25 5 

cost 

= = 

64 8 

π 

Because 0 ≤ t < , cost 

is positive. 

2 

21 π 

sin t = , 0 ≤ t < 

5 2 

2 2 


2 

⎛ 21 ⎞ 

+ 

2 = 

cos t 1 

⎜ 

5 ⎟ 

⎝ ⎠ 

2 21 

cos t = 1− 

25 

4 2 

cost 

= = 

25 5 

π 


2 

⎛ 1 ⎞ 

sin1.7csc1.7 = sin1.7⎜ 

= 1 

sin1.7 

⎟ 

⎝ ⎠ 

⎛ 1 ⎞ 

cos 2.3 sec 2.3 = cos 2.3⎜ 

= 1 

cos 2.3 

⎟ 

⎝ ⎠ 

15 15 

cost 

= = 

64 8 

π 


2 

35. 

36. 

2 π 2 π 

sin + cos = 1 by the Pythagorean identity. 

6 2 

2 π 2 π 

sin + cos = 1 because 

3 3 

2 2 

sin t+ cos t = 1. 

37. 

2 π 2 π 

2 2 

sec − tan = 1 because 1+ tan t = sec t. 

3 3 

38. 

2 π 2 π 

csc − cot = 1 because 

6 6 

2 2 

1+ cot t = csc t. 

501 



39. 

40. 

41. 

42. 

43. 

44. 

45. 

9π ⎛π ⎞ π 2 

cos = cos 2π 

cos 

4 

⎜ + 

4 

⎟= = 

⎝ ⎠ 4 2 

9π ⎛π ⎞ π 

csc = csc 2π 

csc 2 

4 

⎜ + = = 

4 

⎟ 

⎝ ⎠ 4 

⎛ 9π ⎞ ⎛ 9π ⎞ 7π 

2 

sin ⎜− sin 4π 

sin 

4 

⎟= ⎜− + = = − 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 4 2 

⎛ 9π ⎞ ⎛ 9π ⎞ 7π 

sec⎜− sec 4π 

sec 2 

4 

⎟= ⎜− + = = 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 4 

5π ⎛π ⎞ π 

tan = tan π tan 1 

4 

⎜ + = = 

4 

⎟ 

⎝ ⎠ 4 

5π ⎛π ⎞ π 

cot = cot π cot 1 

4 

⎜ + = = 

4 

⎟ 

⎝ ⎠ 4 

⎛ 5π ⎞ ⎛3π ⎞ 3π 

cot ⎜− cot 2π 

cot 1 

4 

⎟= ⎜ − = = − 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 4 

⎛ π ⎞ ⎛ π ⎞ 

51. cos ⎜− − 1000π 

cos 2π 

4 

⎟ = ⎜− + 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 

7π 

= cos 4 

2 

= 

2 

⎛ π ⎞ ⎛ π ⎞ 

52. cos ⎜− − 2000π 

cos 2π 

4 

⎟ = ⎜− + 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 

7π 

= cos 4 

2 

= 

2 

53. a. 

b. 

3π 2 

sin = 4 2 

11π ⎛3π ⎞ 3π 

2 

sin = sin 2π 

sin 

4 

⎜ + 

4 

⎟ = = 

⎝ ⎠ 4 2 

46. 

⎛ 9π ⎞ ⎛ 9π ⎞ 3π 

tan ⎜− tan 3π 

tan 1 

4 

⎟ = ⎜− + = = − 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 4 

54. a. 

3π 2 

cos =− 4 2 

⎛π 

⎞ π 

47. − tan ⎜ + 15π 

=− tan = −1 

4 

⎟ 

⎝ ⎠ 4 

b. 

11π ⎛3π ⎞ 3π 

2 

cos = cos 2π 

cos 

4 

⎜ + 

4 

⎟= =− 

⎝ ⎠ 4 2 

⎛π 

⎞ π 

48. − cot ⎜ + 17π 

=− cot = −1 

4 

⎟ 

⎝ ⎠ 4 

⎛ π ⎞ ⎛ π ⎞ 

49. sin ⎜− − 1000π 

sin 2π 

4 

⎟= ⎜− + 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 

7π 

= sin 4 

2 

=− 

2 

⎛ π ⎞ ⎛ π ⎞ 

50. sin ⎜− − 2000π 

sin 2π 

4 

⎟= ⎜− + 

4 

⎟ 

⎝ ⎠ ⎝ ⎠ 

7π 

= sin 4 

2 

=− 

2 

π 

55. a. cos = 0 

2 

b. 

π 

56. a. sin = 1 

2 

b. 

9π 

⎛π 

⎞ 

cos = cos 4 

2 

⎜ + π 

2 

⎟ 

⎝ ⎠ 

⎡π 

⎤ 

= cos ⎢ + 2(2 π ) 

2 

⎥ 

⎣ ⎦ 

π 

= cos 2 

= 0 

9π ⎛π ⎞ π 

sin = sin 4π 

sin 1 

2 

⎜ + = = 

2 

⎟ 

⎝ ⎠ 2 

502 



0 

π 

57. a. tanπ = = 0 

69. cot ≈ 3.7321 

−1 

12 

b. tan17π = tan( π + 16 π) 

π 

70. cot ≈ 5.6713 

= tan[ π + 8(2 π)] 

18 

= tanπ 

71. sin( −t) − sin t = −sin t− sin t = − 2sin t =− 2a 

= 0 

72. tan( −t) − tan t = −tan t− tan t = − 2 tan t = − 2c 

π 0 

58. a. cot = = 0 

2 1 

73. 4cos( −t) − cos t = 4cost− cost = 3cost = 3b 

15π ⎛π ⎞ π 

b. cot = cot + 7π 

= cot = 0 

2 

⎜ 

2 

⎟ 

74. 3cos( −t) − cost = 3cost− cost = 2cost = 2b 

⎝ ⎠ 2 

75. sin( t+ 2 π) − cos( t+ 4 π) + tan( t+ 

π) 

7π 2 

59. a. sin =− 

= sin( t) − cos( t) + tan( t) 

4 2 

= a− b+ 

c 

47π 

⎛7π 

⎞ 

b. sin = sin + 10π 

76. sin( t+ 2 π) + cos( t+ 4 π) − tan( t+ 

π) 

4 

⎜ 

4 

⎟ 

⎝ ⎠ 

= sin( t) + cos( t) −tan( t) 

⎡7π 

⎤ 

= sin ⎢ + 5( 2π 

) 

= a+ b−c 

4 

⎥ 

⎣ ⎦ 

7π 

77. sin( −t−2 π) −cos( −t−4 π) −tan( −t−π) 

= sin 4 

= − sin( t+ 2 π) − cos( t+ 4 π) + tan( t+ 

π) 

2 

=−sin( t) − cos( t) + tan( t) 

=− 

2 

=−a− b+ 

c 

7π 2 

78. sin( −t− 2 π) + cos( −t−4 π) −tan( −t−π) 

60. a. cos = 4 2 

= − sin( t+ 2 π) + cos( t+ 4 π) + tan( t+ 

π) 

=− sin( t) + cos( t) + tan( t) 

47π ⎛7π ⎞ 7π 

2 

b. cos = cos + 10π 

= cos = 

=− a+ b+ 

c 

4 

⎜ 

4 

⎟ 

⎝ ⎠ 4 2 

79. cost+ cos( t+ 1000 π) −tan t− tan( t+ 

999 π) 

61. sin 0.8 ≈ 0.7174 

− sin t+ 4sin( t−1000 π ) 

= cost+ cost−tan t−tan t− sin t+ 

4sin t 

≈ 62. cos 0.6 0.8253 

= 2cost− 2tant+ 

3sint 

63. tan 3.4 0.2643 

= 3a+ 2b− 

≈ 2c 

64. tan ≈ 3.7 0.6247 

80. − cost+ 7cos( t+ 1000 π) + tan t+ tan( t+ 

999 π) 

65. csc 1 1.1884 

+ sin ≈ t+ sin( t−1000 π ) 

=− cost+ 7 cost+ tan t+ tan t+ sin t+ 

sin t 

66. sec ≈ 1 1.8508 

= 6cost+ 2tant+ 

2sint 

π = 2a+ 6b+ 

2c 

67. cos ≈ 0.9511 

10 

3π 68. sin ≈ 0.8090 

10 

503 



81. a. 

b. 

c. 

82. a. 

b. 

c. 

⎡ 2π 

⎤ 

H = 12 + 8.3sin ⎢ (80 −80) 

365 

⎥ 

⎣ 

⎦ 

= 12 + 8.3sin 0 = 12 + 8.3(0) 

= 12 

There are 12 hours of daylight in Fairbanks on 

March 21. 

⎡ 2π 

⎤ 

H = 12 + 8.3sin ⎢ (172 −80) 

365 

⎥ 

⎣ 

⎦ 

≈ 12 + 8.3sin1.5837 

≈ 20.3 

There are about 20.3 hours of daylight in 

Fairbanks on June 21. 

⎡ 2π 

⎤ 

H = 12 + 8.3sin ⎢ (355 −80) 

365 

⎥ 

⎣ 

⎦ 

≈ 12 + 8.3sin 4.7339 

≈ 3.7 

There are about 3.7 hours of daylight in 

Fairbanks on December 21. 

⎡ 2π 

⎤ 

H = 12 + 24sin ⎢ (80 −80) 

365 

⎥ 

⎣ 

⎦ 

12 + 24sin 0 = 12 + 24(0) = 12 

There are 12 hours of daylight in San Diego on 

March 21. 

⎡ 2π 

⎤ 

H = 12 + 24sin ⎢ (172 −80) 

365 

⎥ 

⎣ 

⎦ 

≈ 12 + 24sin1.5837 

≈ 14.3998 

There are about 14.4 hours of daylight in San 

Diego on June 21. 

⎡ 2π 

⎤ 

H = 12 + 24sin ⎢ (355 −80) 

365 

⎥ 

⎣ 

⎦ 

≈ 12 + 24sin 4.7339 

≈ 9.6 

There are about 9.6 hours of daylight in San 

Diego on December 21. 

83. a. For t = 7, 

π π 

E = sin ⋅ 7 = sin = 1 

14 2 

For t = 14, 

π 

E = sin ⋅ 14 = sinπ 

= 0 

14 

For t = 21, 

π 3π 

E = sin ⋅ 21 = sin = −1 

14 2 

For t = 28, 

π 

E = sin ⋅ 28 = sin 2π 

= sin 0 = 0 

14 

For t = 35, 

π 5π π 

E = sin ⋅ 35 = sin = sin = 1 

14 2 2 

Observations may vary. 

b. Because E(35) = E(7) = 1, the period is 

35 – 7 = 28 or 28 days. 

84. a. At 6 A.M., t = 0. 

π 

H = 10 + 4sin ⋅0 

6 

= 10+ 4sin0= 10+ 4⋅ 0= 

10 

The height is 10 feet. 

At 9 A.M., t = 3. 

π 

H = 10 + 4sin ⋅3 

6 

π 

= 10 + 4sin = 10 + 4(1) = 14 

2 


At noon, t = 6. 

π 

H = 10 + 4sin ⋅6 

6 

= 10 + 4sinπ 

= 10 + 4⋅ 0 = 10 


At 6 P.M., t = 12. 

π 

H = 10 + 4sin ⋅12 

6 

= 10 + 4sin 2π 

= 10 + 4⋅ 0 = 10 


At midnight, t = 18. 

504 



π 

H = 10 + 4sin ⋅18 

6 

= 10 + 4sin 3π 

= 10 + 4sinπ 

= 10 + 4⋅ 0 = 10 


At 3 A.M., t = 21. 

π 

H = 10 + 4sin ⋅21 

6 

7π 

3π 

= 10 + 4sin = 10 + 4sin 

2 2 

= 10 + 4( − 1) = 6 


b. The sine function has a minimum at 3 π . Thus, 

2 

π 3π 

we find a low tide at t = or 

6 2 

t = 9. This value of t corresponds to 3 P.M. For t 

= 9, 

π 

h = 10 + 4sin ⋅9 

6 

3π 

= 10 + 4sin = 10 + 4( − 1) = 6 

2 

The height is 6 feet. From part a, the height at 3 

A.M. is also 6 feet. Thus, low tide is at 

3 A.M. and 3 P.M. 

The sine function has a maximum at 2 

π . Thus, 

π π 

we find a high tide at t = or t = 3. This 

6 2 

value of t corresponds to 9 a.m. From part a, the 

height at 9 A.M. is 14 feet. Because the sine has 

a period of we also find a maximum at 5 π 

2π . 

2 

π 5π 

We find another high tide at t = or t = 15. 

6 2 

This value of t corresponds to 9 P.M. Thus, high 

tide is at 9 A.M. and 9 P.M. 

c. The period of 2π the sine function is or on the 

interval [0, 2 π ] . The cycle of the sine function 

π 5π 

π 

starts at t = or t = 0, and ends at 2 

6 2 

6 t = π 

or t = 12. Thus, the period is 12 hours, which 

means high and low tides occur every 12 hours. 



98. does not make sense; Explanations will vary. 

Sample explanation: sin t cannot be less than − 1. 

10 

Note that − ≈ − 1.58 < − 1. 

2 


Sample explanation: Cosine is not an odd function. 


101. t is in the third quadrant therefore sin t < 0, 

tan t > 0, and cot t > 0. 

Thus, only choice (c) is true. 

102. 

1 

f( x) = sin x and f( a) 

= 

4 

f( a) + f( a+ 2 π) + f( a+ 4 π) + f( a+ 

6 π) 

⎛1 

⎞ 

= 4 f( a) = 4⎜ 

1 because sin x has a 

4 

⎟= 

⎝ ⎠ 

period of 2 π. 

1 

103. f( x) = sin x and f( a) 

= 

4 

f( a) + 2 f( − a) = f( a) −2 f( a) 

1 ⎛1⎞ 

= −2 

4 ⎜ 

4 ⎟ 

⎝ ⎠ 

1 

=− 

4 

f(–a) = –f(a) because sin (–x) = –sin x. 

Sine is an odd function. 

104. The height is given by 

h = 45 + 40 sin(t – 90°) 

h (765 ° ) = 45 + 40sin(765°− 90 ° ) 

≈ 16.7 

You are about 16.7 feet above the ground. 

105. First find the hypotenuse. 

2 2 2 

c = a + b 

2 2 2 

c = 5 + 12 

2 

c = 25 + 144 

2 

c = 169 

c = 13 

Next write the ratio. 

a 5 

c = 13 

505 



106. First find the hypotenuse. 

2 2 2 

c = a + b 

2 2 2 

c = 1 + 1 

2 

c = 1+ 

1 

2 

c = 2 

c = 2 

Next write the ratio and simplify. 

a 1 

c = 2 

1 2 

= ⋅ 

2 2 

2 

= 

2 

2 2 2 2 

+ = + 

2 2 

⎛a⎞ ⎛b⎞ 

a b 

107. ⎜ 

c 

⎟ ⎜ 

c 

⎟ 

⎝ ⎠ ⎝ ⎠ c c 

2 2 

a + b 

= 

2 

c 

2 2 2 

Since c = a + b , continue simplifying by 

2 2 

substituting c for a + b 

2 . 

2 2 2 2 

⎛a⎞ ⎛b⎞ 

a b 

⎜ 

c 

⎟ + ⎜ 

c 

⎟ = + 

2 2 

⎝ ⎠ ⎝ ⎠ c c 

2 2 

a + b 

= 

2 

c 

2 

c 

2 2 

a + b 

= 

2 

c 

2 

c 

= 

2 

c 

= 1 

Section 4.3 


2 2 2 

1. Use the Pythagorean Theorem, c = a + b , to find 

c. 

a = 3, b = 4 

2 2 2 2 2 

c = a + b = 3 + 4 = 9+ 16= 

25 

c = 25 = 5 

Referring to these lengths as opposite, adjacent, and 

hypotenuse, we have 

opposite 3 

sinθ 

= = 

hypotenuse 5 

adjacent 4 

cosθ 

= = 

hypotenuse 5 

opposite 3 

tanθ 

= = 

adjacent 4 

hypotenuse 5 

cscθ 

= = 

opposite 3 

hypotenuse 5 

secθ 

= = 

adjacent 4 

adjacent 4 

cotθ 

= = 

opposite 3 

2. Use the Pythagorean Theorem, 

2 2 2 

c = a + b , to find 

b. 

2 2 2 

a + b = c 

2 2 2 

1 + b = 5 

2 

1+ b = 25 

2 

b = 24 

b = 24 = 2 6 

Note that side a is opposite θ and side b is adjacent 

to θ . 

opposite 1 

sinθ 

= = 

hypotenuse 5 

adjacent 2 6 

cosθ 

= = 

hypotenuse 5 

opposite 1 6 

tanθ 

= = = 

adjacent 2 6 12 

hypotenuse 5 

cscθ 

= = = 5 

opposite 1 

hypotenuse 5 5 6 

secθ 

= = = 

adjacent 2 6 12 

adjacent 2 6 

cotθ 

= = = 2 6 

opposite 1 

506 



3. Apply the definitions of these three trigonometric functions. 

length of hypotenuse 

csc 45°= 

length of side opposite 45° 

2 

= = 2 

1 


sec45°= 

length of side adjacent to 45° 

2 

= = 2 

1 


cot 45°= 


1 

= = 1 

1 

4. 


tan 60°= 


3 

= = 3 

1 


tan 30°= 


1 1 3 3 

= = ⋅ = 

3 3 3 3 

5. a. 

sin 46 

o 

= cos(90 

o 

− 46 

o 

) = cos 44 

o 

π ⎛π π ⎞ 

b. cot = tan ⎜ − ⎟ 

12 ⎝ 2 12 ⎠ 

⎛6π 

π ⎞ 

= tan ⎜ − ⎟ 

⎝ 12 12 ⎠ 

5π 

= tan 12 

6. Because we have a known angle, an unknown opposite side, and a known adjacent side, we select the 

tangent function. 

tan 24 

0 a 

= 

750 

a = 750 tan 24 

0 

a ≈750(0.4452) ≈334 

The distance across the lake is approximately 334 yards. 

507 



7. 

side opposite 14 

tanθ = = 

side adjacent 10 

Use a calculator in degree mode to find θ . 

Many Scientific Calculators 

Many Graphing Calculators 

−1 

TAN ( 14 10 ) ENTER 

÷ TAN ( 14 ÷ 10 ) ENTER 

The display should show approximately 54. Thus, the angle of elevation of the sun is approximately 54°. 


1. 

2. 

c 

= 9 + 12 = 225 

2 2 2 

c = 225 = 15 

opposite 9 3 

sinθ 

= = = 

hypotenuse 15 5 

adjacent 12 4 

cosθ 

= = = 


opposite 9 3 

tanθ 

= = = 

adjacent 12 4 


cscθ 

= = = 

opposite 9 3 


secθ 

= = = 

adjacent 12 4 

adjacent 12 4 

cotθ 

= = = 

opposite 9 3 

c 

= 6 + 8 = 100 

2 2 2 

c = 100 = 10 

opposite 6 3 

sinθ 

= = = 


adjacent 8 4 

cosθ 

= = = 


opposite 6 3 

tanθ 

= = = 

adjacent 8 4 


cscθ 

= = = 

opposite 6 3 


secθ 

= = = 

adjacent 8 4 

adjacent 8 4 

cotθ 

= = = 

opposite 6 3 

508 



3. 

2 2 2 

a + 21 = 29 

5. 

10 + b = 26 

2 2 2 

a 

2 

= 841− 441 = 400 

b 

2 

= 676 − 100 = 576 

a = 400 = 20 

opposite 20 

sinθ 

= = 

hypotenuse 29 

adjacent 21 

cosθ 

= = 

hypotenuse 29 

opposite 20 

tanθ 

= = 

adjacent 21 

hypotenuse 29 

cscθ 

= = 

opposite 20 

hypotenuse 29 

secθ 

= = 

adjacent 21 

adjacent 21 

cotθ 

= = 

opposite 20 

b = 576 = 24 

opposite 10 5 

sinθ 

= = = 


adjacent 24 12 

cosθ 

= = = 


opposite 10 5 

tanθ 

= = = 



cscθ 

= = = 

opposite 10 5 


secθ 

= = = 



cotθ 

= = = 

opposite 10 5 

4. 

2 2 2 

a + 15 = 17 

6. 

2 2 2 

a + 40 = 41 

a 

2 

= 289 − 225 = 64 

a 

2 

= 1681− 1600 = 81 

a = 64 = 8 

opposite 8 

sinθ 

= = 

hypotenuse 17 

adjacent 15 

cosθ 

= = 

hypotenuse 17 

opposite 8 

tanθ 

= = 

adjacent 15 

hypotenuse 17 

cscθ 

= = 

opposite 8 

hypotenuse 17 

secθ 

= = 

adjacent 15 

adjacent 15 

cotθ 

= = 

opposite 8 

a = 81 = 9 

opposite 9 

sinθ 

= = 

hypotenuse 41 

adjacent 40 

cosθ 

= = 

hypotenuse 41 

opposite 9 

tanθ 

= = 

adjacent 40 

hypotenuse 41 

cscθ 

= = 

opposite 9 

hypotenuse 41 

secθ 

= = 

adjacent 40 

adjacent 40 

cotθ 

= = 

opposite 9 

509 



7. 

8. 

a 

+ 21 = 35 

2 2 2 

a 

2 

= 1225 − 441 = 784 

a = 784 = 28 

opposite 28 4 

sinθ 

= = = 


adjacent 21 3 

cosθ 

= = = 


opposite 28 4 

tanθ 

= = = 

adjacent 21 3 


cscθ 

= = = 

opposite 28 4 


secθ 

= = = 

adjacent 21 3 

adjacent 21 3 

cotθ 

= = = 

opposite 28 4 

24 + b = 25 

2 2 2 

b 

2 

= 625 − 576 = 49 

b = 49 = 7 

opposite 24 

sinθ 

= = 

hypotenuse 25 

adjacent 7 

cosθ 

= = 

hypotenuse 25 

opposite 24 

tanθ 

= = 

adjacent 7 

hypotenuse 25 

cscθ 

= = 

opposite 24 

hypotenuse 25 

secθ 

= = 

adjacent 7 

adjacent 7 

cotθ 

= = 

opposite 24 

11. 

12. 


sec 45°= 


2 

= = 2 

1 


csc 45°= 


2 

= = 2 

1 

π 

13. tan = tan 60 ° 

3 


= 


3 

= = 3 

1 

14. 

π length of side adjacent to 60° 

cot = cot 60°= 

3 length of side opposite 60° 

1 1 3 3 

= = ⋅ = 

3 3 3 3 

π π 

15. sin − cos = sin 45°− cos 45° 

4 4 

1 1 

= − = 0 

2 2 

π π 

16. tan + csc = tan 45°+ csc30° 

4 6 

1 2 

= + = 1 + 2 = 3 

1 1 

17. π π π ⎛ 3 ⎞⎛ 2 ⎞ 

sin cos − tan = −1 

3 4 4 ⎜ 

⎟⎜ 2 ⎟⎜ 2 ⎟ 

⎝ ⎠⎝ ⎠ 

9. 


cos30°= 


3 

= 

2 

6 

= − 1 

4 

6 − 4 

= 

4 

10. 


tan 30°= 


18. 

π π π 3 3− 

3 

cos sec − cot = 1− = 

3 3 3 3 3 

1 1 3 3 

= = ⋅ = 

3 3 3 3 

510 



π π π ⎛ 2 ⎞⎛ 3 ⎞ 

+ = +⎜ 

3 4 6 ⎜ 

⎟⎜ 

2 ⎟⎜ 3 ⎟ 

⎝ ⎠⎝ ⎠ 

6 

= 2 3+ 

6 

12 3 + 6 

= 

6 

19. 2 tan cos tan 2( 3 ) 

20. 

π π π ⎛ 3 ⎞⎛2 3 ⎞ 

6tan + sin sec = 6(1) +⎜ 

4 3 6 ⎜ 

⎟⎜ 

2 ⎟⎜ 3 ⎟ 

⎝ ⎠⎝ ⎠ 

6 

= 6 + 6 

= 7 

21. sin 7°= cos(90°− 7 ° ) = cos83° 

22. sin19°= cos( 90°− 19° ) = cos71° 

23. csc 25° = sec(90° – 25 o ) = sec 65° 

24. csc35°= sec(90°− 35 ° ) = sec55° 

π ⎛π π ⎞ 

25. tan = cot ⎜ − ⎟ 

9 ⎝ 2 9 ⎠ 

⎛9π 

2π 

⎞ 

= cot ⎜ − ⎟ 

⎝18 18 ⎠ 

7π 

= cot 18 

26. 

27. 

7 2 5 

tan π ⎛ 

cot π π ⎞ ⎛ 

cot π π ⎞ 

= ⎜ − ⎟= ⎜ − ⎟ = cot 

π 

7 ⎝ 2 7 ⎠ ⎝14 14 ⎠ 14 

2π ⎛π 2π 

⎞ 

cos = sin ⎜ − ⎟ 

5 ⎝ 2 5 ⎠ 

⎛5π 

4π 

⎞ 

= sin ⎜ − ⎟ 

⎝ 10 10 ⎠ 

π 

= sin 10 

3 3 4 3 

28. cos π ⎛ 

sin π π ⎞ ⎛ 

sin π π ⎞ 

= ⎜ − ⎟= ⎜ − ⎟= 

sin 

π 

8 ⎝ 2 8 ⎠ ⎝ 8 8 ⎠ 8 

a 

29. tan 37°= 

250 

a = 250 tan 37° 

a ≈ 250(0.7536) ≈188 cm 

a 

30. tan 61°= 

10 

a = 10 tan 61° 

a ≈10(1.8040) ≈18 cm 

b 

31. cos34°= 

220 

b = 220cos34° 

b ≈220(0.8290) ≈182 in. 

a 

32. sin 34°= 

13 

a = 13sin 34° 

a ≈13(0.5592) ≈7 m 

16 

33. sin 23°= 

c 

16 16 

c = ≈ ≈ 41 m 

sin 23° 

0.3907 

23 

34. tan 44°= 

b 

23 23 

b = ≈ ≈ 24 yd 

tan 44° 

0.9657 

511 



35. Scientific Calculator Graphing Calculator 

.2974 

1 

SIN − − 

SIN 

1 

.2974 ENTER 

Display 

(rounded to the nearest degree) 

17 

If sinθ 

= 0.2974, then θ ≈ 17 ° . 


Display 


.877 COS -1 COS -1 .877 ENTER 29 

If cosθ 

= 0.877, then θ ≈ 29°. 


1 

4.6252 TAN − − 

TAN 

1 

If tanθ 

= 4.6252, then θ ≈ 78 ° . 

4.6252 ENTER 

Display 


78 


Display 


26.0307 TAN -1 TAN -1 26.0307 ENTER 88 

If tanθ 

= 26.0307, then θ ≈ 88 ° . 


1 

.4112 COS − − 

COS 

1 

If cosθ 

= 0.4112, then θ ≈ 1.147 radians. 

.4112 ENTER 

Display 

(rounded to three places) 

1.147 


Display 


.9499 SIN -1 SIN -1 .9499 ENTER 1.253 

If sinθ 

= 0.9499, then θ = 1.253 radians. 


1 

.4169 TAN − − 

TAN 

1 

If tanθ 

= 0.4169, then θ ≈ 0.395 radians. 

.4169 ENTER 

Display 


.395 


.5117 

If tanθ 

= 0.5117, then θ = 0.473 

Display 


1 

TAN − TAN − 1 .5117 ENTER .473 

512 



43. 

π 

tan 

3 1 3 1 

− = − 

2 π 

sec 

2 

6 

1 

π 

cos 6 

3 1 

= − 

2 

1 

3 

2 

3 3 

= − 

2 2 

= 0 

44. 

45. 

46. 

1 2 1 2 

− = − 

π π 

cot csc 

4 6 

1 2 

= − 

1 1 

π π 

tan sin 

4 6 

1 1 

1 1 2 

1 2 

= − 

1 2 

= 1−1 

= 0 

2 2 

1+ sin 40°+ sin 50° 

2 2 

= 1+ sin (90°− 50 ° ) + sin 50° 

2 2 

= 1+ cos 50°+ sin 50° 

= 1+ 

1 

= 2 

2 2 

1− tan 10°+ csc 80° 

2 2 

= 1− cot 80°+ csc 80° 

2 2 

= 1+ csc 80°− cot 80° 

= 1+ 

1 

= 2 

47. csc37° sec53°− tan 53° cot 37° 

= sec53° sec53°− tan 53° tan 53° 

2 2 

= sec 53°− tan 53° 

= 1 

48. cos12° sin 78°+ cos 78° sin12° 

= sin 78° sin 78°+ cos 78° cos 78° 

2 2 

= sin 78°+ cos 78° 

= 1 

513 



49. f ( θ) = 2cosθ − cos2θ 

⎛π ⎞ π ⎛ π ⎞ 

f ⎜ 2cos cos 2 

6 

⎟= − 

6 

⎜ ⋅ 

6 

⎟ 

⎝ ⎠ ⎝ ⎠ 

⎛ 3 ⎞ ⎛π 

⎞ 

= 2 −cos 

⎜ 

2 ⎟ ⎜ 

3 

⎟ 

⎝ ⎠ ⎝ ⎠ 

2 3 1 

= − 

2 2 

2 3 −1 

= 

2 

θ 

50. f ( θ) = 2sinθ 

− sin 2 

π 

⎛π 

⎞ π 

3 

f ⎜ 2sin sin 

3 

⎟ = − 

⎝ ⎠ 3 2 

⎛ 3 ⎞ ⎛π 

⎞ 

= 2 −sin 

⎜ 

2 ⎟ ⎜ 

6 

⎟ 

⎝ ⎠ ⎝ ⎠ 

2 3 1 

= − 

2 2 

2 3 −1 

= 

2 

51. 

52. 

⎛ 

1 

tan 

π ⎞ 

⎜ − θ = cot θ = 

2 

⎟ 

⎝ ⎠ 4 

⎛π 

⎞ 1 1 

csc⎜ 

− θ secθ 

3 

2 

⎟= = = = 

⎝ ⎠ cosθ 

1 

3 

a 

53. tan 40°= 

630 

a = 630 tan 40° 

a ≈630(0.8391) ≈529 

The distance across the lake is approximately 529 yards. 

h 

54. tan 40°= 

35 

h = 35tan 40° 

h ≈35(0.8391) ≈ 29 

The tree’s height is approximately 29 feet. 

55. 

125 

tanθ = 

172 



125 ÷ 172 = TAN −1 


−1 

TAN ( 125 ÷ 172 ) ENTER 

The display should show approximately 36. Thus, the angle of elevation of the sun is approximately 36°. 

514 



56. 

57. 

555 

tan 1320 




555 ÷ 1320 = TAN -1 TAN -1 ( 555 ÷ 1320 ) ENTER 

The display should show approximately 23. Thus, the angle of elevation is approximately 23°. 

500 

sin10°= 

c 

500 500 

c = ≈ ≈ 2880 

sin10° 

0.1736 

The plane has flown approximately 2880 feet. 

a 

58. sin 5°= 

5000 

a = 5000sin 5°≈ 5000(0.0872) = 436 

The driver’s increase in altitude was approximately 436 feet. 

59. 

60 

cosθ = 

75 



60 ÷ 75 = COS −1 


−1 

COS ( 60 ÷ 75 ) ENTER 

The display should show approximately 37. Thus, the angle between the wire and the pole is approximately 37°. 

55 

60. cosθ = 

80 





55 ÷ 80 = COS -1 COS -1 ( 55 ÷ 80 ) ENTER 

The display should show approximately 47. Thus, the angle between the wire and the pole is approximately 47°. 

68. 

θ 0.4 0.3 0.2 0.1 0.01 0.001 0.0001 0.00001 

sinθ 0.3894 0.2955 0.1987 0.0998 0.0099998 

9.999998× 

10 −4 

9.99999998× 

10 −5 

1× 

10 −5 

sinθ 

θ 

0.9736 0.9851 0.9933 0.9983 0.99998 0.9999998 0.999999998 1 

sinθ 

approaches 1 as θ approaches 0. 

θ 

515 



69. 

θ 0.4 0.3 0.2 0.1 0.01 0.001 0.0001 0.00001 

cosθ 0.92106 0.95534 0.98007 0.99500 0.99995 0.9999995 0.999999995 1 

cosθ 

− 1 –0.19735 –0.148878 –0.099667 –0.04996 –0.005 –0.0005 –0.00005 0 

θ 

cosθ 

−1 

approaches 0 as θ approaches 0. 

θ 

70. does not make sense; Explanations will vary. Sample explanation: An increase in the size of a triangle does not affect 

the ratios of the sides. 

71. does not make sense; Explanations will vary. Sample explanation: This value is irrational. Irrational numbers are 

rounded on calculators. 

72. does not make sense; Explanations will vary. Sample explanation: The sine and cosine are cofunctions of each other. 

The sine and cosine are not reciprocals of each other. 


74. false; Changes to make the statement true will vary. A sample change is: 

tan 45 ° ⎛45 

tan 

° ⎞ 

≠ ⎜ ⎟ 

tan15° ⎝15° 

⎠ 

75. true 

76. false; Changes to make the statement true will vary. A sample change is: 

1 1 2 

sin 45°+ cos 45°= + = ≠ 1 

2 2 2 

77. true 

78. In a right triangle, the hypotenuse is greater than either other side. Therefore both 

less than 1 for an acute angle in a right triangle. 

opposite 

hypotenuse 

and 

adjacent 

hypotenuse 

must be 

79. Use a calculator in degree mode to generate the following table. Then use the table to describe what happens to the 

tangent of an acute angle as the angle gets close to 90°. 

θ 60 70 80 89 89.9 89.99 89.999 89.9999 

tanθ 1.7321 2.7475 5.6713 57 573 5730 57,296 572,958 

As θ approaches 90°, tanθ increases without bound. At 90°, tanθ is undefined. 

516 



80. a. Let a = distance of the ship from the lighthouse. 

250 

tan 35°= 

a 

250 250 

a = ≈ ≈357 

tan 35° 

0.7002 

The ship is approximately 357 feet from the 

lighthouse. 

81. a. 

82. a. 

b. Let b = the plane’s height above the lighthouse. 

b 

tan 22°= 

357 

b = 357 tan 22°≈357(0.4040) ≈144 

144 + 250 = 394 

The plane is approximately 394 feet above the 

water. 

y 

r 

2 2 

b. First find r: r = x + y 

2 2 

r = ( − 3) + 4 

r = 5 

y 4 

= , which is positive. 

r 5 

x 

r 

2 2 

b. First find r: r = x + y 

r = 

2 2 

( − 3) + 5 

r = 34 

x −3 −3 34 −3 34 

= = ⋅ = , which is 

r 34 34 34 34 

negative. 

 

83. a. θ′ = 360 − 345 = 15 

b. 

5π 6π 5π π 

θ′ = π − = − = 

6 6 6 6 

Section 4.4 


1. 

2 2 

r = x + y 

2 2 

r = 1 + ( − 3) = 1+ 9 = 10 

Now that we know x, y, and r, we can find the six 

trigonometric functions of θ . 

y −3 3 10 

sinθ 

= = =− 

r 10 10 

x 1 10 

cosθ 

= = = 

r 10 10 

y −3 

tanθ 

= = = −3 

x 1 

r 10 10 

cscθ 

= = = − 

y −3 3 

r 10 

secθ 

= = = 10 

x 1 

x 1 1 

cotθ 

= = = − 

y −3 3 

2. a. θ = 0°= 0 radians 

The terminal side of the angle is on the positive 

x-axis. Select the point 

P = (1,0): x = 1, y = 0, r = 1 

Apply the definitions of the cosine and cosecant 

functions. 

x 1 

cos 0°= cos 0 = = = 1 

r 1 

r 1 

csc0°= csc0 = = , undefined 

y 0 

π 

b. θ = 90°= radians 

2 

The terminal side of the angle is on the positive 

y-axis. Select the point 

P = (0,1): x = 0, y = 1, r = 1 


functions. 

π x 0 

cos90°= cos = = = 0 

2 r 1 

π r 1 

csc90°= csc = = = 1 

2 y 1 

517 



c. θ = 180°= π radians 

The terminal side of the angle is on the negative 

x-axis. Select the point 

P = (–1,0): x =− 1, y = 0, r = 1 


functions. 

x −1 

cos180°= cosπ 

= = =−1 

r 1 

r 1 

csc180°= csc π = = , undefined 

y 0 

d. 

3π 

θ = 270°= radians 

2 

The terminal side of the angle is on the negative 

y-axis. Select the point 

P = (0,–1): x = 0, y = − 1, r = 1 


functions. 

3π 

x 0 

cos 270°= cos = = = 0 

2 r 1 

3π 

r 1 

csc 270°= csc = = =−1 

2 y −1 

3. Because sinθ 

< 0, θ cannot lie in quadrant I; all the 

functions are positive in quadrant I. Furthermore, θ 

cannot lie in quadrant II; sinθ is positive in quadrant 

II. Thus, with sinθ 

< 0, θ lies in quadrant III or 

quadrant IV. We are also given that cosθ < 0 . 

Because quadrant III is the only quadrant in which 

cosine is negative and the sine is negative, we 

conclude that θ lies in quadrant III. 

4. Because the tangent is negative and the cosine is 

negative, θ lies in quadrant II. In quadrant II, x is 

negative and y is positive. Thus, 

1 y 1 

tanθ =− = = 

3 x − 3 

x =− 3, y = 1 

Furthermore, 

2 2 2 2 

r = x + y = ( − 3) + 1 = 9+ 1= 

10 

Now that we know x, y, and r, we can find 

sinθ and secθ . 

y 1 1 10 10 

sinθ 

= = = ⋅ = 

r 10 10 10 10 

r 10 10 

secθ 

= = = − 

x −3 3 

5. a. Because 210° lies between 180° and 270°, it is 

in quadrant III. The reference angle is 

θ ′ = 210°− 180°= 30°. 

b. Because 7 π 3π 

6π 

lies between = and 

4 

2 4 

8π 

2π = , it is in quadrant IV. The reference 

4 

7π 8π 7π π 

angle is θ′ = 2π 

− = − = . 

4 4 4 4 

c. Because –240° lies between –180° and –270°, it 

is in quadrant II. The reference angle is 

θ = 240 − 180 = 60°. 

d. Because 3.6 lies between π ≈ 3.14 and 

3π ≈ 4.71 , it is in quadrant III. The reference 

2 

angle is θ′ = 3.6 −π 

≈ 0.46 . 

6. a. 665°− 360°= 305° 

This angle is in quadrant IV, thus the reference 

angle is θ′ = 360°− 305°= 55°. 

b. 

c. 

15 π 15 8 7 

− 2π 

= π − π = 

π 

4 4 4 4 

This angle is in quadrant IV, thus the reference 

7π 8π 7π π 


− = − = . 

4 4 4 4 

11 11 12 

− π + 2⋅ 2π 

= − π + π = 

π 

3 3 3 3 

This angle is in quadrant I, thus the reference 

π 

angle is θ′ = . 

3 

7. a. 300° lies in quadrant IV. The reference angle is 

θ ′ = 360°− 300°= 60°. 

b. 

3 

sin 60°= 

2 

Because the sine is negative in quadrant IV, 

3 

sin 300°=− sin 60°=− . 

2 

5π lies in quadrant III. The reference angle is 

4 

5 5 4 

θ′ = π − π = π − π = π . 

4 4 4 4 

π 

tan = 1 

4 

Because the tangent is positive in quadrant III, 

5π 

π 

tan =+ tan = 1 . 

4 4 

518 



c. 

8. a. 

b. 

π 

− lies in quadrant IV. The reference angle is 

6 

π 

θ ′ = . 

6 

π 2 3 

sec = 6 3 

Because the secant is positive in quadrant IV, 

⎛ π ⎞ π 2 3 

sec⎜− ⎟= + sec = . 

⎝ 6 ⎠ 6 3 

17 π 17 12 5 

− 2π 

= π − π = π lies in quadrant 

6 6 6 6 

5π 

π 

II. The reference angle is θ′ = π − = . 

6 6 

The function value for the reference angle is 

π 3 

cos = . 6 2 

Because the cosine is negative in quadrant II, 

17π 5π π 3 

cos = cos =− cos = − . 

6 6 6 2 

−22 π −22 24 2 

+ 8π 

= π + π = π lies in 

3 3 3 3 

quadrant II. The reference angle is 

2π 

π 

θ′ = π − = . 

3 3 

The function value for the reference angle is 

π 3 

sin = . 3 2 

Because the sine is positive in quadrant II, 

−22π 2π π 3 

sin = sin = sin = . 

3 3 3 2 


1. We need values for x, y, and r. Because 

P = (–4, 3) is a point on the terminal side of 

θ , x =− 4 and y = 3. Furthermore, 

r x y 

that we know x, y, and r, we can find the six 


y 3 

sinθ 

= = 

r 5 

x −4 4 

cosθ 

= = = − 

r 5 5 

y 3 3 

tanθ 

= = =− 

x −4 4 

r 5 

cscθ 

= = 

y 3 

2 2 2 2 

= + = ( − 4) + 3 = 16+ 9 = 25 = 5Now 

r 5 5 

secθ 

= = =− 

x −4 4 

x −4 4 

cotθ 

= = = − 

y 3 3 

2. We need values for x, y, and r, Because P = (–12, 5) 

is a point on the terminal side of 

θ , x =− 12 and y = 5. Furthermore, 

r x y 

2 2 2 2 

= + = − + = + 

( 12) 5 144 25 

= 169 = 13 



y 5 

sinθ 

= = 

r 13 

x −12 12 

cosθ 

= = = − 

r 13 13 

y 5 5 

tanθ 

= = = − 

x −12 12 

r 13 

cscθ 

= = 

y 5 

r 13 13 

secθ 

= = =− 

x −12 12 

x −12 12 

cotθ 

= = − 

y 5 5 

519 



3. We need values for x, y, and r. Because P = (2, 3) is a 

point on the terminal side of θ , x = 2 and y = 3 . 

Furthermore, 

2 2 2 2 

r = x + y = 2 + 3 = 4+ 9 = 13 



y 3 3 13 3 13 

sinθ 

= = = ⋅ = 

r 13 13 13 13 

x 2 2 13 2 13 

cosθ 

= = = ⋅ = 

r 13 13 13 13 

y 3 

tanθ 

= = 

x 2 

r 13 

cscθ 

= = 

y 3 

r 13 

secθ 

= = 

x 2 

x 2 

cotθ 

= = 

y 3 

4. We need values for x, y, and r, Because 

P = (3, 7) is a point on the terminal side of 

θ , x = 3 and y = 7. Furthermore, 

2 2 2 2 

r = x + y = 3 + 7 = 9+ 49 = 58 



y 7 7 58 7 58 

sinθ 

= = = ⋅ = 

r 58 58 58 58 

x 3 3 58 3 58 

cosθ 

= = = ⋅ = 

r 58 58 58 58 

y 7 

tanθ 

= = 

x 3 

r 58 

cscθ 

= = 

y 7 

r 58 

secθ 

= = 

x 3 

x 3 

cotθ 

= = 

y 7 

5. We need values for x, y, and r. Because P = (3, –3) is 

a point on the terminal side of θ , x = 3 and y =− 3 . 

2 2 2 2 

Furthermore, r = x + y = 3 + ( − 3) = 9+ 

9 

= 18 = 3 2 



y −3 −1 2 2 

sinθ 

= = = − ⋅ = − 

r 3 2 2 2 2 

x 3 1 2 2 

cosθ 

= = = ⋅ = 

r 3 2 2 2 2 

y −3 

tanθ 

= = = −1 

x 3 

r 3 2 

cscθ 

= = = − 2 

y −3 

r 3 2 

secθ 

= = = 2 

x 3 

x 3 

cotθ 

= = = −1 

y −3 

6. We need values for x, y, and r, Because P = (5, –5) is 

a point on the terminal side of θ , x = 5 and y = –5 . 

Furthermore, 

2 2 2 

r = x + y = 5 + ( − 5) = 25 + 25 = 50 

= 5 2 



y −5 −1 2 2 

sinθ 

= = = ⋅ =− 

r 5 2 2 2 2 

x 5 1 2 2 

cosθ 

= = = ⋅ = 

r 5 2 2 2 2 

y −5 

tanθ 

= = = −1 

x 5 

r 5 2 

cscθ 

= = = − 2 

y −5 

r 5 2 

secθ 

= = = 2 

x 5 

x 5 

cotθ 

= = =−1 

y −5 

520 



7. We need values for x, y, and r. Because P = (–2, –5) 


θ , x =− 2 and y =− 5. Furthermore, 

r x y 

2 2 2 2 

= + = ( − 2) + ( − 5) = 4+ 25 = 29Now 

that we know x, y, and r, we can find the six 


y −5 −5 29 5 29 

sinθ 

= = = ⋅ = − 

r 29 29 29 29 

x −2 −2 29 2 29 

cosθ 

= = = ⋅ =− 

r 29 29 29 29 

y −5 5 

tanθ 

= = = 

x −2 2 

r 29 29 

cscθ 

= = = − 

y −5 5 

r 29 29 

secθ 

= = =− 

x −2 2 

x −2 2 

cotθ 

= = = 

y −5 5 

10. θ = π radians 

The terminal side of the angle is on the negative x- 

axis. Select the point P = (–1, 0): x =− 1, y = 0, r = 1 

Apply the definition of the tangent function. 

y 0 

tanπ = = = 0 

x −1 



axis. Select the point P = (–1, 0): 

x =− 1, y = 0, r = 1 Apply the definition of the secant 

function. 

r 1 

secπ = = = −1 

x −1 



axis. Select the point P = (–1, 0): x =− 1, y = 0, r = 1 

Apply the definition of the cosecant function. 

r 1 

csc π = = , undefined 

y 0 

8. We need values for x, y, and r, Because P = (–1, –3) 


θ , x = –1 and y = –3 . Furthermore, 

r x y 

2 2 2 2 

= + = − + − = + = 

( 1) ( 3) 1 9 10 



y −3 −3 10 3 10 

sinθ 

= = = ⋅ = − 

r 10 10 10 10 

x −1 −1 10 10 

cosθ 

= = = ⋅ = − 

r 10 10 10 10 

y −3 

tanθ 

= = = 3 

x −1 

r 10 10 

cscθ 

= = = − 

y −3 3 

r 10 

secθ 

= = =− 10 

x −1 

x −1 1 

cotθ 

= = = 

y −3 3 



axis. Select the point P = (–1, 0): 

x =− 1, y = 0, r = 1 Apply the definition of the 

cosine function. 

x −1 

cosπ 

= = = − 1 

r 1 

13. 

14. 

15. 

3π 

θ = radians 

2 

The terminal side of the angle is on the negative y- 

axis. Select the point P = (0, –1): 

x = 0, y = –1, r = 1 Apply the definition of the 

3π 

y −1 

tangent function. tan = = , undefined 

2 x 0 

3π 

θ = radians 

2 

The terminal side of the angle is on the negative y- 

axis. Select the point P = (0, –1): x = 0, y = − 1, r = 1 

Apply the definition of the cosine function. 

3π x 0 

cos = = = 0 

2 r 1 

π 

θ = radians 

2 

The terminal side of the angle is on the positive y- 

axis. Select the point P = (0, 1): 

x = 0, y = 1, r = 1 Apply the definition of the 

π x 0 

cotangent function. cot = = = 0 

2 y 1 

521 



16. 

π 

θ = radians 

2 

The terminal side of the angle is on the positive y- 

axis. Select the point P = (0, 1): x = 0, y = 1, r = 1 

Apply the definition of the tangent function. 

π y 1 

tan = = , undefined 

2 x 0 


> 0, θ cannot lie in quadrant III or 

quadrant IV; the sine function is negative in those 

quadrants. Thus, with sinθ 

> 0, θ lies in quadrant I 

or quadrant II. We are also given that cosθ > 0 . 

Because quadrant I is the only quadrant in which the 

cosine is positive and sine is positive, we conclude 

that θ lies in quadrant I. 


< 0, θ cannot lie in quadrant I or 

quadrant II; the sine function is positive in those two 

quadrants. Thus, with sinθ 

< 0, θ lies in quadrant III 

or quadrant IV. We are also given that cosθ > 0. 

Because quadrant IV is the only quadrant in which 

the cosine is positive and the sine is negative, we 

conclude that θ lies in quadrant IV. 



quadrant II; the sine function is positive in those two 

quadrants. Thus, with sinθ < 0, θ lies in quadrant 

III or quadrant IV. We are also given that cosθ < 0 . 

Because quadrant III is the only quadrant in which 

the cosine is positive and the sine is negative, we 

conclude that θ lies in quadrant III. 

20. Because tanθ 


quadrant III; the tangent function is positive in those 

two quadrants. Thus, with tanθ 

< 0, θ lies in 

quadrant II or quadrant IV. We are also given that 

sinθ < 0 . Because quadrant IV is the only quadrant 

in which the sine is negative and the tangent is 

negative, we conclude that θ lies in quadrant IV. 

21. Because tanθ 


quadrant III; the tangent function is positive in those 

quadrants. Thus, with tanθ 

< 0, θ lies in quadrant II 

or quadrant IV. We are also given that cosθ < 0 . 

Because quadrant II is the only quadrant in which the 

cosine is negative and the tangent is negative, we 

conclude that θ lies in quadrant II. 

22. Because cotθ 

> 0, θ cannot lie in quadrant II or 

quadrant IV; the cotangent function is negative in 

those two quadrants. Thus, with cotθ 

> 0, θ lies in 

quadrant I or quadrant III. We are also given that 

secθ < 0 . Because quadrant III is the only quadrant 

in which the secant is negative and the cotangent is 

positive, we conclude that θ lies in quadrant III. 

23. In quadrant III x is negative and y is negative. Thus, 

cos = 3 x −3 

θ − = = , x = –3, r = 5. Furthermore, 

5 r 5 

2 2 2 

r = x + y 

2 2 2 

5 = ( − 3) + y 

2 

y = 25 − 9 = 16 

y =− 16 =−4 

Now that we know x, y, and r, we can find the 

remaining trigonometric functions of θ . 

y −4 4 

sinθ 

= = = − 

r 5 5 

y −4 4 

tanθ 

= = = 

x −3 3 

r 5 5 

cscθ 

= = = − 

y −4 4 

r 5 5 

secθ 

= = = − 

x −3 3 

x −3 3 

cotθ 

= = = 

y −4 4 

24. In quadrant III, x is negative and y is negative. Thus, 

12 y −12 

sin θ =− = = , y =− 12, r = 13 . 

13 r 13 

Furthermore, 

2 2 2 

x + y = r 

2 2 2 

x + ( − 12) = 13 

2 

x = 169 − 144 = 25 

x =− 25 =−5 



x −5 5 

cosθ 

= = = − 

r 13 13 

y −12 12 

tanθ 

= = = 

x −5 5 

r 13 13 

cscθ 

= = = − 

y −12 12 

r 13 13 

secθ 

= = =− 

x −5 5 

x −5 5 

cotθ 

= = = 

y −12 12 

522 



25. In quadrant II x is negative and y is positive. Thus, 

5 y 

sin θ = = , y = 5, r = 13 . Furthermore, 

13 r 

2 2 2 

x + y = r 

2 2 2 

x + 5 = 13 

2 

x = 169 − 25 = 144 

x =− 144 =−12 



x −12 12 

cosθ 

= = = − 

r 13 13 

y 5 5 

tanθ 

= = = − 

x −12 12 

r 13 

cscθ 

= = 

y 5 

r 13 13 

secθ 

= = = − 

x −12 12 

x −12 12 

cotθ 

= = =− 

y 5 5 

26. In quadrant IV, x is positive and y is negative. Thus, 

4 x 

cos θ = = , x = 4, r = 5 . Furthermore, 

5 r 

2 2 2 

x + y = r 

2 2 2 

4 + y = 5 

2 

y = 25 − 16 = 9 

y =− 9 =−3 



y −3 3 

sinθ 

= = = − 

r 5 5 

y −3 3 

tanθ 

= = = − 

x 4 4 

r 5 5 

cscθ 

= = =− 

y −3 3 

r 5 

secθ 

= = 

x 4 

x 4 4 

cotθ 

= = = − 

y −3 3 

27. Because 270°< θ < 360 ° , θ is in quadrant IV. In 

quadrant IV x is positive and y is negative. Thus, 

8 x 

cos θ = = , x = 8, 

17 r 

r = 17. Furthermore 

2 2 2 

x + y = r 

2 2 2 

8 + y = 17 

2 

y = 289 − 64 = 225 

y =− 225 =−15 



y −15 15 

sinθ 

= = = − 

r 17 17 

y −15 15 

tanθ 

= = = − 

x 8 8 

r 17 17 

cscθ 

= = =− 

y −15 15 

r 17 

secθ 

= = 

x 8 

x 8 8 

cotθ 

= = = − 

y −15 15 

28. Because 270°< θ < 360 ° , θ is in quadrant IV. In 

quadrant IV, x is positive and y is negative. Thus, 

1 x 

cos θ = = , x = 1, r = 3. Furthermore, 

3 r 

2 2 2 

x + y = r 

2 2 2 

1 + y = 3 

2 

y = 9− 1= 

8 

y =− 8 =−2 2 



y −2 2 2 2 

sinθ 

= = =− 

r 3 3 

y −2 2 

tanθ 

= = = −2 2 

x 1 

r 3 3 2 3 2 

cscθ 

= = = ⋅ = − 

y −2 2 −2 2 2 4 

r 3 

secθ 

= = = 3 

x 1 

x 1 1 2 2 

cotθ 

= = = ⋅ =− 

y −2 2 − 2 2 2 4 

523 



29. Because the tangent is negative and the sine is 

positive, θ lies in quadrant II. In quadrant II, x is 


2 y 2 

tan θ = − = = , x = − 3, y = 2. Furthermore, 

3 x −3 

r x y 

2 2 2 2 

= + = − + = + = 

( 3) 2 9 4 13 



y 2 2 13 2 13 

sinθ 

= = = ⋅ = 

r 13 13 13 13 

x −3 −3 13 3 13 

cosθ 

= = = ⋅ = − 

r 13 13 13 13 

r 13 

cscθ 

= = 

y 2 

r 13 13 

secθ 

= = = − 

x −3 3 

x −3 3 

cotθ 

= = = − 

y 2 2 

30. Because the tangent is negative and the sine is 

positive, θ lies in quadrant II. In quadrant II, x is 


1 y 1 

tan θ =− = = , y = 1, x =−3 

. Furthermore, 

3 x −3 

r x y 

2 2 2 2 

= + = − + = + = 

( 3) 1 9 1 10 



y 1 1 10 10 

sinθ 

= = = ⋅ = 

r 10 10 10 10 

x −3 −3 10 3 10 

cosθ 

= = = ⋅ = − 

r 10 10 10 10 

r 10 

cscθ 

= = = 10 

y 1 

r 10 10 

secθ 

= = = − 

x −3 3 

x −3 

cotθ 

= = =−3 

y 1 

31. Because the tangent is positive and the cosine is 

negative, θ lies in quadrant III. In quadrant III, x is 

4 y −4 

negative and y is negative. Thus, tan θ = = = , 

3 x − 3 

x = –3, y = –4. Furthermore, 

2 2 2 2 

r = x + y = ( − 3) + ( − 4) = 9+ 

16 

= 25 = 5 



y −4 4 

sinθ 

= = = − 

r 5 5 

x −3 3 

cosθ 

= = =− 

r 5 5 

r 5 5 

cscθ 

= = = − 

y −4 4 

r 5 5 

secθ 

= = = − 

x −3 3 

x −3 3 

cotθ 

= = = 

y −4 4 

32. Because the tangent is positive and the cosine is 

negative, θ lies in quadrant III. In quadrant III, x is 

negative and y is negative. Thus, 

5 y −5 

tan θ = = = , x =− 12, y =−5 

. Furthermore, 

12 x −12 

2 2 2 2 

r = x + y = ( − 12) + ( − 5) = 144+ 

25 

= 169 = 13 



y −5 5 

sinθ 

= = =− 

r 13 13 

x −12 12 

cosθ 

= = = − 

r 13 13 

r 13 13 

cscθ 

= = = − 

y −5 5 

r 13 13 

secθ 

= = = − 

x −12 12 

x −12 12 

cotθ 

= = = 

y −5 5 

524 



33. Because the secant is negative and the tangent is 

positive, θ lies in quadrant III. In quadrant III, x is 


r 3 

secθ =− 3 = = , x =− 1, r = 3 . Furthermore, 

x −1 

2 2 2 

x + y = r 

2 2 2 

( − 1) + y = 3 

2 

y = 9− 1= 

8 

y =− 8 =−2 2 



y −2 2 2 2 

sinθ 

= = =− 

r 3 3 

x −1 1 

cosθ 

= = =− 

r 3 3 

y −2 2 

tanθ 

= = = 2 2 

x −1 

r 3 3 2 3 2 

cscθ 

= = = ⋅ = − 

y −2 2 −2 2 2 4 

x −1 1 2 2 

cotθ 

= = = ⋅ = 

y − 2 2 2 2 2 4 

34. Because the cosecant is negative and the tangent is 

positive, θ lies in quadrant III. In quadrant III, x is 


r 4 

cscθ =− 4 = = , y =− 1, r = 4 . Furthermore, 

y −1 

2 2 2 

x + y = r 

2 2 2 

x + ( − 1) = 4 

2 

x = 16 − 1 = 15 

x =− 15 



y −1 1 

sinθ 

= = = − 

r 4 4 

x − 15 15 

cosθ 

= = = − 

r 4 4 

y −1 1 15 15 

tanθ 

= = = ⋅ = 

x − 15 15 15 15 

r 4 4 15 4 15 

secθ 

= = = − ⋅ = − 

x − 15 15 15 15 

x − 15 

cotθ 

= = = 15 

y − 1 

35. Because 160° lies between 90° and 180°, it is in 


θ ′ = 180°− 160°= 20°. 



θ ′ = 180°− 170°= 10°. 


quadrant III. The reference angle is 

θ ′ = 205°− 180°= 25°. 


quadrant III. The reference angle is 

θ ′ = 210°− 180°= 30°. 


quadrant IV. The reference angle is 

θ ′ = 360°− 355°= 5°. 


quadrant IV. The reference angle is 

θ ′ = 360°− 351°= 9°. 

41. Because 7 π 3π 

6π 

8π 

lies between = and 2π = , it 

4 

2 4 

4 

is in quadrant IV. The reference angle is 

7π 8π 7π π 

θ′ = 2π 

− = − = . 

4 4 4 4 

42. Because 5π 4 lies between π = 4π 4 and 3π 2 = 6π 4 , it 

is in quadrant III. The reference angle is 

θ ′ = 5π 4 − π = 5π 4 − 4π 4 = π 4 . 

43. Because 5 π π 3π 

lies between = and π = 

6 

2 6 

in quadrant II. The reference angle is 

5π 6π 5π π 

θ′ = π − = − = . 

6 6 6 6 

6π 

6 

, it is 

44. Because 5 π 10 π 

π 7π 

= lies between = and 

7 14 

2 14 

14π 

π = , it is in quadrant II. The reference angle is 

14 

5π 7π 5π 2π 

θ′ = π − = − = . 

7 7 7 7 

45. − 150°+ 360°= 210° 

Because the angle is in quadrant III, the reference 

angle is θ ′ = 210°− 180°= 30°. 

525 



46. − 250°+ 360°= 110° 

Because the angle is in quadrant II, the reference 

angle is θ′ = 180°− 110°= 70°. 

47. − 335°+ 360°= 25° 

Because the angle is in quadrant I, the reference 

angle is θ′ = 25°. 

48. − 359°+ 360°= 1° 

Because the angle is in quadrant I, the reference 

angle is θ′ = 1°. 

57. 

58. 

11 11 16 5 

− π + 4π 

= − π + π = 

π 

4 4 4 4 


5π 

π 

angle is θ′ = − π = . 

4 4 

17 17 24 7 

− π + 4π 

=− π + π = 

π 

6 6 6 6 


7π 

π 

angle is θ′ = − π = . 

6 6 

49. Because 4.7 lies between π ≈ 3.14 and 3 π ≈ 4.71 , it 

2 

is in quadrant III. The reference angle is 

θ′ = 4.7 −π 

≈ 1.56 . 

50. Because 5.5 lies between 3 π ≈ 4.71 and 2π ≈ 6.28 , 

2 

it is in quadrant IV. The reference angle is 

θ′ = 2π 

−5.5 ≈ 0.78 . 

51. 565°− 360°= 205° 


angle is θ ′ = 205°− 180°= 25°. 

52. 553°− 360°= 193° 


angle is θ′ = 193°− 180°= 13°. 

53. 

54. 

55. 

56. 

17 π 17 12 5 

− 2π 

= π − π = 

π 

6 6 6 6 


5π 

π 

angle is θ′ = π − = . 

6 6 

11 π 11 8 3 

− 2π 

= π − π = 

π 

4 4 4 4 


3π 

π 

angle is θ′ = π − = . 

4 4 

23 π 23 16 7 

− 4π 

= π − π = 

π 

4 4 4 4 

Because the angle is in quadrant IV, the reference 

7π 

π 


− = . 

4 4 

17 π 17 12 5 

− 4π 

= π − π = 

π 

3 3 3 3 


5π 

π 


− = . 

3 3 

59. 

60. 

25 25 36 11 

− π + 6π 

=− π + π = 

π 

6 6 6 6 


11π 

π 


− = . 

6 6 

13 13 18 5 

− π + 6π 

= − π + π = 

π 

3 3 3 3 


5π 

π 


− = . 

3 3 

61. 225° lies in quadrant III. The reference angle is 

θ ′ = 225°− 180°= 45°. 

2 

cos 45°= 

2 

Because the cosine is negative in quadrant III, 

2 

cos 225 ° = − cos 45 ° = − . 

2 

62. 300° lies in quadrant IV. The reference angle is 

θ ′ = 360°− 300°= 60° 

. 

3 

sin 60°= 

2 

Because the sine is negative in quadrant IV, 

3 

sin 300°=− sin 60°=− . 

2 


210 180 30 

θ ′ = °− °= °. 

3 

tan 30°= 

3 

Because the tangent is positive in quadrant III, 

3 

tan 210 ° = tan 30°= . 

3 

526 




θ ′ = 240°− 180°= 60° 

. 

sec60°= 

2 

Because the secant is negative in quadrant III, 

sec 240°=− sec60°− 2 . 

65. 420° lies in quadrant I. The reference angle is 

420 360 60 

θ ′ = °− °= °. 

tan 60°= 

3 

Because the tangent is positive in quadrant I, 

tan 420 ° = tan 60 ° = 3 . 

66. 405° lies in quadrant I. The reference angle is 

θ ′ = 405°− 360°= 45° 

. 

tan 45°= 

1 


tan 405 ° =tan45 ° =1. 

67. 

2π lies in quadrant II. The reference angle is 

3 

2π 3π 2π π 

θ′ = π − = − = . 

3 3 3 3 

π 3 

sin = 3 2 


2π π 3 

sin =sin = . 

3 3 2 

70. 

71. 

72. 

7π lies in quadrant IV. The reference angle is 

4 

7π 8π 7π π 

θ′ = 2 π − = − = . 

4 4 4 4 

π 

cot = 1 

4 

Because the cotangent is negative in quadrant IV, 

7π 

π 

cot = − cot = − 1 . 

4 4 

9π lies in quadrant I. The reference angle is 

4 

9 9 8 

θ′ = π − 2π 

= π − π = π . 

4 4 4 4 

π 

tan = 1 

4 


9π π 

tan =tan = 1 

4 4 

9π lies on the positive y-axis. The reference angle is 

2 

9 9 8 

θ′ = π − 4π 

= π − π = π . 

2 2 2 2 

π 9π Because tan is undefined, tan is also 

2 2 

undefined. 

68. 

69. 

3π lies in quadrant II. The reference angle is 

4 

3π 4π 3π π 

θ′ = π − = − = . 

4 4 4 4 

π 2 

cos = 4 2 

Because the cosine is negative in quadrant II, 

3π π 2 

cos =– cos =− . 

4 4 2 

7π lies in quadrant III. The reference angle is 

6 

7 7 6 

θ′ = π − π = π − π = π . 

6 6 6 6 

π 

csc = 2 

6 

Because the cosecant is negative in quadrant III, 

7π 

π 

csc = − csc =− 2 . 

6 6 

73. –240° lies in quadrant II. The reference angle is 

240 180 60 

θ ′ = °− °= °. 

3 

sin 60°= 

2 


3 

sin( − 240 ° )= sin 60 ° = . 2 

74. –225° lies in quadrant II. The reference angle is 

θ ′ = 225°− 180°= 45 ° . 

2 

sin 45°= 

2 


2 

sin( − 225 ° ) = sin 45°= . 

2 

527 



75. 

76. 

π 


4 

π 

θ ′ = . 

4 

π 

tan = 1 

4 

Because the tangent is negative in quadrant IV, 

⎛ π ⎞ π 

tan ⎜− ⎟= − tan =−1 

⎝ 4 ⎠ 4 

π 


6 

π π 3 

θ = . tan = 

6 6 3 

Because the tangent is negative in quadrant IV, 

⎛ π ⎞ 3 

tan ⎜− ⎟ =− . 

⎝ 6 ⎠ 3 

77. sec 495°= sec135°=− 

2 

78. 

79. 

80. 

2 3 

sec510°= sec150°=− 

3 

19π 

7π 

cot = cot = 3 

6 6 

13π 

π 3 

cot = cot = 

3 3 3 

87. 

π π 3π 

sin cosπ − cos sin 

3 3 2 

⎛ 3 ⎞ ⎛1⎞ 

= ( −1) − ( −1 

⎜ 

) 

2 ⎟ ⎜ 

2 

⎟ 

⎝ ⎠ ⎝ ⎠ 

3 1 

=− + 

2 2 

1− 

3 

= 

2 

π π 

88. sin cos 0 − sin cosπ 

4 6 

⎛ 2 ⎞ ⎛1⎞ 

= () 1 − ( −1 

⎜ 

) 

2 ⎟ ⎜ 

2 

⎟ 

⎝ ⎠ ⎝ ⎠ 

2 1 

= + 

2 2 

2 + 1 

= 

2 

89. 

11π 5π 11π 5π 

sin cos + cos sin 

4 6 4 6 

⎛ 2 ⎞⎛ 3 ⎞ ⎛ 2 ⎞⎛1⎞ 

= − + − 

⎜ 2 ⎟⎜ 2 ⎟ ⎜ 2 

⎟⎜⎟ 2 

⎟ 

⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 

6 2 

=− − 

4 4 

6 + 2 

=− 

4 

81. 

82. 

83. 

84. 

23π 

7π 

2 

cos = cos = 

4 4 2 

35π 

11π 

3 

cos = cos = 

6 6 2 

⎛ 17π 

⎞ 7π 

3 

tan ⎜− tan 

6 

⎟= = 

⎝ ⎠ 6 3 

⎛ 11π 

⎞ π 

tan ⎜− = tan = 1 

4 

⎟ 

⎝ ⎠ 4 

90. 

17π 5π 17π 5π 

sin cos + cos sin 

3 4 3 4 

⎛ 3 ⎞⎛ 2 ⎞ ⎛1⎞⎛ 2 ⎞ 

= − − + − 

⎜ 2 ⎟⎜ 2 ⎟ ⎜ 

2 

⎟ 

⎜ 2 ⎟ 

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ 

6 2 

= − 

4 4 

6 − 2 

= 

4 

85. 

⎛ 17π 

⎞ π 3 


3 

⎟ = = 

⎝ ⎠ 3 2 

86. 

⎛ 35π 

⎞ π 1 


6 

⎟ = = 

⎝ ⎠ 6 2 

528 



91. 

92. 

93. 

94. 

3 15 5 

sin π ⎛ 

tan π ⎞ ⎛ 

cos 

π ⎞ 

2 

⎜− − − 

4 

⎟ ⎜ 

3 

⎟ 

⎝ ⎠ ⎝ ⎠ 

⎛1 

⎞ 

= ( −1)( 1) 

− ⎜ 

2 ⎟ 

⎝ ⎠ 

1 

=−1− 

2 

2 1 

=− − 

2 2 

3 

=− 

2 

3 8 5 

sin π ⎛ 

tan π ⎞ ⎛ 

cos 

π ⎞ 

2 

⎜− + − 

3 

⎟ ⎜ 

6 

⎟ 

⎝ ⎠ ⎝ ⎠ 

⎛ 3 ⎞ 

= ( − 1)( 3) 

+ ⎜ 

− 

2 ⎟ 

⎝ ⎠ 

3 

=− 3 − 

2 

2 3 3 

=− − 

2 2 

3 3 

=− 

2 

⎛4 4 

f π π ⎞ ⎛ 

f π ⎞ ⎛ 

f 

π ⎞ 

⎜ + 

3 6 

⎟+ ⎜ + 

3 

⎟ ⎜ 

6 

⎟ 

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 

⎛4π π ⎞ 4π π 

= sin ⎜ + sin sin 

3 6 

⎟+ + 

⎝ ⎠ 3 6 

3π 4π π 

= sin + sin + sin 

2 3 6 

⎛ 3 ⎞ ⎛1⎞ 

= ( − 1) 

+ − + 

⎜ 

2 ⎟ ⎜ 

2 

⎟ 

⎝ ⎠ ⎝ ⎠ 

3 + 1 

=− 

2 

⎛5 5 

g π π ⎞ ⎛ 

g π ⎞ ⎛ 

g 

π ⎞ 

⎜ + 

6 6 

⎟+ ⎜ + 

6 

⎟ ⎜ 

6 

⎟ 

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 

⎛5π π ⎞ 5π π 

= cos ⎜ + cos cos 

6 6 

⎟+ + 

⎝ ⎠ 6 6 

5π 

π 

= cosπ 

+ cos + cos 

6 6 

⎛ 3 ⎞ ⎛ 3 ⎞ 

= ( − 1) 

+ − + 

⎜ 2 ⎟ ⎜ 2 ⎟ 

⎝ ⎠ ⎝ ⎠ 

=−1 

95. ( ) 

⎛17π 

⎞ ⎛ ⎛17π 

⎞⎞ 

h g ⎜ = h⎜g 

⎟ 

3 

⎟ ⎜ 

3 

⎟ 

⎝ ⎠ ⎝ ⎝ ⎠⎠ 

⎛ ⎛17π 

⎞⎞ 

= 2⎜cos⎜ 

3 

⎟⎟ 

⎝ ⎝ ⎠⎠ 

⎛1 

⎞ 

= 2 ⎜ 2 ⎟ 

⎝ ⎠ 

= 1 

⎛11π 

⎞ ⎛ ⎛11π 

⎞⎞ 

h f ⎜ = h⎜ f ⎟ 

4 

⎟ ⎜ 

4 

⎟ 

⎝ ⎠ ⎝ ⎝ ⎠⎠ 

⎛ ⎛11π 

⎞⎞ 

= 2⎜sin⎜ 

⎟ 

4 

⎟ 

⎝ ⎝ ⎠⎠ 

96. ( ) 

⎛ 2 ⎞ 

= 2 

⎜ 

2 ⎟ 

⎝ ⎠ 

= 2 

97. The average rate of change is the slope of the line 

, ( ) x , f( x ) 

through the points ( x f x ) and ( ) 

m = 

f( x2) − f( x1) 

x − x 

2 1 

1 1 

⎛3π 

⎞ ⎛5π 

⎞ 

sin ⎜ sin 

2 

⎟− 

⎜ 

4 

⎟ 

= 

⎝ ⎠ ⎝ ⎠ 

3π 

5π 

− 

2 4 

⎛ 2 ⎞ 

−1− − 

⎜ 

2 ⎟ 

= 

⎝ ⎠ 

π 

4 

2 

− 1+ 

= 2 

π 

4 

⎛ 2 ⎞ 

4 − 1+ 

⎜ 

2 ⎟ 

= 

⎝ ⎠ 

⎛π 

⎞ 

4⎜ 

4 

⎟ 

⎝ ⎠ 

2 2 − 4 

= 

π 

2 2 

529 



98. The average rate of change is the slope of the line 

through the points ( x1, g( x 

1) 

) and ( x2, g( x 

2) 

) 

gx ( 

2) − gx ( 

1) 

m = 

x2 − x1 

⎛3π 

⎞ 

cos ( π ) − cos ⎜ 

4 

⎟ 

= 

⎝ ⎠ 

3π 

π − 

4 

⎛ 2 ⎞ 

−1− ⎜ 

− 

2 ⎟ 

= 

⎝ ⎠ 

π 

4 

⎛ 2 ⎞ 

4 − 1+ 

⎜ 

2 ⎟ 

= 

⎝ ⎠ 

⎛π 

⎞ 

4⎜ 

4 

⎟ 

⎝ ⎠ 

2 2 − 4 

= 

π 

99. 

2 

π 

sinθ = when the reference angle is and θ is 

2 

4 

in quadrants I or II. 

QI 

QII 

π 

π 

θ = θ = π − 

4 4 

3π 

= 

4 

π 3π 

θ = , 

4 4 

1 

π 

100. cosθ = when the reference angle is and θ is in 

2 

3 

quadrants I or IV. 

QI 

QIV 

π 

π 

θ = θ = 2π 

− 

3 3 

5π 

= 

3 

π 5π 

θ = , 

3 3 

2 

π 

101. sinθ =− when the reference angle is and 

2 

4 

θ is in quadrants III or IV. 

QIII 

QIV 

π 

π 

θ = π + θ = 2π 

− 

4 4 

5π 

7π 

= = 

4 4 

5π 

7π 

θ = , 

4 4 

1 

π 

102. cosθ =− when the reference angle is and θ is 

2 

3 

in quadrants II or III. 

QII 

QIII 

π 

π 

θ = π − θ = π + 

3 3 

2π 

4π 

= = 

3 3 

2π 

4π 

θ = , 

3 3 

π 

103. tanθ =− 3 when the reference angle is and θ is 3 

in quadrants II or IV. 

QII 

QIV 

π 

π 

θ = π − θ = 2π 

− 

3 3 

2π 

5π 

= = 

3 3 

2π 

5π 

θ = , 

3 3 

3 

π 

104. tanθ =− when the reference angle is and 

3 

6 

θ is in quadrants II or IV. 

QII 

QIV 

π 

π 

θ = π − θ = 2π 

− 

6 6 

5π 

11π 

= = 

6 6 

5π 

11π 

θ = , 

6 6 



Sample explanation: Sine is defined for all values of 

the angle. 

530

Chapter 4 Trigonometric Functions

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