Chapter 4 Trigonometric Functions

Trigonometric Functions
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
Check Point Exercises
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
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.