=
5
6
because cot
5
6
= − √ 3 and
5
6
is in (0) (the range of cot −1 ).
(b) sec −1 2 =
3
because sec
3
= 2 and
3
is in
0
2
∪
3
2
(the range of sec −1 ).
74. (a) arcsin(sin(54)) = arcsin −1 √ 2
= −
4
because sin −
4
= −1 √ 2 and −
4
is in
−
2
2
.
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
58
¤
CHAPTER 1 FUNCTIONSAND MODELS
(b) Let = sin −1
5
13
[see the figure].
cos 2 sin −1
5
13
= cos2 = cos 2 − sin 2
=
12
13
2
−
5
13
2
=
144
169
−
25
169
=
119
169
75. Let = sin −1 . Then −
2
≤ ≤
2
⇒ cos ≥ 0, so cos(sin −1 ) = cos =
1 − sin 2 =
√ 1 − 2 .
76. Let = sin −1 . Then sin = , so from the triangle (which
illustrates the case 0), we see that
tan(sin −1 ) = tan =
√ 1 − 2 .
77. Let = tan −1 . Then tan = , so from the triangle (which
illustrates the case 0), we see that
sin(tan −1 ) = sin =
√ 1 + 2 .
78. Let = arccos. Then cos = , so from the triangle (which
illustrates the case 0), we see that
sin(2arccos) = sin2 = 2sin cos
= 2( √ 1 − 2 )() = 2
√ 1 − 2
79. The graph of sin −1 is the reflection of the graph of
sin about the line = .
80. The graph of tan −1 is the reflection of the graph of
tan about the line = .
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
CHAPTER 1 REVIEW
¤
59
81. () = sin −1 (3 + 1).
Domain () = { | −1 ≤ 3 + 1 ≤ 1} = { | −2 ≤ 3 ≤ 0} =
| − 2
3
≤ ≤ 0 =
− 2
3 0
.
Range () =
| −
2
≤ ≤
2
=
−
2
2
.
82. (a) () = sin sin −1
Since one function undoes what the other one does, we get the
identity function, = , on the restricted domain −1 ≤ ≤ 1.
(b) () = sin −1 (sin)
This is similar to part (a), but with domain .
Equations for on intervals of the form
−
2
+
2
+ , for any integer , can be
found using () = (−1) + (−1) +1 .
The sine function is monotonic on each of these intervals, and hence, so is (but in a linear fashion).
83. (a) If the point () is on the graph of = (), then the point ( − ) is that point shifted units to the left. Since
is 11, the point () is on the graph of = −1 () and the point corresponding to ( − ) on the graph of is
( − ) on the graph of −1 . Thus, the curve’s reflection is shifted down the same number of units as the curve itself is
shifted to the left. So an expression for the inverse function is −1 () = −1 () − .
(b) If we compress (or stretch) a curve horizontally, the curve’s reflection in the line = is compressed (or stretched)
vertically by the same factor. Using this geometricprinciple, we see that the inverse of () = () can be expressed as
−1 () = (1) −1 ().
1 Review
1. False. Let () = 2 , = −1, and = 1. Then ( + ) = (−1 + 1) 2 = 0 2 = 0, but
() + () = (−1) 2 + 1 2 = 2 6= 0 = ( + ).
2. False. Let () = 2 . Then (−2) = 4 = (2), but −2 6= 2.
3. False. Let () = 2 . Then (3) = (3) 2 = 9 2 and 3() = 3 2 . So (3) 6= 3().
4. True. The inverse function −1 of a onetoone function is defined by −1 () = ⇔ () = .