Section 7.1
The Inverse Sine, Cosine, and Tangent Functions
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Review of Properties of Functions and Their Inverses
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5 is not in the interval , so 8 2 2
we need an angle in the interval 5, for which sin sin .
2 2 8
π π π
π π πθ
−
− =
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1 5(a) cos cos6π−
( )1(b) cos cos 0.2−
1 5(c) cos cos4π−
( )1(d) cos cos 2−
[ )5 5 = since is in the interval 0, .6 6π π π
[ ] = 0.2 since 0.2 is in the interval 1,1 .−
[ )5 Since is not in the interval 0, we find an angle that has the same 4
5 3cosine value that is in that interval. cos cos4 4
π π
π π =
1 3 3 = cos cos4 4π π− =
[ ]This is undefined since 2 is not the interval 1,1 .−
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−2 −1 1 2
−π/2
−π/4
π/4
π/2
1tan−=y xCopyright © 2013 Pearson Education, Inc. All rights reserved
−3 −2 −1
−π/2
−π/3
−π/6
π/6
π/3
1tan−=y x
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( )1
1
Find the inverse function of 3cos 1, .2 2
Find the range of and the domain and range of .
f f x x x
f f
π π−
−
= + − ≤ ≤
3cos 1y x= +
3cos 1x y= + interchange x and y
3cos 1y x= −
1cos3
xy −=
( )1 11cos3
xy f x− −− = =
Recall that the domain of f is the range of f-1 and the range of f is the domain of f-1.
1 1Domain of is 1 13
xf − −− ≤ ≤
3 1 3x− ≤ − ≤ 2 4x− ≤ ≤
1
So the range of is 2 4
and the range of is , .2 2
f y
f π π−
− ≤ ≤
−
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1Solve the equati 2coson: 2
x π− =
1os4
c x π− =
s4
cox π=
22
x =
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