MATH 135: PRECALCULUS
INVERSE TRIGONOMETRY I
π
(i.)
1. Graph each pair of functions on a single pair of axes: (i.) cos x and cos−1 x,
(ii.) sin x and sin−1 x, and (iii.) tan x
and tan−1 x.
π
2
π
2
(ii.)
π
2
− π2
(iii.)
π
2
− π2
π
π
2
− π2
π
2
− π2
2. Solve the following equations for the given variable (θ or x in the appropriate
restricted domain).
i. cos θ =
1
2
√
sin x
− 3/2
=
− 3=
1/2
cos x
√
√
ii. tan x = − 3
iii. sin θ = 3
θ=
π
3
x=−
π
3
No solution, as 3 is outside the range of sin x
iv. tan2 (2x) = 1, where 2x is an angle in Quadrant IV
tan(2x) = −1
Take the negative square root since 2x is in Quadrant IV
tan−1 (tan(2x)) = tan−1 (−1)
2x = −
π
4
x=−
π
8
Take the inverse tangent
Divide by 2
3. Evaluate the following expressions.
i. cos cos
−1
√ !!
3
2
√
3
2
In this case, cosine and its inverse cancel.
ii. sin−1 (sin(2π))
sin−1 (0) = 0
h π πi
They don’t cancel since 2π is not in − , , so first you compute sin(2π).
2 2
iii. tan
−1
tan
Same as ( ii.).
5π
3
√ π
tan−1 − 3 = −
3