Calculus- Early Transcendentals, 2021a

2.6. Inverse Trigonometric Functions 67
Example 2.27: The Triangle Technique
Rewrite the expression cos(sin −1 x) without trig functions. Note that the domain of this function is
all x ∈ [−1,1].
Solution. Let θ = sin −1 x. We need to compute cosθ. Taking the sine of both sides gives sinθ =
sin(sin −1 (x)) = x by the cancellation rule. We then draw a right triangle using sinθ = x/1:
If z is the remaining side, then by the Pythagorean Theorem:
√
z 2 + x 2 = 1 → z 2 = 1 − x 2 → z = ± 1 − x 2
and hence z =+ √ 1 − x 2 since θ ∈ [−π/2, π/2]. Thus, cosθ =
cos(sin −1 x)=
√
1 − x 2 by SOH CAH TOA, so,
√
1 − x 2 . ♣
Example 2.28: The Triangle Technique 2
For x ∈ (0,1), rewrite the expression sin(2cos −1 x). Compute sin(2cos −1 (1/2)).
Solution. Let θ = cos −1 x so that cosθ = x. The question now asks for us to compute sin(2θ). Wethen
draw a right triangle using cosθ = x/1:
To find sin(2θ) we use the double angle formula sin(2θ)=2sinθ cosθ. Butsinθ = √ √
1 − x 2 ,forθ ∈
√
[0,π],andcosθ = x. Therefore, sin(2cos −1 x)=2x 1 − x 2 .Whenx = 1/2wehavesin(2cos −1 (1/2)) =
3
2 . ♣
Exercises for 2.6
Exercise 2.6.1 Compute the following: