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Question

Mathematics Question on Inverse Trigonometric Functions

sin⁡−1(cos⁡π6)\sin^{-1} \left(\cos \frac{\pi}{6}\right) is equal to:

A

π2\frac{\pi}{2}

B

π6\frac{\pi}{6}

C

π3\frac{\pi}{3}

D

3π2\frac{3\pi}{2}

Answer

π3\frac{\pi}{3}

Explanation

Solution

Let sin⁡−1[cos⁡π6]=θ\sin^{-1} \left[\cos \frac{\pi}{6} \right] =\theta ⇒cos⁡π6=sin⁡θ⇒32=sin⁡θ\Rightarrow \cos \frac{\pi}{6} = \sin\theta \Rightarrow \frac{\sqrt{3}}{2} = \sin\theta Now, cos⁡θ=1−sin⁡2θ=1−(32)2 \cos\theta = \sqrt{1 - \sin^{2} \theta} = \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^{2}} =1−34=12 = \sqrt{1 - \frac{3}{4}} = \frac{1}{2} ⇒cos⁡θ=cos⁡π3⇒θ=π3 \Rightarrow \cos\theta = \cos \frac{\pi}{3} \Rightarrow \theta = \frac{\pi}{3} ∴sin⁡−1[cos⁡π6]=π3\therefore \sin^{-1} \left[\cos \frac{\pi}{6}\right] = \frac{\pi}{3}