Question

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

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{6}$
• C. $\frac{\pi}{3}$
• D. $\frac{3\pi}{2}$

Answer 1

Answer: (C)

Let $\sin^{-1} \left[\cos \frac{\pi}{6} \right] =\theta $
$\Rightarrow \cos \frac{\pi}{6} = \sin\theta \Rightarrow \frac{\sqrt{3}}{2} = \sin\theta$
Now, $ \cos\theta = \sqrt{1 - \sin^{2} \theta} = \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^{2}}$
$ = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$
$ \Rightarrow \cos\theta = \cos \frac{\pi}{3} \Rightarrow \theta = \frac{\pi}{3}$
$ \therefore \sin^{-1} \left[\cos \frac{\pi}{6}\right] = \frac{\pi}{3} $