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