Question

Find the simplified value of the following expressions ${\sin }^{-1}\left[\cos \left\{{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}\right]$.

Answer 1

${\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\theta$ such that $\sin \theta =\frac{\sqrt{3}}{2}$

$\sin$ $\theta =\frac{\sqrt{3}}{2}=\sin$ $\frac{\pi }{3}$

$\theta =\frac{\pi }{3}$

i.e. ${\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{3}$

Now Put $[{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{3}]$ in given equation, we get

${\sin }^{-1}\left[\cos (\frac{\pi }{3})\right]$

$⟹{\sin }^{-1}\left(\frac{1}{2}\right)$ Because $\cos \frac{\pi }{3}=\frac{1}{2}$

Now,

${\sin }^{-1}\left(\frac{1}{2}\right)=\theta$ such that $\sin \theta =\frac{1}{2}$

$\sin$ $\theta =\frac{1}{2}=\sin$ $\frac{\pi }{6}$

$\theta =\frac{\pi }{6}$

i.e. ${\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}$

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