#### State True or False for the statement The principal value of $\sin^{-1}\left [ \cos\left ( \sin^{-1}\frac{1}{2} \right ) \right ]$ is $\frac{\pi}{3}$

True

Principal value of sin-1 x is $\left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]$

Principal value of cos-1 x is [0, π]

We have, $\sin^{-1}\left [ \cos \left [ \sin^{-1}\left (\frac{1}{2} \right ) \right ] \right ]$

As, $\sin\frac{\pi}{6}=\frac{1}{2}$  so

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

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

As, $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ so,

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

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