Write the value
Question:

Write the value of $\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$.

Solution:

We have

$\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}$

$=\cos ^{-1}\left(\cos \frac{\pi}{3}\right)+2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right)$

$\left[\because\right.$ The range of sine is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] ; \frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and the range of cosine is $[0, \pi] ; \frac{\pi}{3} \in[0, \pi]$

$=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right)$

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

$=\frac{2 \pi}{3}$

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