Find the values of

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

We know that $\sin ^{-1}(\sin x)=x$ if $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, which is the principal value branch of $\sin ^{-1} x .$

Here, $\frac{2 \pi}{3} \notin\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Now, $\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ can be written as:

$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\sin ^{-1}\left[\sin \left(\pi-\frac{2 \pi}{3}\right)\right]=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)$ where $\frac{\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

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