Write the principal value

Let $y=\sin ^{-1}\left(-\frac{1}{2}\right)$

Then,

$\sin y=-\frac{1}{2}=\sin \left(-\frac{\pi}{6}\right)$

$y=-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Here, $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is the range of the principal value branch of the inverse sine function.

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