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

Solveing the above equation, we get

$\sin \mathrm{y}=-1 / 2$

since, $\sin \pi / 6=1 / 2$

therefore, $\sin y=-\sin$

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

Since range is $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$\sin ^{-1}\left(-\frac{1}{2}\right)$ is $-\pi / 6$.