Answer
$arcsin(-\frac{1}{2})=-\frac{\pi}{6}$
Work Step by Step
The range of $arcsin~x$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$. So,
$arcsin(-\frac{1}{2})=-\frac{\pi}{6}$ because $sin(-\frac{\pi}{6})=-\frac{1}{2}$ and $-\frac{\pi}{2}\leq-\frac{\pi}{6}\leq\frac{\pi}{2}$
