Answer
$-\dfrac{\pi}{6}$
Work Step by Step
We need to find: $\sin^{-1} (-\dfrac{1}{2})$
The inverse sine function has the range $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$. We are looking for the angle whole sine value equals $\dfrac{1}{2}$.
We know from the unit circle that
$\sin (\dfrac{-\pi}{6})=-\dfrac{1}{2}$
Since $\dfrac{ -\pi}{6}$ lies in $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$, then we have:
$\sin^{-1} (-\dfrac{1}{2}) =-\dfrac{\pi}{6}$
