# How do you evaluate sin^-1(1/2)?

$\frac{\pi}{6}$
$\textcolor{red}{{\sin}^{-} 1 \left(\sin x\right) = x , \forall x \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]}$
${\sin}^{-} 1 \left(\frac{1}{2}\right) = {\sin}^{-} 1 \left(\sin \left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6} , w h e r e , \frac{\pi}{6} \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$
So, ${\sin}^{-} 1 \left(\frac{1}{2}\right) = \frac{\pi}{6}$