Precalculus: Mathematics for Calculus, 7th Edition

$\Rightarrow \sin ^{-1}\left( \sin \left( \dfrac {7 \pi }{6}\right) \right) =-\dfrac {\pi }{6}$
$\sin ^{-1}\left( \sin x\right) =x\left( -\dfrac {\pi }{2}\leq x\leq \dfrac {\pi }{2}\right)$ $\dfrac {7\pi }{6} > \dfrac {\pi }{2}$ so we need to convert it $\sin \dfrac {7\pi }{6}=\sin (-\dfrac {\pi }{6})$. ($\sin \alpha =\sin \left( \pi -\alpha \right)$) $\Rightarrow \sin ^{-1}\left( \sin \left( \dfrac {7 \pi }{6}\right) \right) =\sin ^{-1}\left( \sin \left( -\dfrac {\pi }{6}\right) \right) =-\dfrac {\pi }{6}$