Answer
$\sin ^{-1}\left( \sin \left( \dfrac {5\pi }{6}\right) \right) =\sin ^{-1}\left( \sin \left( \dfrac {\pi }{6}\right) \right) =\dfrac {\pi }{6}$
Work Step by Step
$\sin ^{-1}\left( \sin x\right) =x\left( -\dfrac {\pi }{2}\leq x\leq \dfrac {\pi }{2}\right) $
$\dfrac {5\pi }{6} > \dfrac {\pi }{2}$ so we need to convert it
$
\sin \dfrac {5\pi }{6}=\sin \dfrac {\pi }{6}$
$\Rightarrow \sin ^{-1}\left( \sin \left( \dfrac {5\pi }{6}\right) \right) =\sin ^{-1}\left( \sin \left( \dfrac {\pi }{6}\right) \right) =\dfrac {\pi }{6}$
