#### Answer

$-\dfrac {\sqrt {2}}{4}\left( 1+\sqrt {3}\right) $

#### Work Step by Step

$\sin \left( -\dfrac {5\pi }{12}\right) =-\sin \left( \dfrac {5\pi }{12}\right) =-\sin \left( \dfrac {\pi }{4}+\dfrac {\pi }{6}\right) =-\left( \sin \dfrac {\pi }{4}\cos \dfrac {\pi }{6}+\cos \dfrac {\pi }{4}\sin \dfrac {\pi }{6}\right) =$
$=-(\dfrac {\sqrt {2}}{2}\dfrac {\sqrt {3}}{2}+\dfrac {\sqrt {2}}{2}\dfrac {1}{2})=-\dfrac {\sqrt {2}}{4}\left( 1+\sqrt {3}\right) $