#### Answer

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

#### Work Step by Step

$\sin \left( -\dfrac {7\pi }{12}\right) = -\sin \dfrac {7\pi }{12}=-\sin \left( \dfrac {\pi }{3}-\dfrac {\pi }{4}\right) =-(\sin \dfrac {\pi }{3}\cos \dfrac {\pi }{4}-\cos \dfrac {\pi }{3}\sin \dfrac {\pi }{4})=-(\dfrac {\sqrt {3}}{2}\dfrac {\sqrt {2}}{2}-\dfrac {1}{2}\dfrac {\sqrt {2}}{2})=-\dfrac {\sqrt {2}}{4}\left( 1+\sqrt {3}\right) $