Answer
$\dfrac {\sqrt {2}}{4}\left( 1+\sqrt {3}\right) $
Work Step by Step
$\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) $