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