#### Answer

$\sin \left( \dfrac {\pi }{3}+\theta \right) -\sin \left( \dfrac {\pi }{3}-\theta \right) =\left( \sin \dfrac {\pi }{3}\cos \theta +\cos \dfrac {\pi }{3}\sin \theta \right) -\left( \sin \dfrac {\pi }{3}\cos \theta -\cos \dfrac {\pi }{3}\sin \theta \right)=2\cos \dfrac {\pi }{3}\sin \theta =2\times \left( \dfrac {1}{2}\right) \times \sin \theta =\sin \theta $