Answer
$ \sin ^{2}\dfrac {\pi }{12}=\dfrac {2-\sqrt {3}}{4}\approx 0.067$
Work Step by Step
$\sin \alpha \cos \theta -\cos \alpha \sin \theta =\sin \left( \alpha -\theta \right) \Rightarrow \left( \sin \dfrac {\pi }{3}\cos \dfrac {\pi }{4}-\cos \dfrac {\pi }{3}\sin \dfrac {\pi }{4}\right) ^{2}=\left( \sin \left( \dfrac {\pi }{3}-\dfrac {\pi }{4}\right) \right) ^{2}=\left( \sin \dfrac {\pi }{12}\right) ^{2}$
$\cos {2}\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha =\left( 1-\sin ^{2}\alpha \right) -\sin ^{2}\alpha =1-2\sin ^{2}\alpha \Rightarrow \sin ^{2}\alpha =\dfrac {1-\cos {2}\alpha }{2}$
$\Rightarrow \sin ^{2}\dfrac {\pi }{12}=\dfrac {1-\cos \left( 2\times \dfrac {\pi }{12}\right) }{2}=\dfrac {1-\cos \dfrac {\pi }{6}}{2}=\dfrac {1-\dfrac {\sqrt {3}}{2}}{2}=\dfrac {2-\sqrt {3}}{4}\approx 0.067$
