Question

Prove that: $\sin \alpha \sin \left({60}^{\circ }-\alpha \right)\sin \left({60}^{\circ }+\alpha \right)=\frac{1}{4}\sin 3\alpha$

Answer 1

$L.H.S.$

$\sin \alpha \sin ({60}^0-\alpha )\sin ({60}^0+\alpha )$

$=\sin \alpha ({\sin }^2{60}^0-{\sin }^2\alpha )since(\sin (A-B)\sin (A+B)=({\sin }^{2}A-{\sin }^{2}B))$

$=\sin \alpha ({\left(\frac{\sqrt{3}}{2}\right)}^2-{\sin }^2\alpha )$

$=\sin \alpha (\frac{3}{4}-{\sin }^2\alpha )$

$=\frac{1}{4}(3\sin \alpha -4{\sin }^3\alpha )$

$=\frac{1}{4}\sin 3\alpha$

$R.H.S.$