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Prove: $\sin A . \sin (60-A) . \sin (60+A)=\frac{1}{4} \sin 3 A$
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Solution:

$\begin{aligned} L \cdot H . S . &=\sin A \sin (60-A) \cdot \sin (60+A) \\ &=\sin A(\sin 60 \cos A-\cos 60 \sin A)(\sin 60 \cos A+\cos 60 \sin A) \\ &=\sin A\left[\frac{\sqrt{3}}{2} \cos A-\frac{1}{2} \sin A\right]\left[\frac{\sqrt{3}}{2} \cos A+\frac{1}{2} \sin A\right] \\ & =\sin A\left[\left(\frac{\sqrt{3}}{2} \cos A\right)^{2}-\left(\frac{1}{2} \sin A\right)^{2}\right] \\ & =\sin A\left[\frac{3}{4} \cos ^{2} A-\frac{1}{4} \sin ^{2} A\right] \\ & =\frac{1}{4} \sin A\left[3 \cos ^{2} A-\sin ^{2} A\right]\\ &=\frac{1}{4}\left[3 \sin A-3 \sin ^{3} A-\sin ^{3} A\right]\\ & =\frac{1}{4}\left[3 \sin A-4 \sin ^{3} A\right]\\ & =\frac{1}{4} \sin 3 A=R . H . S \end{aligned}$

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