Question
Prove that:
$\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5}=\frac{5}{16}$

Answer

 $\text{LHS}=\sin36^\circ.\sin72^\circ.\sin108^\circ.\sin144^\circ$ $\begin{bmatrix}\because\sin144^\circ=\sin(180^\circ-366^\circ)=\sin36^\circ\\\sin108^\circ=\sin(180^\circ-78^\circ)=\sin72^\circ\end{bmatrix}$
$\sin36^\circ.\sin72^\circ.\sin72^\circ.\sin36^\circ$
$=\frac{1}{4}(2\sin36.\sin72^\circ)^2$
$=\frac{1}{4}(2\sin36.\cos18^\circ)^2$ $[\because\sin72^\circ=108^\circ]$
$=\frac{4}{4}\Bigg(\frac{\sqrt{10-2\sqrt{5}}}{4}.\frac{\sqrt{10+2\sqrt{5}}}{4}\Bigg)^2$
$=\frac{1}{64}\big(10-2\sqrt{5}\big)\big(10+2\sqrt{5}\big)$
$=\frac{100-20}{64\times4}$
$=\frac{80}{256}$
$=\frac{5}{16}$
$=\text{RHS}$

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