Prove that 

 $\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }=\frac{3}{16}$

Substituting  $A={20}^{\circ }$ in (i)

$\mathrm{Sin}{20}^{\circ }\mathrm{Sin}\left({60}^{\circ }+{20}^{\circ }\right)\mathrm{Sin}\left({60}^{\circ }-{20}^{\circ }\right)$

$=\frac{1}{4}\sin 3\times {20}^{\circ }$

$\Rightarrow \mathrm{Sin}{20}^{\circ }\mathrm{Sin}{80}^{\circ }\mathrm{Si}n{40}^{\circ }=\frac{1}{4}\mathrm{Sin}{60}^{\circ }$

Multiply both sides.  By Sin ${60}^{\circ }$

$\mathrm{Sin}{20}^{\circ }\mathrm{Sin}{40}^{\circ }\mathrm{Sin}{60}^{\circ }\mathrm{Sin}{80}^{\circ }$

$=\frac{1}{4}{\mathrm{Sin}}^2{60}^{\circ }=\frac{1}{4}\cdot \frac{3}{4}=\frac{3}{16}$