Prove that
$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=\frac{3}{16}$
L.H.S
$=\frac{1}{2}\left(2 \sin 80^{\circ} \sin 20^{\circ}\right) \sin 40^{\circ} \frac{\sqrt{3}}{2}$
$=\frac{\sqrt{3}}{4}\left\{\cos \left(80^{\circ}-20^{\circ}\right)-\cos \left(80^{\circ}+20^{\circ}\right)\right\} \sin 40^{\circ}$
$=\frac{\sqrt{3}}{4}\left\{\cos 60^{\circ} \sin 40^{\circ}-\cos 100^{\circ} \sin 40^{\circ}\right\}$
$\left.=\frac{\sqrt{3}}{4}\left\{\frac{1}{2} \sin 40^{\circ}-\cos 100^{\circ} \sin 40^{\circ}\right\}\right\}$
$=\frac{\sqrt{3}}{8}\left\{\sin 40^{\circ}-2 \cos 100^{\circ} \sin 40^{\circ}\right\}$
$=\frac{\sqrt{3}}{8}\left\{\sin 40^{\circ}-\left(\sin \left(100^{\circ}+40^{\circ}\right)-\sin \left(100^{\circ}-40^{\circ}\right)\right\}\right.$
$=\frac{\sqrt{3}}{8}\left\{\sin 40^{\circ}-\sin 140^{\circ}+\sin 60^{\circ}\right\}$
$=\frac{\sqrt{3}}{8}\left\{\sin 40^{\circ}-\sin 140^{\circ}+\frac{\sqrt{3}}{2}\right\}$
$=\frac{\sqrt{3}}{8}\left\{\sin 40^{\circ}-\sin \left(180^{\circ}-40^{\circ}\right)+\frac{\sqrt{3}}{2}\right\}$
$=\frac{\sqrt{3}}{8}\left\{\sin 40^{\circ}-\sin 40^{\circ}+\frac{\sqrt{3}}{2}\right\}$
$=\frac{3}{16}$
$=$ R.H.S