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Prove that \(\sin 20^{\circ} \sin 30^{\circ} \sin 40^{\circ} \sin 80^{\circ}=\frac{\sqrt{3}}{16}.\)

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\(\operatorname{Sin} 20^{\circ} \operatorname{Sin} 30^{\circ} \operatorname{Sin} 40^{\circ} \operatorname{Sin} 80^{\circ}\)  

\(=\sin 20^{\circ} \times \frac{1}{2} \times \sin 40^{\circ} \sin 80^{\circ} \)

\(=\frac{1}{2} \times \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ} \)

\(=\frac{1}{2} \times \sin 20^{\circ} \sin \left(60^{\circ}-20^{\circ}\right) \sin \left(60^{\circ}+20^{\circ}\right) \)

\(=\frac{1}{2} \times \frac{1}{4} \sin \left(3 \times 20^{\circ}\right) \quad\left[\because \sin x \sin \left(60^{\circ} -x\right) \sin \left(60^{\circ}+x\right)=\frac{1}{4} \sin 3x\right] \)

\(=\frac{1}{8} \sin 60^{\circ} \)

\(=\frac{1}{8} \times \frac{\sqrt{3}}{2} \)

\(=\frac{\sqrt{3}}{16} .\)

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