Prove the following : 20^ 40^ 60^ 80^ =3 / 16 - Vidyadip

Question

Prove the following : $\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=3 / 16$

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Answer

$\begin{aligned}
& \text { L.H.S. }=\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 60^{\circ} \cdot \sin 80^{\circ} \\
& =\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot\left(\frac{\sqrt{3}}{2}\right) \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{2} \cdot \frac{1}{2}\left(2 \cdot \sin 40^{\circ} \cdot \sin 20^{\circ}\right) \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{4}\left[\cos \left(40^{\circ}-20^{\circ}\right)-\cos \left(40^{\circ}+20^{\circ}\right)\right] \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{4}\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{4} \cdot \cos 20^{\circ} \cdot \sin 80^{\circ}-\frac{\sqrt{3}}{4} \cos 60^{\circ} \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{2 \times 4}\left(2 \sin 80^{\circ} \cdot \cos 20^{\circ}\right)-\frac{\sqrt{3}}{4}\left(\frac{1}{2}\right) \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \cdot\left[\sin \left(80^{\circ}+20^{\circ}\right)+\sin \left(80^{\circ}-20^{\circ}\right)\right]
\end{aligned}$
$-\frac{\sqrt{3}}{8} \cdot \sin 80^{\circ}$
$\begin{aligned}
& =\frac{\sqrt{3}}{8} \cdot\left(\sin 100^{\circ}+\sin 60^{\circ}\right)-\frac{\sqrt{3}}{8} \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \sin 100^{\circ}+\frac{\sqrt{3}}{8} \sin 60^{\circ}-\frac{\sqrt{3}}{8} \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \cdot \sin \left(180^{\circ}-80^{\circ}\right)+\frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{8} \cdot \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \sin 80^{\circ}+\frac{3}{16}-\frac{\sqrt{3}}{8} \sin 80^{\circ} \\
& =\frac{3}{16}=\text { R.H.S. } \quad \ldots\left[\because \sin \left(180^{\circ}-\theta\right)=\sin \theta\right]
\end{aligned}$

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