Prove that: 20^ 40^ 80^ = 3 8

Question

Prove that: $\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}=\frac{\sqrt{3}}{8}$

✓

Answer

Given, $\ce{LHS} =\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}$
$=\frac{1}{2}\left[2 \sin 20^{\circ} \cdot \sin 40^{\circ}\right] \sin 80^{\circ}[\text { multiplying and dividing by } 2]$
$=\frac{1}{2}\left[\cos \left(20^{\circ}-40^{\circ}\right)-\cos \left(20^{\circ}+40^{\circ}\right)\right] \cdot \sin 80^{\circ}[\because 2 \sin x \cdot \sin y=\cos ( x - y )-\cos ( x + y )]$
$=\frac{1}{2}\left[\cos \left(-20^{\circ}\right)-\cos 60^{\circ}\right] \sin 80^{\circ}$
$=\frac{1}{2}\left[\cos 20^{\circ}-\frac{1}{2}\right] \cdot \sin 80^{\circ}\left[\because \cos (-\theta)=\cos \theta \text { and } \cos 60^{\circ}=\frac{1}{2}\right]$
$=\frac{1}{2} \times \frac{1}{2}\left[2\left(\cos 20^{\circ}-\frac{1}{2}\right) \cdot \sin 80^{\circ}\right] \text { [again multiplying and dividing by 2] }$
$=\frac{1}{4}\left[2 \cos 20^{\circ} \cdot \sin 80^{\circ}-\sin 80^{\circ}\right]$
$=\frac{1}{4}\left[\sin \left(20^{\circ}+80^{\circ}\right)-\sin \left(20^{\circ}-80^{\circ}\right)-\sin 80^{\circ}\right][\because 2 \cos x \cdot \sin y=\sin (x+y)-\sin (x-y)]$
$=\frac{1}{4}\left[\sin 100^{\circ}-\sin \left(-60^{\circ}\right)-\sin 80^{\circ}\right]$
$=\frac{1}{4}\left[\sin 100^{\circ}+\sin 60^{\circ}-\sin 80^{\circ}\right][\because \sin (-\theta)=-\sin \theta]$
$=\frac{1}{4}\left[\sin \left(180^{\circ}-80^{\circ}\right)+\sin 60^{\circ}-\sin 80^{\circ}\right]\left[\because \sin 100^{\circ}=\sin \left(180^{\circ}-80^{\circ}\right)\right]$
$=\frac{1}{4}\left[\sin 80^{\circ}+\sin 60^{\circ}-\sin 80^{\circ}\right][\because \sin (\pi-\theta)=\sin \theta]$
$=\frac{1}{4} \times \sin 60^{\circ}=\frac{1}{4} \times \frac{\sqrt{3}}{2}\left[\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$
$=\frac{\sqrt{3}}{8}= \ce{RHS} $
Hence proved.