Question

Prove the following:

sin 20° sin 40° sin 80° = ${\displaystyle \frac{\sqrt{3}}{8}}$

Answer 1

L.H.S. = sin 20° sin 40° sin 80° 

= sin 20°. sin 40°. sin 80°

= ${\displaystyle \frac{1}{2}(2.{\sin 40}^{\circ }.{\sin 20}^{\circ }).{\sin 80}^{\circ }}$

= ${\displaystyle \frac{1}{2}[\cos ({40}^{\circ }-{20}^{\circ })-\cos ({40}^{\circ }+{20}^{\circ })].{\sin 80}^{\circ }}$

= ${\displaystyle \frac{1}{2}\left({\cos 20}^{\circ }-{\cos 60}^{\circ }\right){\sin 80}^{\circ }}$

= ${\displaystyle \frac{1}{2}{\cos 20}^{\circ }.{\sin 80}^{\circ }-\frac{1}{2}{\cos 60}^{\circ }.{\sin 80}^{\circ }}$

= ${\displaystyle \frac{1}{2\times 2}\left(2{\sin 80}^{\circ }.{\cos 20}^{\circ }\right)-\frac{1}{2}\left(\frac{1}{2}\right).{\sin 80}^{\circ }}$

= ${\displaystyle \frac{1}{4}.[\sin ({80}^{\circ }+{20}^{\circ })+\sin ({80}^{\circ }-{20}^{\circ })]-\frac{1}{4}.{\sin 80}^{\circ }}$

= ${\displaystyle \frac{1}{4}.\left({\sin 100}^{\circ }+{\sin 60}^{\circ }\right)-\frac{1}{4}.{\sin 80}^{\circ }}$

= ${\displaystyle \frac{1}{4}{\sin 100}^{\circ }+\frac{1}{4}{\sin 60}^{\circ }-\frac{1}{4}{\sin 80}^{\circ }}$

= ${\displaystyle \frac{1}{4}.\sin ({180}^{\circ }-{80}^{\circ })+\frac{1}{4}\times \frac{\sqrt{3}}{2}-\frac{1}{4}.{\sin 80}^{\circ }}$

= ${\displaystyle \frac{1}{4}{\sin 80}^{\circ }+\frac{\sqrt{3}}{8}-\frac{1}{4}{\sin 80}^{\circ }}$ ...[∵ sin (180° – θ) = sin θ]

= ${\displaystyle \frac{\sqrt{3}}{8}}$

= R.H.S.