Question

Prove that:
 sin 20° sin 40° sin 60° sin 80° = \[\frac{3}{16}\]

 

Answer 1

\[LHS = \sin 20^\circ \sin 40^\circ\sin 60^\circ \sin 80^\circ\sin 60^\circ \left[ 2\sin 20^\circ \sin 40^\circ \right]\sin 80^\circ\]
\[ = \frac{1}{2} \times \frac{\sqrt{3}}{2}\left[ \cos \left( 20^\circ - 40^\circ \right) - \cos \left( 20^\circ + 40^\circ \right) \right]\sin 80^\circ\]
\[ = \frac{\sqrt{3}}{4}\left[ \cos 20^\circ - \frac{1}{2} \right]\sin 80^\circ\]
\[ = \frac{\sqrt{3}}{4}\sin 80^\circ\left[ \cos 20^\circ - \frac{1}{2} \right]\]
\[ = \frac{\sqrt{3}}{4}\sin 80^\circ \cos 20^\circ - \frac{\sqrt{3}}{8}\sin 80^\circ\]
\[ = \frac{\sqrt{3}}{4}\sin \left( 90^\circ - 10^\circ \right)\cos 20^\circ - \frac{\sqrt{3}}{8}\sin 80^\circ\]
\[ = \frac{\sqrt{3}}{4}\cos 10^\circ \cos 20^\circ - \frac{\sqrt{3}}{8}\sin\left( 80^\circ \right)\]
\[= \frac{\sqrt{3}}{8}\left[ 2\cos 10^\circ \cos 20^\circ \right] - \frac{\sqrt{3}}{8}\sin 80^\circ\]
\[ = \frac{\sqrt{3}}{8}\left[ \cos \left( 10^\circ + 20^\circ \right) + \cos \left( 10^\circ - 20^\circ \right) \right] - \frac{\sqrt{3}}{8}\sin 80^\circ\]
\[ = \frac{\sqrt{3}}{8}\left[ \cos 30^\circ + \cos \left( - 10^\circ \right) \right] - \frac{\sqrt{3}}{8}\sin 80^\circ\]
\[ = \frac{\sqrt{3}}{8}\left[ \cos 30^\circ + \cos \left( 90^\circ - 80^\circ \right) \right] - \frac{\sqrt{3}}{8}\sin 80^\circ\]
\[ = \frac{3}{16} + \frac{\sqrt{3}}{8}\sin 80^\circ - \frac{\sqrt{3}}{8}\sin 80^\circ \left[ \because \cos \left( 90^\circ - 80^\circ \right) = \sin 80^\circ \right]\]
\[ = \frac{3}{16} = RHS\]