Question

Prove that:
sin 10° sin 50° sin 60° sin 70° = \[\frac{\sqrt{3}}{16}\]

 

Answer 1

\[LHS = \sin 10^\circ \sin 50^\circ \sin 60^\circ \sin 70^\circ\]
\[ = \frac{1}{2}\sin 60^\circ \left[ 2\sin 10^\circ \sin 50^\circ \right]\sin 70^\circ\]
\[ = \frac{1}{2} \times \frac{\sqrt{3}}{2}\left[ \cos \left( 10^\circ - 50^\circ \right) - \cos \left( 10^\circ + 50^\circ \right) \right]\sin 70^\circ\]
\[ = \frac{\sqrt{3}}{4}\left[ \cos \left( - 40^\circ \right) - \frac{1}{2} \right]\sin 70^\circ\]
\[ = \frac{\sqrt{3}}{4}\sin 70^\circ\left[ \cos 40^\circ - \frac{1}{2} \right]\]
\[ = \frac{\sqrt{3}}{4}\sin 70^\circ \cos 40^\circ - \frac{\sqrt{3}}{8}\sin 70^\circ\]
\[ = \frac{\sqrt{3}}{4}\sin \left( 90^\circ - 20^\circ \right) \cos 40^\circ - \frac{\sqrt{3}}{8}\sin 70^\circ\]
\[ = \frac{\sqrt{3}}{4}\cos 20^\circ \cos 40^\circ - \frac{\sqrt{3}}{8}\sin 70^\circ\]
\[= \frac{\sqrt{3}}{8}\left[ 2\cos 20^\circ\cos 40^\circ \right] - \frac{\sqrt{3}}{8}\sin 70^\circ\]
\[ = \frac{\sqrt{3}}{8}\left[ \cos \left( 20^\circ + 40^\circ \right) + \cos \left( 20^\circ - 40^\circ \right) \right] - \frac{\sqrt{3}}{8}\sin 70^\circ\]
\[ = \frac{\sqrt{3}}{8}\left[ \cos 60^\circ + \cos \left( - 20^\circ \right) \right] - \frac{\sqrt{3}}{8}\sin 70^\circ\]
\[ = \frac{\sqrt{3}}{8}\left[ \cos 60^\circ + \cos \left( 90^\circ - 70^\circ \right) \right] - \frac{\sqrt{3}}{8}\sin 70^\circ\]
\[ = \frac{\sqrt{3}}{16} + \frac{\sqrt{3}}{8}\sin 70^\circ - \frac{\sqrt{3}}{8}\sin 70^\circ \left[ \because \cos \left( 90^\circ - 70^\circ \right) = \sin 70^\circ \right]\]
\[ = \frac{\sqrt{3}}{16} = RHS\]