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Question
Prove that:
sin 20° sin 40° sin 80° = \[\frac{\sqrt{3}}{8}\]
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Solution
\[LHS = \sin 20^\circ\sin 40^\circ \sin 80^\circ\]
\[ = \frac{1}{2}\left[ 2\sin 20^\circ \sin 40^\circ \right]\sin 80^\circ\]
\[ = \frac{1}{2}\left[ \cos \left( 20^\circ - 40^\circ \right) - \cos \left( 20^\circ + 40^\circ \right) \right] \sin 80^\circ\]
\[ = \frac{1}{2}\left[ \cos 20^\circ - \frac{1}{2} \right] \sin 80^\circ\]
\[ = \frac{1}{2}\sin 80^\circ\left[ \cos 20^\circ - \frac{1}{2} \right]\]
\[ = \frac{1}{2}\sin 80^\circ \cos 20^\circ - \frac{1}{4}\sin 80^\circ\]
\[ = \frac{1}{2}\sin \left( 90^\circ - 10^\circ \right) \cos 20^\circ - \frac{1}{4}\sin 80^\circ\]
\[ = \frac{1}{2}\cos 10^\circ \cos 20^\circ - \frac{1}{4}\sin 80^\circ\]
\[= \frac{1}{4}\left[ 2\cos 10^\circ cos 20^\circ \right] - \frac{1}{4}\sin 80^\circ\]
\[ = \frac{1}{4}\left[ \cos \left( 10^\circ + 20^\circ \right) + \cos \left( 10^\circ - 20^\circ \right) \right] - \frac{1}{4}\sin 80^\circ\]
\[ = \frac{1}{4}\left[ \cos 30^\circ + \cos \left( - 10^\circ \right) \right] - \frac{1}{4}\sin 80^\circ\]
\[ = \frac{1}{4}\left[ \cos 30^\circ + \cos \left( 90^\circ - 80^\circ \right) \right] - \frac{1}{4}\sin 80^\circ\]
\[ = \frac{\sqrt{3}}{8} + \frac{1}{4}\sin 80^\circ - \frac{1}{4}\sin 80^\circ \left[ \because \cos \left( 90^\circ - 80^\circ \right) = \sin 80^\circ \right]\]
\[ = \frac{\sqrt{3}}{8} = RHS\]
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