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Question
Prove that:
cos 20° cos 40° cos 80° = \[\frac{1}{8}\]
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Solution
\[LHS = \cos 20^\circ \cos 40^\circ \cos 80^\circ\]
\[ = \frac{1}{2}\left[ 2\cos 20^\circ \cos 40^\circ \right] \cos 80^\circ\]
\[ = \frac{1}{2}\left[ \cos \left( 20^\circ + 40^\circ \right) + \cos\left( 20^\circ - 40^\circ \right) \right] \cos 80^\circ\]
\[ = \frac{1}{2}\left[ \cos 60^\circ + \cos \left( - 20^\circ \right) \right] \cos 80^\circ\]
\[ = \frac{1}{2}\cos 80^\circ\left[ \frac{1}{2} + \cos 20^\circ \right]\]
\[ = \frac{1}{4}cos 80^\circ + \frac{1}{2}\cos 80^\circ \cos 20^\circ\]
\[= \frac{1}{4}\cos 80^\circ + \frac{1}{4}\left[ 2\cos 80^\circ \cos 20^\circ \right]\]
\[ = \frac{1}{4}\cos 80^\circ + \frac{1}{4}\left[ \cos \left( 80^\circ + 20^\circ \right) + \cos \left( 80^\circ - 20^\circ \right) \right]\]
\[ = \frac{1}{4}\cos 80^\circ + \frac{1}{4}\left[ \cos 100^\circ + \cos 60^\circ \right]\]
\[ = \frac{1}{4}\cos 80^\circ + \frac{1}{4}\left[ \cos \left( 180^\circ - 80^\circ \right) + \frac{1}{2} \right]\]
\[ = \frac{1}{4}\cos 80^\circ - \frac{1}{4}\cos 80^\circ + \frac{1}{8} \left\{ \because \cos \left( 180^\circ - 80^\circ \right) = - \cos 80^\circ \right\}\]
\[ = \frac{1}{8} = RHS\]
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