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Question
Prove that:
cos 40° cos 80° cos 160° = \[- \frac{1}{8}\]
Solution
\[ = \frac{1}{2}\left[ 2\cos 40^\circ \cos 80^\circ \right] \cos 160^\circ\]
\[ = \frac{1}{2}\left[ \cos \left( 40^\circ + 80^\circ \right) + \cos \left( 40^\circ - 80^\circ \right) \right] \cos 160^\circ\]
\[ = \frac{1}{2}\left[ \cos 120^\circ + \cos \left( - 40^\circ \right) \right] \cos 160^\circ\]
\[ = \frac{1}{2}\cos \left( 160^\circ \right)\left[ - \frac{1}{2} + \cos 40^\circ \right]\]
\[ = - \frac{1}{4}\cos 160^\circ + \frac{1}{2}\cos 160^\circ \cos 40^\circ\]
\[= - \frac{1}{4}\cos 160^\circ + \frac{1}{4}\left[ 2\cos 160^\circ \cos 40^\circ \right]\]
\[ = - \frac{1}{4}\cos 160^\circ + \frac{1}{4}\left[ \cos \left( 160^\circ + 40^\circ \right) + \cos \left( 160^\circ - 40^\circ \right) \right]\]
\[ = - \frac{1}{4}\cos 160^\circ + \frac{1}{4}\left[ \cos 200^\circ + \cos 120^\circ \right]\]
\[ = - \frac{1}{4}\cos 160^\circ + \frac{1}{4}\left[ \cos \left( 360^\circ - 160^\circ \right) - \frac{1}{2} \right]\]
\[ = - \frac{1}{4}\cos 160^\circ + \frac{1}{4}\cos 160^\circ - \frac{1}{8} \left[ \because \cos \left( 360^\circ - 160^\circ \right) = \cos 160^\circ \right]\]
\[ = - \frac{1}{8} = RHS\]
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