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Question
Prove that: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
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Solution
\[LHS = \cos78° \cos42° \cos36° \]
\[ = \frac{\left( 2\cos78° \cos42° \right)}{2} \cos36° \]
\[ = \frac{\cos\left( 78° + 42° \right) + \cos\left( 78° - 42° \right)}{2} \times \cos36° \]
\[ \left[ 2\text{ cos } A\text{ cos } B = \cos\left( A + B \right) + \cos\left( A - B \right) \right]\]
\[ = \frac{1}{2}\left( \cos120° + \cos36° \right) \cos36° \]
\[= \frac{1}{2}\left( - \cos\left( 180° - 120° \right) + \cos36° \right) \cos36° \]
\[ = \frac{1}{2}\left( - \cos60° + \cos36° \right) \cos36° \]
\[ = \frac{1}{2}\left( - \frac{1}{2} + \frac{\sqrt{5} + 1}{4} \right)\frac{\sqrt{5} + 1}{4}\]
\[ = \frac{1}{2} \times \frac{\sqrt{5} - 1}{4} \times \frac{\sqrt{5} + 1}{4}\]
\[ = \frac{1}{8}\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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