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Question
Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78° = \frac{1}{16}\]
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Solution
\[LHS = \cos36° \cos42° \cos60° \cos78° \]
\[ = \frac{1}{2}\cos36° \cos60° \left( 2\cos42° \cos78° \right)\]
\[ \left[ 2\text{ cos } A\text{ cos } B = \cos\left( A + B \right) + \cos\left( A - B \right) \right]\]
\[ = \frac{1}{2}\left( \frac{\sqrt{5} + 1}{4} \right) \times \frac{1}{2}\left( \cos120° + \cos36° \right) \]
\[ = \left( \frac{\sqrt{5} + 1}{16} \right)\left( - \frac{1}{2} + \frac{\sqrt{5} + 1}{4} \right)\]
\[ = \frac{\left( \sqrt{5} + 1 \right) \left( \sqrt{5} - 1 \right)}{64}\]
\[ = \frac{5 - 1}{64}\]
\[ = \frac{1}{16}\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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