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Question
Prove that:
cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
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Solution
Consider LHS:
\[ \cos 3A + cos 5A + \cos 7A + \cos 15A\]
\[ = 2\cos \left( \frac{3A + 5A}{2} \right) \cos \left( \frac{3A - 5A}{2} \right) + 2\cos \left( \frac{7A + 15A}{2} \right) \cos \left( \frac{7A - 15A}{2} \right) \left\{ \because \cos A + \cos B = 2\cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\cos 4A \cos\left( - A \right) + 2\cos 11A \cos\left( - 4A \right)\]
\[= 2\cos 4A \cos A + 2\cos 11A \cos 4A\]
\[ = 2\cos 4A \left\{ \cos A + \cos 11A \right\}\]
\[ = 2\cos 4A \times \left\{ 2\cos \left( \frac{A + 11A}{2} \right) \cos \left( \frac{A - 11A}{2} \right) \right\}\]
\[ = 4\cos 4A \cos 6A \cos\left( - 5A \right)\]
\[ = 4\cos 4A \cos 5A \cos 6A\]
= RHS
Hence, LHS = RHS
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