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