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Question
Prove that:
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Solution
Consider LHS:
\[ \frac{\cos 3A + 2\cos 5A + \cos 7A}{\cos A + 2\cos 3A + \cos 5A}\]
\[ = \frac{\cos 3A + \cos 7A + 2\cos 5A}{\cos A + \cos 5A + 2\cos 3A}\]
\[ = \frac{2\cos \left( \frac{3A + 7A}{2} \right) \cos \left( \frac{3A - 7A}{2} \right) + 2\cos 5A}{2\cos \left( \frac{A + 5A}{2} \right) \cos \left( \frac{A - 5A}{2} \right) + 2\cos 3A}\]
\[ \]
\[ = \frac{2\cos 5A \cos \left( - 2A \right) + 2\cos 5A}{2\cos 3A \cos \left( - 2A \right) + 2\cos 3A}\]
\[ = \frac{2\cos 5A \cos 2A + 2\cos 5A}{2\cos 3A \cos 2A + 2\cos 3A}\]
\[ = \frac{2\cos 5A \left[ \cos 2A + 1 \right]}{2\cos 3A \left[ \cos 2A + 1 \right]}\]
\[ = \frac{\cos 5A}{\cos 3A}\]
= RHS
Hence, RHS = LHS.
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