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Question
Solution
\[\text{ LHS }= \cos x \cos\frac{x}{2} - \cos 3x \cos\frac{9x}{2}\]
\[ = \frac{1}{2}\left[ 2\cos x \cos\frac{x}{2} - 2\cos 3x \cos\frac{9x}{2} \right]\]
\[ = \frac{1}{2}\left[ \cos\left( x + \frac{x}{2} \right) + \cos\left( x - \frac{x}{2} \right) - \cos\left( 3x + \frac{9x}{2} \right) - \cos\left( 3x - \frac{9x}{2} \right) \right]\]
\[ = \frac{1}{2}\left[ \cos\frac{3x}{2} + \cos\frac{x}{2} - \cos\frac{15x}{2} - \cos\frac{3x}{2} \right]\]
\[ = \frac{1}{2}\left[ \cos\frac{x}{2} - \cos\frac{15x}{2} \right]\]
\[ = \frac{1}{2}\left[ - 2\sin\left( \frac{x + 15x}{4} \right)\sin\left( \frac{x - 15x}{4} \right) \right]\]
\[ = \frac{1}{2}\left[ - 2\sin\left( 4x \right)\sin\left( - \frac{7x}{2} \right) \right]\]
\[ = \sin\left( 4x \right)\sin\left( \frac{7x}{2} \right) = \text{ RHS }\]
Hence, LHS = RHS
Notes
Disclaimer: The given question is incorrect. The correct question should be
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