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Question
Solution
\[\text{ LHS }= 4\cos x \cos \left( \frac{\pi}{3} + x \right) \cos \left( \frac{\pi}{3} - x \right)\]
\[ = 2\cos x\left[ 2 \cos \left( \frac{\pi}{3} + x \right) \cos \left( \frac{\pi}{3} - x \right) \right]\]
\[ = 2\cos x\left[ \cos \left( \frac{\pi}{3} + x + \frac{\pi}{3} - x \right) + \cos \left( \frac{\pi}{3} + x - \frac{\pi}{3} + 2x \right) \right] \left[ \because 2\cos A \cos B = \cos (A + B) + \cos (A - B) \right]\]
\[ = 2\cos x\left[ \cos \frac{2\pi}{3} + \cos 2x \right]\]
\[ = 2\cos x\left[ - \frac{1}{2} + \cos 2x \right]\]
\[ = - \cos x + 2\cos x \cos 2x\]
\[ = - \cos x + \cos \left( x + 2x \right) + \cos \left( x - 2x \right)\]
\[ = - \cos x + \cos 3x + \cos\left( - x \right)\]
\[ = - \cos x + \cos 3x + \cos x\]
\[ = \cos 3x\]
\[\text{ RHS }= \cos 3x\]
Hence, LHS = RHS
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