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$4 \cos 6 θcos 4 θcos 2 θ =$

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$\cos 12 θ + \cos 8 θ + \cos 4 θ + 1$

$\cos 12 θ - \cos 8 θ + \cos 4 θ + 1$

$\cos 12 θ - \cos 8 θ - \cos 4 θ - 1$

$\cos 12 θ + \cos 8 θ - \cos 4 θ + 1$

answer is A.

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Detailed Solution

$\begin{array}{l} = 2 (\cos 6 θcos 4 θ) \cos 2 θ \\ = 2 (\cos 10 θ + \cos 2 θ) \cos 2 θ \\ = 2 \cos 10 θcos 2 θ + 2 \cos^{2} 2 θ \\ = \cos 12 θ + \cos 8 θ + \cos 4 θ + 1 \end{array}$

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