The value of $2 \cos x - \cos 3 x - \cos 5 x - 16 \cos^{3} x \sin^{2} x$ is
(a) 2
(b) 1
(c) 0
(d) −1

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Solution

(c) 0

$We have, 2 \cos x - \cos 3 x - \cos 5 x - 16 \cos^{3} x \sin^{2} x = 2 \cos x - \cos 3 x - \cos 5 x - 16 [\frac{\cos 3 x + 3 \cos x}{4} \times \frac{(1 - \cos 2 x)}{2}] = 2 \cos x - \cos 3 x - \cos 5 x - 2 [(\cos 3 x + 3 \cos x) (1 - \cos 2 x)] = 2 \cos x - \cos 3 x - \cos 5 x - 2 [\cos 3 x - \cos 3 x \cos 2 x + 3 \cos x - 3 \cos x \cos 2 x] = 2 \cos x - \cos 3 x - \cos 5 x - 2 [\cos 3 x + 3 \cos x] + 2 \cos 3 x \cos 2 x + 3 [2 \cos x \cos 2 x] = 2 \cos x - \cos 3 x - \cos 5 x - 2 [\cos 3 x + 3 \cos x] + \cos 5 x + \cos x + 3 \cos 3 x + 3 \cos x [2 cosAcosB = \cos (A + B) + \cos (A - B)] = 2 \cos x - \cos 3 x - \cos 5 x - 2 \cos 3 x - 6 \cos x + \cos 5 x + \cos x + 3 \cos 3 x + 3 \cos x = 0$

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