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The value of 2 cosx

Question:

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

Solution:

(c) 0

We have,

$2 \cos x-\cos 3 x-\cos 5 x-16 \cos ^{3} x \sin ^{2} x$

$=2 \cos x-\cos 3 x-\cos 5 x-16\left[\frac{\cos 3 x+3 \cos x}{4} \times \frac{(1-\cos 2 x)}{2}\right]$

$=2 \cos x-\cos 3 x-\cos 5 x-2[(\cos 3 x+3 \cos x)(1-\cos 2 x)]$

$=2 \cos x-\cos 3 x-\cos 5 x-2[\cos 3 x-\cos 3 x \cos 2 x+3 \cos x-3 \cos x \cos 2 x]$

$=2 \cos x-\cos 3 x-\cos 5 x-2[\cos 3 x+3 \cos x]+2 \cos 3 x \cos 2 x+3[2 \cos x \cos 2 x]$

$=2 \cos x-\cos 3 x-\cos 5 x-2[\cos 3 x+3 \cos x]+\cos 5 x+\cos x+3 \cos 3 x+3 \cos x$

$[2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$

$=2 \cos x-\cos 3 x-\cos 5 x-2 \cos 3 x-6 \cos x+\cos 5 x+\cos x+3 \cos 3 x+3 \cos x=0$

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