cos x cdot cos 7 x - cos 5 x cdot cos 13 x = | vedclass

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(B) We use the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.$\cos x \cos 7 x - \cos 5 x \cos 13 x = \frac{1}{2} [2 \cos 7 x \cos x - 2 \cos 13 x

Vedclass · Accepted Answer

$2 \sin ^{2} 6 x \cdot \cos 6 x$

Vedclass · Answer

(B) We use the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.$\cos x \cos 7 x - \cos 5 x \cos 13 x = \frac{1}{2} [2 \cos 7 x \cos x - 2 \cos 13 x \cos 5 x]$$= \frac{1}{2} [(\cos 8 x + \cos 6 x) - (\cos 18 x + \cos 8 x)]$$= \frac{1}{2} [\cos 6 x - \cos 18 x]$Using the formula $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$:$= \frac{1}{2} [-2 \sin \frac{6 x + 18 x}{2} \sin \frac{6 x - 18 x}{2}]$$= - \sin 12 x \sin(-6 x)$$= \sin 12 x \sin 6 x$Since $\sin 12 x = 2 \sin 6 x \cos 6 x$,we have:$= (2 \sin 6 x \cos 6 x) \sin 6 x = 2 \sin ^{2} 6 x \cos 6 x$.

$\cos x \cdot \cos 7 x - \cos 5 x \cdot \cos 13 x = $

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