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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 = $

Similar Questions

If $\tan \left(\frac{x}{2}\right) = \frac{m}{n}$,then the value of $m \sin (x) + n \cos (x)$ is equal to

$\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = $

The value of $\tan \left(\frac{7 \pi}{8}\right)$ is

If $\cos A = \frac{3}{4}$,then $32 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) = $

$\cot 5^o - \tan 5^o - 2 \tan 10^o - 4 \tan 20^o - 8 \cot 40^o$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module