Question

solve:

$\cos (4x+3x)-\cos (4x-3x)$

Answer 1

We have,

$\cos (4x+3x)-\cos (4x-3x)$

$=\cos 7x-\cos x$

We know that

$\cos C-\cos D=2\sin \left(\frac{C+D}{2}\right).\sin \left(\frac{D-C}{2}\right)$

Therefore,

$=2\sin \left(\frac{7x+x}{2}\right).\sin \left(\frac{x-7x}{2}\right)$

$=2\sin \left(\frac{8x}{2}\right).\sin \left(\frac{-6x}{2}\right)$

$=-2\sin \left(4x\right).\sin \left(3x\right)$

Hence, this is the answer.