Question

Please solve no. 10 with appropriate process

Answer 1

Given,

$\cos^{2}2x - \cos^{2}6x \\ \\ = (\cos 2x + \cos 6x)(\cos 2x -\cos 6x) \\ \\ = (2 \cos \frac{2x +6x}{2} . \cos \frac{2x -6x}{2}) (2\sin \frac{2x+6x}{2} . \sin \frac{6x-2x}{2}) \\ \\ = 2\cos 4x. \cos (-2x) * 2 \sin4x \sin2x \\ \\ =2\cos 4x. \cos2x * 2 \sin4x \sin2x \\ \\ =(2\sin2x . \cos2x) (2\sin4x .\cos4x) \\ \\ = \sin4x . \sin8x \\ \\ Note;- \\ \\ 1. \ \ \cos C + \cos D = 2 \cos \frac{C +D}{2}. \cos \frac{C -D}{2} \\ \\ 2. \ \ \cos C - \cos D = 2\sin \frac{C+D}{2} . \sin \frac{D-C}{2}$