Question -
Answer -
Let us consider LHS:
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= (sin 3x) (sin x) + sin2 x + (cos 3x)(cos x) – cos2 x
= [(sin 3x) (sin x) + (cos 3x) (cos x)] + (sin2 x– cos2 x)
= [(sin 3x) (sin x) + (cos 3x) (cos x)] – (cos2 x– sin2 x)
= cos (3x – x) – cos 2x
We know, cos 2x = cos2 x –sin2 x
cos A cos B + sin A sin B = cos(A – B)
So,
= cos 2x – cos 2x
= 0
= RHS
Hence Proved.