Prove that (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
We know that
Replacing x with 3x and y with x, we get-
∴ sin 3x + sin x = 2 sin 2x cos x …(1)
Similarly,
We know that
Replacing x with 3x and y with x, we get-
∴ cos 3x - cos x = -2 sin 2x sin x …(2)
Now,
L.H.S = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= (2 sin 2x cos x) sin x + (-2 sin 2x sin x) cos x
= 2 sin 2x cos x sin x - 2 sin 2x sin x cos x
= 0
= R.H.S
Hence, L.H.S = R.H.S…Proved