Prove that:

(sin 3x + sin x) sin x + (cos 3x-cos x) cos x = 0

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Solution

We have L.H.S

= (sin 3x+ sin x) sin x + (cos 3x - cos x) cosx

= $[2 sin (\frac{3 x + x}{2}) cos (\frac{3 x - x}{2})]$ sin x

+ $[- 2 sin (\frac{3 x + x}{2}) sin (\frac{3 x - x}{2})]$ cos x

= [2 sin 2x cosx] sin x + [-2 sin 2x sin x] cosx

= 2 sin 2x cosx sinx - 2 sin2x cos x sin x = 0

= R.H.S.

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