Prove the following:
sin 2x + 2 sin 4x + sin 6x = 4 ${cos}^{2}$ x sin 4x

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Solution

We have

L.H.S = sin 2x + 2 sin 4x + sin 6x

= [sin 4x+ sin 2x] + [sin 6x + sin 4x]

= 2 sin $(\frac{4 x + 2 x}{2}) cos (\frac{4 x - 2 x}{2}) + 2 sin (\frac{6 x + 4 x}{2}) cos (\frac{6 x - 4 x}{2})$

$[∵ sin C + sin D = 2 sin (\frac{C + D}{2}) cos (\frac{C - D}{2})]$

= 2 sin 3x cos x+ 2 sin 5x cos x
= 2 cos x [sin 3x+ sin 5x]

$= 2 cos x [2 sin (\frac{5 x + 3 x}{2}) . cos (\frac{5 x - 3 x}{2})]$

= 2 cos x [2 sin 4x. cos x]

= 4 ${cos}^{2}$ x sin 4x = R.H.S.

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