Prove that:

sin x + sin 3x +sin 5x + sin 7x = 4 cos x cos 2x sin 4x

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Solution

We have L.H.S.

= sin x + sin 3x + sin 5x + sin 7x

= (sin 7x+ sin x) + (sin 5x + sin 3x)

= $[2 sin (\frac{7 x + x}{2}) cos (\frac{7 x - x}{2})] + [2 sin (\frac{5 x + 3 x}{2}) cos (\frac{5 x - 3 x}{2})]$

2 sin 4x cos 3x + 2 sin 4x cos x

$[∵ sin C + sin D = 2 s i n \frac{C + D}{2} . cos \frac{C - D}{2}]$

= 2 sin 4x [ cos 3x + cos x]

= 2 sin 4x $[2 cos (\frac{3 x + x}{2}) cos (\frac{3 x - x}{2})]$

$[∵ cos C + cos D = 2 cos \frac{C + D}{2} . cos \frac{C - D}{2}]$

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

= 4 cos x cos 2x sin 4x = R.H.S

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