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prove that sin x+sin3x+sin5x+sin7x=4cosx cos2x sin 4x

 sinx+sin3x+sin5x+sin7x

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

2 sin 4x . cos x +  2 sin 4x . cos 3x

2 sin 4x (cos x+ cos 3x)

2 sin 4x (2 cos 2x . cos x)

2 cos 2x . cos x . sin 4x

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LHS= sin x+sin 3x+sin 5x+sin 7x = (sinx+sin 7x)+(sin 5x+sin 3x) =2 sin 4x.cos x+sin 4x.cos 3x =2 sin 4x[cos 3x+cos x] =2 sin 4x[2 cos 2x+cos x] =4 sin 4x.cos2x+cos x hence proved

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We first need to arrange numbers so that there sum can be easily divisible by 2 .

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= sin x+sin 3x+sin 5x+sin 7x = (sinx+sin 7x)+(sin 5x+sin 3x) =2 sin 4x.cos x+sin 4x.cos 3x =2 sin 4x[cos 3x+cos x] =2 sin 4x[2 cos 2x+cos x] =4 sin 4x.cos2x+cos x
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LHS= sinx + sin3x + sin5x +sin7x
= (sin3x + sin5x) + (sin7x+sinx)
=2sin4x . cos x + 2sin4x . cos3x
​= 2sin4x(cosx + cos3x)
= 2sin4x ( 2cos2x.cosx)
= 4 cos2x. cosx . sin4x
 
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