# Q(5)  Prove that  $\small \sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos2x \sin4x$

we know that
$sinA + sinB =2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this identity in our problem
If we notice we need sin4x in our final result so it is better if we made a combination of sin7x and sin x , sin3x and  sin5x tp get sin4x

$(sin7x + sinx) + (sin5x + sin3x) = 2\sin\frac{7x+x}{2}\cos\frac{7x-x}{2}$ $+2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2}$
$=$$2\sin4x\cos3x + 2\sin4x\cos x$
take 2sin4x common
= 2sin4x(cos3x + cosx)
Now,
We know that
$cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this
$cos3x + cosx =2\cos\frac{3x+x}{2}\cos\frac{3x-x}{2}$
= $2\cos2x\cos x$
= 2sin4x($2\cos2x\cos x$)
= 4cosxcos2xsin4x = R.H.S.

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