Prove that: $\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x$

We use the formula: $\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$.
$L.H.S. = (\sin 7x + \sin x) + (\sin 5x + \sin 3x)$
$= 2 \sin \left( \frac{7x+x}{2} \right) \cos \left( \frac{7x-x}{2} \right) + 2 \sin \left( \frac{5x+3x}{2} \right) \cos \left( \frac{5x-3x}{2} \right)$
$= 2 \sin 4x \cos 3x + 2 \sin 4x \cos x$
$= 2 \sin 4x (\cos 3x + \cos x)$
Using $\cos C + \cos D = 2 \cos \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right)$:
$= 2 \sin 4x \left[ 2 \cos \left( \frac{3x+x}{2} \right) \cos \left( \frac{3x-x}{2} \right) \right]$
$= 2 \sin 4x [ 2 \cos 2x \cos x ]$
$= 4 \cos x \cos 2x \sin 4x = R.H.S.$