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

It is known that $\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)$.

$\therefore$ L.H.S. $=\sin x+\sin 3 x+\sin 5 x+\sin 7 x$

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

$=2 \sin \left(\frac{x+5 x}{2}\right) \cdot \cos \left(\frac{x-5 x}{2}\right)+2 \sin \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)$

$=2 \sin 3 x \cos (-2 x)+2 \sin 5 x \cos (-2 x)$

$=2 \sin 3 x \cos 2 x+2 \sin 5 x \cos 2 x$

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

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

$=2 \cos 2 x[2 \sin 4 x \cdot \cos (-x)]$

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