Question:
Prove that
$\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$
Solution:
=Sin3x+sin2x-sinx
$=(\sin 3 x-\sin x)+\sin 2 x$
$=\left(2 \cos \frac{3 x+x}{2} \sin \frac{3 x-2 x}{2}\right)+\sin 2 x$
$=2 \cos 2 x \sin x+\sin 2 x$
$=2 \cos 2 x \sin x+2 \sin x \cos x$
$=2 \sin x(\cos 2 x+\cos x)$
$=2 \sin x\left(2 \cos \frac{2 x+x}{2} \cos \frac{2 x-x}{2}\right)$
$=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$
Using the formula,
$\sin A-\sin B=2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$
$\cos A+\cos B=2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$
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