Prove that


Prove that

$\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$




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

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

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

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

$=2 \sin x(\cos 2 x+\cos x)$

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

$=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$

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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