Question

Prove that: $\sin 3x+\sin 2x-\sin x=4\sin x\cos \frac{x}{2},\cos \frac{3x}{2}$.

Answer 1

$\sin 3x+\sin 2x-\sin x$

$\sin x+si2x-\sin x$

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

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

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

divide by $\frac{x}{2}$

$\sin \frac{2x}{2}=2\sin \frac{x}{2}.\cos \frac{x}{2}$

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

$\sin 3x=2\sin \frac{3y}{2}\cos \frac{3y}{2}\to$

$=2\sin \frac{3x}{2}\cos \frac{3x}{2}+\left[2\cos \frac{3x}{2}\sin \frac{x}{2}\right]$

$\Rightarrow 2\cos \frac{3x}{2}[\sin \frac{3x}{2}+\sin \frac{x}{2}]$

$x=\frac{3y}{2}$ & $y=\frac{x}{2}$

Wegel,

$\Rightarrow 2\cos \frac{3x}{2}\left[2\sin \frac{\left(4\frac{x}{2}\right)}{2}\cos \frac{\left(\frac{2x}{2}\right)}{2}\right]$

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

Hence proved