Question

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

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sin 3x+\sin 2x-\sin x \\ =\sin 3x+2\sin \left(\frac{2x-x}{2}\right)\cos \left(\frac{2x+x}{2}\right)\left[\because \sin A-\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)\right] \\ =\sin 3x+2\sin \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right) \\ =2\sin \left(\frac{3x}{2}\right)\cos \left(\frac{3x}{2}\right)+2\sin \left(\frac{3x}{2}\right)\cos \frac{x}{2}\left[\because \sin 2A=2\sin A\cos A\right] \\ =2\cos \left(\frac{3x}{2}\right)\left[\sin \left(\frac{3x}{2}\right)\cos \left(\frac{x}{2}\right)\right] \\ =2\cos \left(\frac{3x}{2}\right)\left[2\sin \left(\frac{{\displaystyle \frac{3x}{2}}+{\displaystyle \frac{x}{2}}}{2}\right)\cos \left(\frac{{\displaystyle \frac{3x}{2}}-{\displaystyle \frac{x}{2}}}{2}\right)\right]\left[\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\right] \\ =2\cos \left(\frac{3x}{2}\right)\left[2\sin \left(x\right)\cos \left(\frac{x}{2}\right)\right] \\ =4\sin \left(x\right)\cos \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right)=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$