Question

Prove that $\cos 2x\cos \frac{x}{2}-\cos 3x\cos \frac{9x}{2}=\sin 5x\sin \frac{5x}{2}$

Answer 1

$L.H.S.=\cos 2x\cos \frac{x}{2}-\cos 3x\cos \frac{9x}{2}$
$=\frac{1}{2}(\cos (2x+\frac{x}{2})+\cos (2x-\frac{x}{2}))-\frac{1}{2}(\cos (3x+\frac{9x}{2})+\cos (3x-\frac{9x}{2}))=\frac{1}{2}(\cos \frac{5x}{2}+\cos \frac{3x}{2}-\cos \frac{15x}{2}-\cos \frac{3x}{2})=\frac{1}{2}(\cos \frac{5x}{2}-\cos \frac{15x}{2})=\frac{1}{2}(-2\sin \left(\frac{\frac{5x}{2}+\frac{15x}{2}}{2}\right)\sin \left(\frac{\frac{5x}{2}-\frac{15x}{2}}{2}\right))=\sin \left(\frac{20x}{4}\right)\sin \left(\frac{10x}{4}\right)=\sin \left(5x\right)\sin \left(\frac{5x}{2}\right)=R.H.S$

Hence proved