Question

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

Answer 1

Prove that

$\begin{array}{c}\cos 2x\cos \frac{x}{2}-\cos 3x\cos \frac{9x}{2}=\sin 5x.\sin \frac{5x}{2}\\ L.H.S.\\ \cos 2x\cos \frac{x}{2}-\cos 3x\cos \frac{9x}{2}\end{array}$

Multiply and divide by 2, we get

$\frac{1}{2}\left[2\cos 2x\cos \frac{x}{2}-2\cos 3x\cos \frac{9x}{2}\right]$

using formulae

$\begin{array}{c}2\cos x\cos y=\cos \left(x+y\right)+\cos \left(x-y\right)weget\\ =\frac{1}{2}\left[\cos \left(2x+\frac{x}{2}\right)+\cos \left(2x-\frac{x}{2}\right)-\cos \left(3x+\frac{9x}{2}\right)-\cos \left(3x-\frac{9x}{2}\right)\right]\\ =\frac{1}{2}\left[\cos \frac{5x}{2}+\cos \frac{3x}{2}-\cos \frac{15x}{2}-\cos \frac{3x}{2}\right]\\ =\frac{1}{2}\left[\cos \frac{5x}{2}-\cos \frac{15x}{2}\right]\\ =\frac{1}{2}\left[-\left(-2\sin \frac{5x}{2}.\sin 5x\right)\right]\\ =\frac{1}{2}2\sin 5x.\sin \frac{5x}{2}\\ =\sin 5x.\sin \frac{5x}{2}R.H.S.\end{array}$

Proved.