Question

Prove that $\cos 2x.\mathrm{co}s\left(\frac{x}{2}\right)-\cos 3x.\cos \left(\frac{9x}{2}\right)=\sin 5x.\sin \left(\frac{5x}{2}\right)$

Answer 1

$L.H.S=\cos 2x.\mathrm{co}s\left(\frac{x}{2}\right)-\cos 3x.\cos \left(\frac{9x}{2}\right)$

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

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

$=\frac{1}{2}\left\{\left[\cos \frac{5x}{2}+\cos \frac{3x}{2}\right]-\left[\cos \frac{15x}{2}+\cos \frac{3x}{2}\right]\right\}$

$=\frac{1}{2}\left[\cos \frac{5x}{2}+\cos \frac{3x}{2}-\cos \frac{15x}{2}-\cos \frac{3x}{2}\right]$

$=\frac{1}{2}\left[\cos \frac{5x}{2}-\cos \frac{15x}{2}\right]$

$=\frac{1}{2}\left(2.\sin 5x.\sin \frac{5x}{2}\right)$……………………$\left(2\sin a\sin b=\cos \left(a+b\right)-\cos \left(a-b\right)\right)$

$=\sin 5x.\sin \frac{5x}{2}$.

$=R.H.S.$

Hence, it is proved that, $\cos 2x.\mathrm{co}s\left(\frac{x}{2}\right)-\cos 3x.\cos \left(\frac{9x}{2}\right)=\sin 5x.\sin \left(\frac{5x}{2}\right)$