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सिद्ध कीजिए: $\cos 2 x \cos _{2}^{x}-\cos 3 x \cos \frac{9 x}{2}=\sin 5 x \sin \frac{5 x}{2}$
Solution
We have
${\text{L}}{\text{.H}}{\text{.S}}{\text{. }} = \frac{1}{2}\left[ {2\cos 2x\cos \frac{x}{2} – 2\cos \frac{{9x}}{2}\cos 3x} \right]$
$ = {1}{2}[ \cos \left( {2x + \frac{x}{2}} \right) + \cos \left( {2x – \frac{x}{2}} \right)$
$ – \cos \left( {\frac{{9x}}{2} + 3x} \right) – \cos \left( {\frac{{9x}}{2} – 3x} \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 \left\{ {\frac{{\frac{{5x}}{2} + \frac{{15x}}{2}}}{2}} \right\}\sin \left\{ {\frac{{\frac{{5x}}{2} – \frac{{15x}}{2}}}{2}} \right\}} \right]$
$ = – \sin 5x\sin \left( { – \frac{{5x}}{2}} \right)$
$ = \sin 5x\sin \frac{{5x}}{2} = R.H.S.$
