Prove that: 2 x x 2 - 3 x 9 x 2 = 5 x 5 x 2 - Vidyadip

Question

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

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Answer

LHS $=\cos 2 x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}$
$\Rightarrow \mathrm{LHS}=\frac{1}{2}\left\{2 \cos 2 x \cos \frac{x}{2}-2 \cos 3 x \cos \frac{9 x}{2}\right\}$
$\Rightarrow \mathrm{LHS}=\frac{1}{2}\left[\cos \left\{\left(2 x+\frac{x}{2}\right)+\cos \left(2 x-\frac{x}{2}\right)\right\}\right.$ $-\left\{\cos \left(3 x+\frac{9 x}{2}\right)+\cos \left(\frac{9 x}{2}-3 x\right)\right\} ]$[Using: 2 cos A cos B = cos (A + B) + cos (A - B)]
$\Rightarrow \mathrm{LHS}=\frac{1}{2}\left\{\cos \frac{5 x}{2}+\cos \frac{3 x}{2}-\cos \frac{15 x}{2}-\cos \frac{3 x}{2}\right\}$
$\Rightarrow \mathrm{LHS}=\frac{1}{2}\left\{\cos \frac{5 x}{2}-\cos \frac{15 x}{2}\right\}$
$\left. { \Rightarrow \quad {\text{LHS}} = \frac{1}{2}\left\{ {2\sin \left( {\frac{{\frac{{5x}}{2} + \frac{{15x}}{2}}}{2}} \right)\sin \left( {\frac{{\frac{{15x}}{2} - \frac{{5x}}{2}}}{2}} \right)} \right.} \right\}$ $\left[\because \cos C-\cos D=2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}\right]$
$\Rightarrow $ LHS = sin (5x) sin ($\frac {5x} {2}$) = RHS
Hence proved.

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