Solving problem with Factor and sum formulae

$\displaystyle \text{Prove}\ \text{that}\ \ \ \frac{{\cos 8x-\cos 7x}}{{1+2\cos 5x}}=\cos 3x-\cos 2x\ .$

Solution

$\displaystyle \ \ \ \frac{{\cos 8x-\cos 7x}}{{1+2\cos 5x}}=\frac{{-2\sin \frac{{15x}}{2}\sin \frac{x}{2}}}{{1+2\left( {2{{{\cos }}^{2}}\frac{{5x}}{2}-1} \right)}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{-2\sin \frac{{15x}}{2}\sin \frac{x}{2}}}{{4{{{\cos }}^{2}}\frac{{5x}}{2}-1}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{-2\sin \frac{{15x}}{2}\sin \frac{x}{2}}}{{4{{{\cos }}^{2}}\frac{{5x}}{2}-1}}\ \ \times \ \ \frac{{-2{{{\sin }}^{2}}\frac{{5x}}{2}}}{{-2{{{\sin }}^{2}}\frac{{5x}}{2}}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\left( {-2\sin \frac{{5x}}{2}\sin \frac{x}{2}} \right)\left( {-2\sin \frac{{15x}}{2}\sin \frac{{5x}}{2}} \right)}}{{2{{{\sin }}^{2}}\frac{{5x}}{2}-8{{{\sin }}^{2}}\frac{{5x}}{2}{{{\cos }}^{2}}\frac{{5x}}{2}}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\left( {\cos 3x-\cos 2x} \right)\left( {\cos 10x-\cos 5x} \right)}}{{2{{{\sin }}^{2}}\frac{{5x}}{2}-2{{{\left( {2\sin \frac{{5x}}{2}\cos \frac{{5x}}{2}} \right)}}^{2}}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\left( {\cos 3x-\cos 2x} \right)\left( {\cos 10x-\cos 5x} \right)}}{{2{{{\sin }}^{2}}\frac{{5x}}{2}-2{{{\sin }}^{2}}5x}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\left( {\cos 3x-\cos 2x} \right)\left( {\cos 10x-\cos 5x} \right)}}{{\left( {1-\cos 5x} \right)-\left( {1-\cos 10x} \right)}}\ \ \ \ \ \ \left[ {\because \cos 2\theta =1-2{{{\sin }}^{2}}\theta } \right]$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\left( {\cos 3x-\cos 2x} \right)\left( {\cos 10x-\cos 5x} \right)}}{{\cos 10x-\cos 5x}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos 3x-\cos 2x$