Solving problem with double angle formula and factor formula

Find the value of
$\displaystyle (\ \text{i}\ )\ \ \ {{\sin }^{4}}\frac{\pi }{8}+{{\cos }^{4}}\frac{\pi }{8}+{{\sin }^{4}}\frac{{7\pi }}{8}+{{\cos }^{4}}\frac{{7\pi }}{8}$

$\displaystyle (\ ii\ )\ \frac{3}{{{{{\sin }}^{2}}{{{20}}^{\circ }}}}-\frac{1}{{{{{\cos }}^{2}}{{{20}}^{\circ }}}}+64{{\sin }^{2}}{{20}^{\circ }}$

Solution

$\displaystyle (\ \text{i}\ )\ \ \ {{\sin }^{4}}\frac{\pi }{8}+{{\cos }^{4}}\frac{\pi }{8}+{{\sin }^{4}}\frac{{7\pi }}{8}+{{\cos }^{4}}\frac{{7\pi }}{8}$

$\displaystyle \ \ \ \ \ ={{\left( {{{{\sin }}^{2}}\frac{\pi }{8}+{{{\cos }}^{2}}\frac{\pi }{8}} \right)}^{2}}-2{{\sin }^{2}}\frac{\pi }{8}{{\cos }^{2}}\frac{\pi }{8}+{{\left( {{{{\sin }}^{2}}\frac{{7\pi }}{8}+{{{\cos }}^{2}}\frac{{7\pi }}{8}} \right)}^{2}}-2{{\sin }^{2}}\frac{{7\pi }}{8}{{\cos }^{2}}\frac{{7\pi }}{8}$ 

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

$\displaystyle \ \ \ \ \ =2-\frac{1}{2}{{\sin }^{2}}\frac{\pi }{4}-\frac{1}{2}{{\sin }^{2}}\frac{{7\pi }}{4}$

$\displaystyle \ \ \ \ \ =2-\frac{1}{2}\left( {{{{\sin }}^{2}}\frac{\pi }{4}+{{{\sin }}^{2}}\frac{{7\pi }}{4}} \right)$

$\displaystyle \ \ \ \ \ =2-\frac{1}{2}\left( {{{{\sin }}^{2}}\frac{\pi }{4}+{{{\sin }}^{2}}\left( {2\pi -\frac{\pi }{4}} \right)} \right)$

$\displaystyle \ \ \ \ \ =2-\frac{1}{2}\left( {{{{\sin }}^{2}}\frac{\pi }{4}+{{{\sin }}^{2}}\frac{\pi }{4}} \right)\ \ \ \ \left[ {\because \sin \left( {2\pi -\theta } \right)=-\sin \theta \ ,\ {{{\sin }}^{2}}\left( {2\pi -\theta } \right)={{{\sin }}^{2}}\theta } \right]$

$\displaystyle \ \ \ \ \ =2-{{\sin }^{2}}\frac{\pi }{4}$

$\displaystyle \ \ \ \ \ =2-\frac{1}{2}=\frac{3}{2}$ 

$\displaystyle (\ ii\ )\ \frac{3}{{{{{\sin }}^{2}}{{{20}}^{\circ }}}}-\frac{1}{{{{{\cos }}^{2}}{{{20}}^{\circ }}}}+64{{\sin }^{2}}{{20}^{\circ }}$ 

$\displaystyle \ \ \ \ =\frac{{3{{{\cos }}^{2}}{{{20}}^{\circ }}-{{{\sin }}^{2}}{{{20}}^{\circ }}+64{{{\sin }}^{4}}{{{20}}^{\circ }}{{{\cos }}^{2}}{{{20}}^{\circ }}}}{{{{{\sin }}^{2}}{{{20}}^{\circ }}{{{\cos }}^{2}}{{{20}}^{\circ }}}}$ 

$\displaystyle \ \ \ \ =\frac{{3\left( {1-{{{\sin }}^{2}}{{{20}}^{\circ }}} \right)-{{{\sin }}^{2}}{{{20}}^{\circ }}+16{{{\sin }}^{2}}{{{20}}^{\circ }}\left( {4{{{\sin }}^{2}}{{{20}}^{\circ }}{{{\cos }}^{2}}{{{20}}^{\circ }}} \right)}}{{\frac{1}{4}\left( {4{{{\sin }}^{2}}{{{20}}^{\circ }}{{{\cos }}^{2}}{{{20}}^{\circ }}} \right)}}$

$\displaystyle \ \ \ \ =\frac{{3-3{{{\sin }}^{2}}{{{20}}^{\circ }}-{{{\sin }}^{2}}{{{20}}^{\circ }}+16{{{\sin }}^{2}}{{{20}}^{\circ }}{{{\left( {2\sin {{{20}}^{\circ }}\cos {{{20}}^{\circ }}} \right)}}^{2}}}}{{\frac{1}{4}{{{\left( {2\sin {{{20}}^{\circ }}\cos {{{20}}^{\circ }}} \right)}}^{2}}}}$

$\displaystyle \ \ \ \ =\frac{{3-4{{{\sin }}^{2}}{{{20}}^{\circ }}+16{{{\sin }}^{2}}{{{20}}^{\circ }}{{{\sin }}^{2}}{{{40}}^{\circ }}}}{{\frac{1}{4}{{{\sin }}^{2}}{{{40}}^{\circ }}}}$ 

$\displaystyle \ \ \ \ =\frac{{3-4{{{\sin }}^{2}}{{{20}}^{\circ }}+{{{\left( {4\sin {{{40}}^{\circ }}\sin {{{20}}^{\circ }}} \right)}}^{2}}}}{{\frac{1}{4}{{{\sin }}^{2}}{{{40}}^{\circ }}}}$ 

$\displaystyle \ \ \ \ =\frac{{3-4{{{\sin }}^{2}}{{{20}}^{\circ }}+{{{\left( {2\left( {\cos {{{20}}^{\circ }}-\cos {{{60}}^{\circ }}} \right)} \right)}}^{2}}}}{{\frac{1}{4}{{{\sin }}^{2}}{{{40}}^{\circ }}}}$ 

$\displaystyle \ \ \ \ =\frac{{3-4{{{\sin }}^{2}}{{{20}}^{\circ }}+4{{{\left( {\cos {{{20}}^{\circ }}-\cos {{{60}}^{\circ }}} \right)}}^{2}}}}{{\frac{1}{4}{{{\sin }}^{2}}{{{40}}^{\circ }}}}$ 

$\displaystyle \ \ \ \ =\frac{{12-16{{{\sin }}^{2}}{{{20}}^{\circ }}+16{{{\left( {\cos {{{20}}^{\circ }}-\frac{1}{2}} \right)}}^{2}}}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{12-16{{{\sin }}^{2}}{{{20}}^{\circ }}+16\left( {{{{\cos }}^{2}}{{{20}}^{\circ }}-\cos {{{20}}^{\circ }}+\frac{1}{4}} \right)}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{12-16{{{\sin }}^{2}}{{{20}}^{\circ }}+16{{{\cos }}^{2}}{{{20}}^{\circ }}-16\cos {{{20}}^{\circ }}+4}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{16\left( {1+{{{\cos }}^{2}}{{{20}}^{\circ }}-{{{\sin }}^{2}}{{{20}}^{\circ }}-\cos {{{20}}^{\circ }}} \right)}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$ 

$\displaystyle \ \ \ \ =\frac{{16\left( {1+\cos {{{40}}^{\circ }}-\cos {{{20}}^{\circ }}} \right)}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{16\left( {1-2\sin {{{30}}^{\circ }}\sin {{{10}}^{\circ }}} \right)}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{16\left( {\sin {{{90}}^{\circ }}-\sin {{{10}}^{\circ }}} \right)}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{32\cos {{{50}}^{\circ }}\sin {{{40}}^{\circ }}}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$ 

$\displaystyle \ \ \ \ =\frac{{32{{{\sin }}^{2}}{{{40}}^{\circ }}}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}\ \ \ \ \ \left[ {\because \cos {{{50}}^{\circ }}=\sin {{{40}}^{\circ }}} \right]$

$\displaystyle \ \ \ \ =32$