Exercise:
Show that, $\frac{\cos \left( \frac{2\pi }{13} \right)}{\sec \left( \frac{6\pi }{13} \right)}\cos \left( \frac{8\pi }{13} \right)=\frac{3-\sqrt{13}}{16}$
Solution: Let $\theta =\frac{2\pi }{13}\in \left[ 0,\frac{\pi }{2} \right]$ and we know that $\sec \frac{6\pi }{13}=\frac{1}{\cos \frac{6\pi }{13}}$ , then
$p=\frac{\cos \left( \frac{2\pi }{13} \right)\cos \left( \frac{8\pi }{13} \right)}{\sec \left( \frac{6\pi }{13} \right)}=\cos \left( \frac{2\pi }{13} \right)\cos \left( \frac{6\pi }{13} \right)\cos \left( \frac{8\pi }{13} \right)$
Let $z=\cos \frac{2\pi }{13}+i\sin \frac{2\pi }{13}\Leftrightarrow z+\frac{1}{z}={{e}^{i\frac{2\pi }{13}}}+{{e}^{-i\frac{2\pi }{13}}}=2\cos \frac{2\pi }{13}$
Observe that, $\cos \frac{6\pi }{13}=\cos \left( 3\frac{2\pi }{13} \right)$
So by using De Movie Law we get ${{z}^{3}}+\frac{1}{{{z}^{3}}}={{e}^{i\frac{6\pi }{13}}}+{{e}^{-i\frac{6\pi }{13}}}=2\cos \frac{6\pi }{13}$
Thus $\cos \frac{2n\pi }{13}=\frac{1}{2}\left( {{z}^{n}}+\frac{1}{{{z}^{n}}} \right)\,\,\,\,\,,\,\,n\in \mathbb{N}$ i.e $\cos \frac{6\pi }{13}=\frac{1}{2}\left( {{z}^{3}}+\frac{1}{{{z}^{3}}} \right)$ and $\cos \frac{8\pi }{13}=\frac{1}{2}\left( {{z}^{4}}+\frac{1}{{{z}^{4}}} \right)$
Therefore, $p=\frac{1}{8}\left( z+\frac{1}{z} \right)\left( {{z}^{3}}+\frac{1}{{{z}^{3}}} \right)\left( {{z}^{4}}+\frac{1}{{{z}^{4}}} \right)$ but ${{z}^{13}}=\cos \left( 2\pi \right)+i\sin \left( 2\pi \right)=1$
So $p=\frac{1}{8}\left( z+\frac{{{z}^{13}}}{z} \right)\left( {{z}^{3}}+\frac{{{z}^{13}}}{{{z}^{3}}} \right)\left( {{z}^{4}}+\frac{{{z}^{13}}}{{{z}^{4}}} \right)=\frac{1}{8}\left( z+{{z}^{12}} \right)\left( {{z}^{3}}+{{z}^{10}} \right)\left( {{z}^{4}}+{{z}^{9}} \right)$
Hence $p=\frac{1}{8}\left( {{z}^{31}}+{{z}^{26}}+{{z}^{24}}+{{z}^{20}}+{{z}^{19}}+{{z}^{15}}+{{z}^{13}}+{{z}^{8}} \right)$ but ${{z}^{13}}={{e}^{i2\pi }}=1$
$\Rightarrow p=\frac{1}{8}{{z}^{13}}\left( {{z}^{18}}+{{z}^{13}}+{{z}^{11}}+{{z}^{7}}+{{z}^{6}}+{{z}^{2}}+{{z}^{-5}}+1 \right)$ but ${{z}^{-5}}={{z}^{8}}={{e}^{-i\frac{3\pi }{13}}}$
$\Rightarrow p=\frac{1}{8}\left( {{z}^{18}}+{{z}^{11}}+{{z}^{7}}+{{z}^{6}}+{{z}^{8}}+{{z}^{2}}+2 \right)$ but ${{z}^{18}}={{z}^{5}}.{{z}^{13}}={{z}^{5}}$
Hence $p=\frac{1}{18}\left( {{z}^{11}}+{{z}^{8}}+{{z}^{7}}+{{z}^{6}}+{{z}^{5}}+{{z}^{2}}+2 \right)$ (*)
Let $X={{z}^{11}}+{{z}^{8}}+{{z}^{7}}+{{z}^{6}}+{{z}^{5}}+{{z}^{2}}\,\,\And \,\,Y={{z}^{12}}+{{z}^{10}}+{{z}^{9}}+{{z}^{4}}+{{z}^{3}}+z$
Observe that $z+{{z}^{12}}={{e}^{i\frac{2\pi }{13}}}+{{e}^{i\frac{24\pi }{13}}}$ but $\frac{24\pi }{13}=\frac{26\pi -2\pi }{13}=2\pi -\frac{2\pi }{13}$
Hence ${{e}^{i\frac{24\pi }{13}}}=\cos \frac{24\pi }{13}+i\sin \frac{24\pi }{13}=\cos \left( 2\pi -\frac{2\pi }{13} \right)+i\sin \left( 2\pi -\frac{2\pi }{13} \right)={{e}^{-i\frac{2\pi }{13}}}$
So $z+{{z}^{12}}=2\cos \frac{2\pi }{13}$ also ${{z}^{2}}+{{z}^{11}}=2\cos \frac{4\pi }{13}$ , ${{z}^{3}}+{{z}^{10}}=2\cos \frac{6\pi }{13}$
${{z}^{4}}+{{z}^{9}}=2\cos \frac{8\pi }{13}$ , ${{z}^{5}}+{{z}^{8}}=2\cos \frac{10\pi }{13}$ and ${{z}^{6}}+{{z}^{7}}=2\cos \frac{12\pi }{13}$
So \[X+Y=z+{{z}^{2}}+...+{{z}^{11}}+{{z}^{12}}=\sum\nolimits_{n=0}^{11}{{{z}^{n+1}}}=2\sum\nolimits_{n=1}^{6}{\cos \frac{2n\pi }{13}}=-1\] why ??
Since \[\sum\nolimits_{n=1}^{6}{\cos n\theta =\operatorname{Re}\left( \sum\nolimits_{n=1}^{6}{{{e}^{inx}}} \right)}\] notice that this series is geometric series of ratio ${{e}^{ix}}$
So $\operatorname{Re}\left( \sum\nolimits_{n=1}^{6}{{{e}^{inx}}} \right)=\operatorname{Re}\left( {{e}^{ix}}\frac{{{e}^{i6x}}-1}{{{e}^{ix}}-1} \right)=\operatorname{Re}\left( {{e}^{ix}}\frac{{{e}^{i\frac{6x}{2}}}\left( {{e}^{i\frac{6x}{2}}}-{{e}^{-i\frac{6x}{2}}} \right)}{{{e}^{\frac{ix}{2}}}\left( {{e}^{i\frac{x}{2}}}-{{e}^{-\frac{ix}{2}}} \right)} \right)=\operatorname{Re}\left( {{e}^{i\frac{x\left( 1+6 \right)}{2}}}\frac{\sin 3x}{\sin x/2} \right)$
Hence $\sum\nolimits_{n=1}^{6}{\cos n\theta =\cos \left( \frac{7\theta }{2} \right).\frac{\sin 3\theta }{\sin \frac{\theta }{2}}}$ thus $\sum\nolimits_{n=1}^{6}{\cos \frac{2n\pi }{13}}=\cos \frac{14\pi }{26}\frac{\sin \frac{6\pi }{13}}{\sin \frac{13\pi }{4}}=-1$
Also $XY={{z}^{3}}+{{z}^{5}}+2{{z}^{6}}+{{z}^{7}}+2{{z}^{8}}+3{{z}^{9}}+2{{z}^{10}}+3{{z}^{11}}+3{{z}^{12}}+3z{{z}^{13}}$
$+3{{z}^{2}}{{z}^{13}}+2{{z}^{3}}{{z}^{13}}+3{{z}^{4}}{{z}^{13}}+2{{z}^{5}}{{z}^{13}}+{{z}^{6}}{{z}^{13}}+2{{z}^{7}}{{z}^{13}}+{{z}^{8}}{{z}^{13}}+{{z}^{10}}{{z}^{13}}$
So $XY=3\left( z+{{z}^{2}}+{{z}^{3}}+{{z}^{4}}+{{z}^{5}}+...+{{z}^{11}}+{{z}^{12}} \right)=3\left( X+Y \right)=-3$
Now we can form an equation using sum-product roots X and Y as follows:
\[\Rightarrow {{x}^{2}}-\left( X+Y \right)x+XY=0\Leftrightarrow {{x}^{2}}+x-3=0\Leftrightarrow X=\frac{-1+\sqrt{13}}{2}\,\,\,\,\,\,or\,\,\,Y=\frac{-1-\sqrt{13}}{2}\]
We have $X={{z}^{11}}+{{z}^{8}}+{{z}^{7}}+{{z}^{6}}+{{z}^{5}}+{{z}^{2}}\,=\left( {{z}^{6}}+{{z}^{7}} \right)+\left( {{z}^{5}}+{{z}^{8}} \right)+\left( {{z}^{2}}+{{z}^{11}} \right)$
$=\left( {{z}^{6}}+\frac{{{z}^{13}}}{{{z}^{6}}} \right)+\left( {{z}^{5}}+\frac{{{z}^{13}}}{{{z}^{5}}} \right)+\left( {{z}^{2}}+\frac{{{z}^{13}}}{{{z}^{2}}} \right)=\left( {{z}^{6}}+\frac{1}{{{z}^{6}}} \right)+\left( {{z}^{5}}+\frac{1}{{{z}^{5}}} \right)+\left( {{z}^{2}}+\frac{1}{{{z}^{2}}} \right)$
$=\frac{1}{2}\left[ \cos \left( \frac{12\pi }{13} \right)+\cos \left( \frac{10\pi }{13} \right)+\cos \left( \frac{4\pi }{13} \right) \right]=\frac{1}{2}\left( -\cos \frac{\pi }{13}-\cos \frac{3\pi }{13}+\cos \frac{4\pi }{13} \right)<0$
So $Y=\frac{-1-\sqrt{13}}{2}$ Therefore, $P=\frac{Y+2}{8}=\frac{1}{8}\left( \frac{3-\sqrt{13}}{2} \right)=\frac{3-\sqrt{13}}{16}$
*__________________
Idea of solution Credit to Tantalus Chokesiriphol
Idea of solution Credit to Tantalus Chokesiriphol
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