<s>[INST] Calculation of $ \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right)$ Calculation of $\displaystyle \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right)$ and $\displaystyle \cos\left(\frac{2\pi}{7}\right)\times \cos\left(\frac{4\pi}{7}\right) \times \cos\left(\frac{6\pi}{7}\right)$ is 
$\bf{My\; Try::}$ Let $\displaystyle \frac{2\pi}{7}=\phi\;,$ Then $3\phi = 2\pi-4\phi\Rightarrow \cos \left(3\phi\right) = \cos \left(2\pi-4\phi\right)=\cos \left(4\phi\right)$
So $4\cos^3 \phi -3\cos \phi = 2\left[2\cos^2 \phi - 1\right]^2-1=2\left[4\cos^4 \phi+1-4\cos^2 \phi\right]-1 = 8\cos^4 \phi-8\cos^2 \phi+1$
So $8\cos^4\phi-4\cos^3 \phi-8\cos^2 \phi+3\cos \phi+1=0$
Now I did not understand how can i solve after that
Help me
Thanks 
 [/INST] The three corresponding nodes in the unit radius tetradecagon (14-gon) can be regarded as the tops of three isosceles with top angle $\alpha = \frac{\pi}{7}$, due to the equal chords:

Summing the bases of these isosceles yields the 'radius' of the 14-gon: 
$$ \color{green}{g} \color{black}{+} \color{blue}{b} \color{black}{+} \color{red}{r} \color{black}{ = 1}$$ 
But we also have that: 
$$\begin{align}\cos\left(\color{blue}{ \frac{2\pi}{7} } \right) &= \color{green}{g}\color{black}{ \,\,\,+ \,\,\,\frac{1}{2}}\color{blue}{b} \\ \cos\left(\color{green}{\frac{4\pi}{7} } \right) &= -\frac{1}{2}\color{green}{g} \\  \cos\left(\color{red}{\frac{6\pi}{7} } \right)&= -\color{green}{g} \color{black}{\,\,\,- \,\,\,}  \color{blue}{b}\color{black}{ \,\,\,- \,\,\, \frac{1}{2}}\color{red}{r} \end{align}$$
Adding these gives 
$$\cos\left(\color{blue}{ \frac{2\pi}{7} } \right) + \cos\left(\color{green}{\frac{4\pi}{7} } \right) +   \cos\left(\color{red}{\frac{6\pi}{7} } \right) = -\frac{1}{2}\left(\color{green}{g} \color{black}{+} \color{blue}{b} \color{black}{+} \color{red}{r}\right)= -\frac{1}{2}$$
</s>