<s>[INST] Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$ 
Prove that 
  $$\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$$

My attempt 
\begin{align}
\text{LHS}&=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{8\pi}7\\
&=-2\cos\frac{4\pi}7\cos\frac\pi7+2\cos^2\frac{4\pi}7-1\\
&=-2\cos\frac{4\pi}7\left(\cos\frac\pi7-\cos\frac{4\pi}7\right)-1
\end{align}
Now, please help me to complete the proof.
 [/INST] Here a solution solving a cubic equation in order to extend a bit the statement of the post.
$$\cos x+\cos 2x+\cos 4x=\cos x+(2\cos 3x\cos x)=\cos x(1+2\cos 3x)$$
When $x=\frac{2\pi}{7}$ one has $$\cos 3x=\cos(\pi-\frac{\pi}{7})=-\cos(\frac{\pi}{7})$$ so the equality becomes $$\cos(\frac{2\pi}{7})(1-2\cos(\frac{\pi}{7}))=-\frac 12\\(2\cos^2(\frac{\pi}{7})-1)(1-2\cos(\frac{\pi}{7}))=-\frac 12$$ Puting now $X=\cos x$ we get the equation $$8X^3-4X^2-4X+1=0$$ The roots are
$$X_1=-0.623489801819=\cos(\frac{5\pi}{7})\\X_2=0.222520933956=\cos(\frac{3\pi}{7})\\\color{red}{X_3=0.900968867902=\cos(\frac{\pi}{7})}$$ It follows that we also could have put, instead of the given equality $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$ the two following ones
$$\cos\frac {5\pi}{7}+ \cos\frac {10\pi}{7}+ \cos\frac {20\pi}{7}=-\frac{1}{2}\\\cos\frac {3\pi}{7}+ \cos\frac {6\pi}{7}+ \cos\frac {12\pi}{7}=-\frac{1}{2}$$
(Or even put $-X_i+n\pi; i=1,2,3$ instead of $X_i$)
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