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.
 A: $cos(2\pi/7)$+$cos(4\pi/7)$+$cos(8\pi/7)$
= $cos(2\pi/7)$+$cos(4\pi/7)$+$cos(6\pi/7)$ (angles add to give $2\pi$, thus one is $2\pi$ minus the other)
At this point, we'll make an observation
$cos(2\pi/7)$$sin(\pi/7)$ = $\frac{sin(3\pi/7) - sin(\pi/7)}{2}$ ..... (A)
$cos(4\pi/7)$$sin(\pi/7)$ = $\frac{sin(5\pi/7) - sin(3\pi/7)}{2}$ ..... (B)
$cos(6\pi/7)$$sin(\pi/7)$ = $\frac{sin(7\pi/7) - sin(5\pi/7)}{2}$ ..... (C)
Now, add (A), (B) and (C) to get
$sin(\pi/7)*(cos(2\pi/7)+cos(4\pi/7)+cos(6\pi/7))$ = $\frac{sin(7\pi/7) - sin(\pi/7)}{2}$ = -$sin(\pi/7)/2$
The $sin(\pi/7)$ cancels out from both sides to give you your answer.
A: Recall the identity $$\sin \alpha \cos \beta = \frac{1}{2} \left( \sin(\alpha+\beta) + \sin(\alpha - \beta) \right).$$  Then
$$\begin{align*} \sin \frac{2\pi}{7} \left( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} \right)
&= \frac{1}{2} \left( \sin \frac{4\pi}{7} +  \sin \frac{6\pi}{7} + \sin \frac{-2\pi}{7} + \sin \frac{10\pi}{7} + \sin \frac{-6\pi}{7} \right), \\
\end{align*}$$
and observing that $$\sin \frac{10\pi}{7} = \sin \left(2\pi - \frac{4\pi}{7}\right) = \sin \frac{-4\pi}{7} = -\sin \frac{4\pi}{7}, \\
\sin \frac{-2\pi}{7} = - \sin \frac{2\pi}{7}, \\
\sin \frac{-6\pi}{7} = - \sin \frac{6\pi}{7},$$
we have
$$\sin \frac{2\pi}{7} \left( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} \right) = -\frac{1}{2} \sin \frac{2\pi}{7},$$ where upon dividing both sides by $\sin 2\pi/7$, we obtain the desired result.
A: Consider the polynomial $x^6+x^5+\dots+x+1$. This has roots at $e^{i2n\pi/7} = cos(2n\pi/7)+ isin(2n\pi/7)$ for $n = 1,2,\dots,7$.
Since $cos(x) = cos(2\pi - x)$ the sum you want is just half of the real part of the sum of the roots of this polynomial.  Now by Vieta's formula the sum of the roots is $-1$, hence the sum is $-1/2$.
A: Here a solution solving a cubic equation in order to extend a bit the statement of the post.
$$\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 $X_i$)
