Simplify $2\cos(6\pi/7) + 2\cos(2\pi/7) + 2\cos(4\pi/7) + 1$ . I am trying to simplify 
$$2\cos(6\pi/7) + 2\cos(2\pi/7) + 2\cos(4\pi/7) + 1$$ 
However if I plug this in the calculator the answer is zero. Is there a way to keep on simplifying without the calculator?
I know the identity $2\cos(\theta) = (e^{i\theta} +e^{-i\theta}) $ but I think that might make it worst.
Thanks!
 A: using $\cos x = \frac{e^{ix}+e^{-ix}}2$ we have
$$
S=\sum_{n=-3}^3e^\frac{2ni\pi}7
$$
this is the sum of the roots of $x^7-1=0$ which is zero
A: HINT:
Your sum equals
$$\sum_{k=0}^6 \cos\frac{2 k \pi}{7}$$
because $\cos (\theta) = \cos (2\pi - \theta)$, so for instance $\cos\frac{4 \pi}{7} = \cos\frac{10 \pi}{7}$.  Now this sum equals the real part of 
$$\sum_{k=0}^6 (\cos\frac{2 k \pi}{7}+ i \sin \frac{2 k \pi}{7}) = \sum_{k=0}^6  (\cos\frac{2  \pi}{7}+ i \sin \frac{2 \pi}{7})^k$$
and the last sum equals $0$ ( sum of a geometric progression, and use $(\cos\frac{2  \pi}{7}+ i \sin \frac{2 \pi}{7})^7 = 1$)
A: Using Lagrange's trigonometric identity (https://en.wikipedia.org/wiki/List_of_trigonometric_identities):
\begin{equation}
\sum_{n=1}^N \cos(n\theta)= -\frac{1}{2} + \frac{\sin\left(\left(N+\frac{1}{2}\right)\theta\right)}{2 \sin \left(\frac{\theta}{2}\right)} \,,
\end{equation}
with $\theta = 2\pi /7$ and $N=3$, we find that the first three terms of the quantity you want to simplify are equal to $-1$ since the second term in the above formula is zero:
\begin{equation}
\sin\left(\left(N+\frac{1}{2}\right)\theta\right) = \sin \left(\frac{7}{2} \frac{2\pi}{7}\right) =\sin\pi=0
\end{equation}
A: $$2\cos\left(\frac{6\pi}{7}\right)+2\cos\left(\frac{2\pi}{7}\right)+2\cos\left(\frac{4\pi}{7}\right)+1=$$ 
$$2\left(\cos\left(\frac{6\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{2\pi}{7}\right)\right)+1=$$ 
$$2\left(\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right)\right)+1=$$ 
$$2\left(\sum_{n=1}^{3}\cos\left(\frac{2n\pi}{7}\right)\right)+1=$$ 
$$2\left(\frac{\csc\left(\frac{\pi}{7}\right)\sin\left(\frac{\pi+6\pi}{7}\right)-1}{2}\right)+1=$$ 
$$2\left(\frac{0-1}{2}\right)+1=$$ 
$$2\left(-\frac{1}{2}\right)+1=$$ 
$$-1+1=0$$ 
