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!

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 $(\cos\frac{2  \pi}{7}+ i \sin \frac{2 \pi}{7})^7 = 1$)
