How to show that $\cos\frac{2\pi}{n} + \cos\frac{4\pi}{n} + \ldots+ \cos\frac{2\pi(n-1)}{n} = -1$ for all positive integers $n$? It doesn't make sense to me, how we will even show it works for $n =1$  and we can work out induction from there but how can I show even the base case.
Thanks in Advance.
 A: The result is not true for $n=1$ but is true for $n>1$.
Let $\zeta = \text{cos}(\frac{2\pi}{n}) + i\text{sin}(\frac{2\pi}{n})$.
Then $\zeta\neq 1$ and $\zeta^n = 1$ by De Moivre. These two facts tell us that $1+\zeta+\zeta^2...+\zeta^{n-1} = 0$.
Taking real parts and using De Moivre gives the result.
A: If you want to prove the formula without complex analysis, here is the trick:
Let 
$$S:= \cos\frac{2\pi}{n} + \cos\frac{4\pi}{n} + \ldots+ \cos\frac{2\pi(n-1)}{n}$$
Then
$$\sin(\frac{\pi}{n}) S= \sin(\frac{\pi}{n})  \cos\frac{2\pi}{n} +\sin(\frac{\pi}{n})  \cos\frac{4\pi}{n} + \ldots+ \sin(\frac{\pi}{n}) \cos\frac{2\pi(n-1)}{n}$$
Now, as
$$\sin(\frac{\pi}{n}) \cos\frac{2 k\pi}{n} = \frac{1}{2} \left( \sin(\frac{(2k+1)\pi}{n}) -  \sin(\frac{(2k-1)\pi}{n}) \right) $$ 
you get a telescopic sum.
A: Move the one to the left hand side, and you get the sum of the x-components of vectors arranged symmetrically around the unit circle. Of course, this sum is 0.
A: Complex numbers with unit length are written as $e^{i\theta} = i*sin(\theta) + cos(\theta)$  
The roots of the equation $x^n = 1$ are 

$e^{(2k\pi/n)i}$ where $0\le k\lt n$

Each of these n values solve the equation $x^n=1$. So the summation of the roots is 0 (co-efficient of the $x^{n-1}$ multiplied by -1) .

$\sum\limits_{i=0}^{n-1} e^{(2k\pi/n)i}$ = 0
  => $ 1 + \sum\limits_{i=1}^{n-1} e^{(2k\pi/n)i}$ = 0

By equating the real part of both sides the required result is obtained.
For the given problem when $n=1$ the equation becomes $x=1$ and that is the only root. For an nth degree polynomial the summation of roots of the equation is the co-efficient of the $x^{n-1}$ multiplied with -1. In this case the summation is -1 and the equation does not hold. 
From $n=2$ this co-efficient becomes zero and the equation holds.
