Prove that $\cos(2\pi/n)+\cos(4\pi/n)+\cdots+\cos(2(k-1)\pi/n)=-1$ How do you prove that 
$$\cos(2\pi/n)+\cos(4\pi/n)+\cdots+\cos(2(n-1)\pi/n)=-1,$$ where $n \geq 2$? 
 A: Hint: This identity is a real part of an analogous identity between complex numbers. 


*

*First sub-hint: $\mathrm e^{\mathrm i\theta}=\cos(\theta)+\mathrm i\sin(\theta)$.

*Second sub-hint: $1+z+\cdots+z^{n-1}=\frac{1-z^n}{1-z}$ for every complex number $z\ne1$.

A: Will Jagy's solution is the classic, but if you don't want to use complex numbers, then one can argue for the same result by thinking of $\cos(2k\pi/n)$ as the $x$-coordinates of a set of $n$ points equidistributed about the circle. Then their sum is equal to $n$ times their average value; but since they're equidistributed then the average is 0. Ultimately, of course, this reasoning is no different from the complex approach, but sometimes I like pictures.
A: Using the following identity: $$\cos(a)+\cos(a+b)+\cos(a+2b)+...+\cos(a+(n-1)b)=\frac{\cos\big(a+(n-1)\frac{b}{2}\big)\sin\big(\frac{nb}{2}\big)}{\sin\big(\frac{b}{2}\big)}$$ and taking $a=0$ and $b=2\pi/n$; your statement can be proved. This way is very similar to Saha's spotless answer.
A: A proof using trigonometric identity proceeds by multiplying the identity with $2 \sin\left(\frac{\pi}{n}\right)$ and using $2 \cos(\alpha) \sin(\beta) = \sin(\alpha+\beta) - \sin(\alpha-\beta)$:
$$ \begin{eqnarray}
   \sum_{m=1}^{n-1} 2 \sin\left(\frac{\pi}{n} \right) \cos\left(\frac{2 \pi}{n} m\right) &=& \sum_{m=1}^{n-1} \left(\underbrace{\sin\left( \frac{\pi}{n} (2m+1) \right)}_{f_{m+1}} - \underbrace{\sin\left( \frac{\pi}{n} (2m-1) \right)}_{f_m} \right) = f_n-f_1 \\ 
  &=& \sin\left( \frac{\pi}{n} \left(2n-1\right) \right) - \sin\left( \frac{\pi}{n} \right) \\
   &=& \sin\left( 2\pi - \frac{\pi}{n}  \right) - \sin\left( \frac{\pi}{n} \right) = -2 \sin\left(\frac{\pi}{n}\right)
\end{eqnarray}
$$
We thus proved:
$$
  2 \sin\left(\frac{\pi}{n} \right) \left\{ \cos\left(\frac{2\pi}{n}\right) + \cos\left(\frac{4\pi}{n}\right) + \cdots +  \cos\left(\frac{2\pi}{n} (n-1)\right)  \right\} = -2 \sin\left(\frac{\pi}{n} \right)
$$
Choosing $n>1$,  $\sin\left(\frac{\pi}{n} \right) \not=0$, establishing the identity.
A: Take $\omega = \cos 2 \pi / n + i \sin 2 \pi / n. $ Then $\omega^n = 1,$ or
$ \omega^n - 1 = 0,$ or
$$ (\omega - 1) ( 1 + \omega + \omega^2 + \omega^3 + \cdots + \omega^{n-1} )= 0. $$ As $\omega \neq 1,$ it follows that 
$$  1 + \omega + \omega^2 + \omega^3 + \cdots + \omega^{n-1} = 0.   $$
A: If $p_i,\ 1\leq i\leq n$ on unit circle forms a regular $n$-gon, then define $R$ to be $\frac{2\pi}{n}$-rotation. Hence $p_1+\cdots + p_n$ is invariant under $R$ so that it is $0$-vector.
