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 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.
