Prove that $\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$ 
Prove that $$\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$$

My attempt, 
Let an equation $x^{2n+1}-1=0$, which has roots $$\cos \frac{2\pi}{2n+1}+i \sin \frac{2\pi}{2n+1}, \cos \frac{4\pi}{2n+1}+i \sin \frac{4\pi}{2n+1},...,\cos \frac{4n\pi}{2n+1}+i \sin \frac{4n\pi}{2n+1}$$
$$(1+\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+...+\cos \frac{4n\pi}{2n+1})+(\sin \frac{2\pi}{2n+1}+\sin \frac{4\pi}{2n+1}+...+\sin \frac{4n\pi}{2n+1})i=0$$
$$\cos \frac{2\pi}{2n+1}+ \cos \frac{4\pi}{2n+1}+...+ \cos \frac{4n\pi}{2n+1}=-1$$
Since $$\cos \frac{2\pi}{2n+1}=\cos \frac{4n\pi}{2n+1}$$
$$\cos \frac{4\pi}{2n+1}=\cos \frac{(4n-2)\pi}{2n+1}$$
So, $$2(\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+...+ \cos \frac{2n\pi}{2n+1})=-1$$
which proves that $$\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$$
My question:
Is my attempt with using complex number too tedious and long? Is there another way to solve this question?

Let $\theta=\dfrac {2\pi}{2n+1}$.
Consider the summation 
$$\sum_{r=1}^{2n+1}e^{ir\theta}$$
as plotted out on the Argand diagram for each step of the summation. 
The summation traces out, starting from $(0,0)$, a regular polygon with $(2n+1)$ sides and unit side length. The last segment is from $(-1,0)$ to $(0,0)$. The polygon is symmetrical about $x=-\frac 12$, and has $n$ non-horizontal segments on either side of this line. 
The topmost point of the polygon has coordinates
 $$\left(\sum_{r=1}^n \cos r\theta, \sum_{r=1}^n \sin r\theta\right)=\left(-\frac12, H\right)$$ 
by symmetry. 
Hence $$\sum_{r=1}^n \cos r\theta=\sum_{r=1}^n \cos \frac {2\pi r}{2n+1}=\color{red}{-\frac 12}$$
An example is shown below for the case where $n=3$ (i.e. $2n+1=7$).


Note
This method can also be used to find $\displaystyle\sum_{r=1}^n\sin\frac {2\pi r}{2n+1}$. 
Assume that the polygon has radius  $r$ and interior angle $2\alpha$. 
Given that the exterior angle is $\theta$, we have
$\alpha=\frac {\pi}2-\frac \theta2$. Hence, $\sin\alpha=\cos\frac\theta 2$ and $\cos\alpha=\sin\frac\theta 2$. Also, we know that $r\cos\alpha=\frac 12$. 
It can be shown by simple geometry that 
$$\small H=r(1+\sin\alpha)=\frac {1+\sin\alpha}{2\cos\alpha}=\frac {1+\cos\frac\theta 2}{2\sin\frac \theta 2}=\frac {2\cos^2\frac \theta4}{4\sin\frac\theta 4\cos\frac\theta 4}=\frac{\cos\frac\theta 4}{\sin\frac\theta 4}=\frac 12\cot\frac\theta 4=\frac 12 \cot\frac{\pi}{2(2n+1)}$$
Hence
$$\sum_{r=1}^n \sin r\theta=\sum_{r=1}^n\sin\frac {2\pi r}{2n+1}=\frac 12\cot\frac {\pi}{2(2n+1)}$$
