Proving $\sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =\sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0. $ Is there anybody who can help me show the following? 

$$
\sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =0 
\qquad\hbox{and}\qquad 
\sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0
$$

I can only prove this claim for the sine expression when $k=0$, by grouping the output values such that they all cancel out (for both $n$ even and odd). I have thought about using complex numbers in exponential form to prove it, but I can't do it ... :P 
I would really appreciate an algerbraic proof.   

Imagine that you have $n$ oxen, all equally strong, pulling a stone on the ground.
Somebody, misguidedly as we will soon see, attached the harnesses of the oxen equally separated around the stone in such a way that the first ox is pulling in the direction $x$, the next in the direction $x+2\pi/n$, the next in the direction $x+2\cdot2\pi/n$ et cetera. When the oxen started pulling, the stone did not move at all.
A group theoretically minded passer-by observed the plight, and explained the problem. Let $\vec{F}$ be the net force applied by the oxen to the stone. It is the sum of $n$ equal length vectors 
$$\vec{F}_k=F(\cos(x+\frac{2\pi k}n)\vec{i}+\sin(x+\frac{2\pi k}n)\vec{j}),$$
where $\vec{F}_k$ is the force applied by the ox number $k$. The explanation proceeds. Let $R$ be the rotation by the angle $2\pi/n$ about the stone. We see that $R(\vec{F}_k)=\vec{F}_{k+1}$ for $k=1,2,\ldots,n-1$, and 
$R(\vec{F}_n)=\vec{F}_1$. Therefore the sum vector $\vec{F}$ is stable under the rotation $R$.
But if a plane vector is stable under rotation by an angle other than a multiple of $2\pi$ that vector must be the zero vector, for otherwise its direction changes. So, the group theorist concluded, we have $\vec{F}=\vec{0}$.
Studying the $\vec{i}$ and $\vec{j}$ components of the result 
$\vec{F}=\sum_{k=1}^n\vec{F}_k=\vec{0}$ gives your claim.

Here is what the scene could look like when $n=7$. A stone in the middle, and the arrows indicating the directions the oxen are pulling.

You see that if we rotate the image by the angle $2\pi/7$ effectively the oxen just trade places cyclically, and, as they were equally strong, nothing changes in the big picture.
