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.
Let $k$ be an arbitrary real number. Consider the trigonometric identity
$$2 \cos A\, \sin B = \sin(A + B) - \sin(A - B)$$
Setting $A = k + x(\frac{2\pi}{n})$ and $B = \frac{\pi}{n}$, we get
$$2\cos\left(k + x\frac{2\pi}{n}\right) \sin\left(\frac{\pi}{n}\right) = \sin\left[k + \left(x + \frac{1}{2}\right)\frac{2\pi}{n}\right] - \sin\left[k + \left(x - \frac{1}{2}\right)\frac{2\pi}{n}\right]$$
Summing from $x = 0$ to $x = n-1$ and letting $u_x = k + \left(x - \frac{1}{2}\right)\frac{2\pi}{n}$, we obtain
\begin{align}&\sum_{x = 0}^{n-1} 2\cos\left(k + x\frac{2\pi}{n}\right)\sin\left(\frac{\pi}{n}\right) \\ &= \sum_{x = 0}^{n-1}\, [\sin(u_{x+1}) - \sin(u_x)] \\ &= (\sin(u_1) - \sin(u_{0})) + (\sin(u_2) - \sin(u_1)) +\cdots + (\sin(u_n) - \sin(u_{n-1}))\\ &= \sin(u_n) - \sin(u_0)\\ &= 0\end{align}
Since $\sin(\frac{\pi}{n}) \neq 0$, we deduce that
$$\sum_{x = 0}^{n-1} \cos\left(k + x\frac{2\pi}{n}\right) = 0.\tag{*}\label{eq1}$$
If we take equation \eqref{eq1} and replace $k$ with $k - \pi/2$, we get
$$\sum_{x = 0}^{n-1}\sin\left(k + x\frac{2\pi}{n}\right) = 0.$$
These are the $x$ and $y$ coordinates of the sum of the vertices of a regular polygon with $n$ sides (treated as unit vectors in $\mathbb{R}^2$). Since the set of vertices is invariant with respect to certain nontrivial rotations, so is their sum, but the only individual vector which is invariant with respect to a nontrivial rotation is the origin, so the sum is zero, hence so too with the coordinates.
The same idea can be cast in the complex plane for sums of roots of unity, where rotation is achieved by multiplying by said roots of unity (in which case the desired equality follows by taking real and imaginary parts).
