<s>[INST] 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.   
 [/INST] 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 
$$\sum_{x = 0}^{n-1}\sin\left(k + x\frac{2\pi}{n}\right) = 0.$$
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