Prove that $\sum\limits_{i=1}^{k} \cos^2\left(\frac{2i\pi}{k}\right) = \frac{k}{2}$ I was looking at the solid of revolution generated by revolving $\cos(x)$ about the $x$-axis on the interval $[0, 2\pi]$, and I noticed that when the volume of the solid was approximated with $3$ or more cylinders via the disk method the approximation would equal the true volume. To prove this, I deduced that it suffices to show $$\sum_{i=1}^{k} \cos^2\left(\frac{2i\pi}{k}\right) = \frac{k}{2} \text{, } k \geq 3$$ which Wolfram Alpha confirms to be true. However, after a lot of trial and error, I am still unable to prove this. Is anyone familiar with the type of sum? Thanks in advance.
 A: One has
$$\cos^2{x}={1+\cos{2x}\over 2}$$
So the sum rewrites as
$$\sum_{i=1}^k\cos^2{2i\pi\over k}={k\over 2}+{1\over 2}\sum_{i=1}^k\cos{4i\pi\over k}$$
relabel the index so there is no confusion with $i=\sqrt{-1}$
$$\sum_{m=1}^k\cos{4m\pi\over k}=\sum_{m=1}^k\mathfrak{R}(e^{4im\pi\over k})=\mathfrak{R}\left(\sum_{m=1}^ke^{4im\pi\over k}\right)$$
Put $q=e^{4i\pi\over k}$. The sum is one of a geometric sequence of first term $q$ and reason $q$. So the sum for $q\neq 1$ i.e $k\gt 2$ is
$$q{1-q^k\over 1-q}=0$$
A: It is clearly false. Let $k=1$. Then the RHS equals $1/2$, while the LHS is reduced to $\cos^2(2\pi)=1.$
A: Hint: As already pointed out by other users the given result seems to be false without further conditions. In order to determine the value of the sum, you could use Euler's Identity
$$\cos x = \dfrac{\exp(ix)+\exp(-ix)}{2}.$$
For your problem $x = \dfrac{2\pi}{k}$. Together with the finite geometric sum
$$S = 1 +q +q^2 + \ldots + q^n = \dfrac{1-q^{n+1}}{1-q}$$
you will be able to determine a closed form solution to the sum. You can use this method for arbitrary powers of cosine and sine.
A: Not a very elegant solution, but it becomes easy if you use: $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$
Then $\cos(x)^2=\frac{e^{2ix}+2+e^{-2ix}}{4}=\frac{e^{2ix}}{2}+\frac{1}{2}+\frac{e^{-2ix}}{2}$.
Your sum becomes $$\sum_{n=1}^{k}\frac{e^{2ix}}{2}+\sum_{n=1}^{k}\frac{1}{2}+\sum_{n=1}^{k}\frac{e^{-2ix}}{2}$$ when $x=\frac{2\pi n}{k}$, and $k\geq 2$, this sum reduces to $\frac{k}{2}$ since $$\sum_{n=1}^{k}e^{\frac{2i\pi n}{k}}=\sum_{n=0}^{k-1}e^{\frac{2i\pi n}{k}}=\frac{1-1}{e^{\frac{2\pi i}{k}}-1}$$ Notice that we would be dividing by $0$ if $k=1$.
