Prove that $\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}\right)+\ldots+\cos\left(\frac{2(n-1)\pi}{n}\right)=-1$ May you help on how to start, or where to look for the following question?
By using the $n$-th roots of the unity, show that:


*

*$\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}\right)+\ldots+\cos\left(\frac{2(n-1)\pi}{n}\right)=-1$

*$\sin\left(\frac{2\pi}{n}\right)+\sin\left(\frac{4\pi}{n}\right)+\ldots+\sin\left(\frac{2(n-1)\pi}{n}\right)=0$


Can we prove the above by induction? Thank you very much in advance.
 A: Hint:
Compute 
$$e^{i\frac{2\pi}{n}} + \cdots + e^{i\frac{2(n - 1)\pi}{n}}$$
then compare real and imaginary parts.
A: An answer not (explicitly) using roots of unity, which shows the geometry of what is going on.
Consider the points around a circle, $P_k=(\cos 2k\pi/n,\sin 2k\pi/n)$, $k=0,\dots,n-1$ (note, we include the case $k=0$, which is not in your sum.)
They form a regular $n$-gon, and a simple rotational symmetry argument shows that the center of mass:
$$\frac{1}{n}\left(P_0+P_1+\cdots + P_{n-1}\right)$$
is a point on the plane fixed by a rotation of $2\pi/n$ around $(0,0)$, and there is no such point other than $(0,0)$ (if $n>1$. If $n=1$...)
This means that $$P_1 + P_2+\cdots + P_{n-1} = -P_0=(-1,0).$$
This also explains why it is hard to find an inductive proof of this - the transformation step from a regular $n$-gon to a regular $n+1$-gon is not simply "adding a point." You'd have to do a lot of moving around of the points, which makes it hard to see how you could use the result for $n$ points to get a result for $n+1$ points.
(The division by $n$ is an unnecessary part above, but it lets us evoke the intuitive term "center of mass.")
A: Consider $n$-th roots of the unity $z$, we have $1+z+z^2+\cdots+z^{n-1}=0$, i.e. $$1+(\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n})+(\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n})^2+\cdots+(\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n})^{n-1}=0,$$ by De Moivre's formula, we have $$1+\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}\right)+\ldots+\cos\left(\frac{2(n-1)\pi}{n}\right)+i\left(\sin\left(\frac{2\pi}{n}\right)+\sin\left(\frac{4\pi}{n}\right)+\ldots+\sin\left(\frac{2(n-1)\pi}{n}\right)\right)=0,$$ which implies that $$1+\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}\right)+\ldots+\cos\left(\frac{2(n-1)\pi}{n}\right)=0,$$ and $$\sin\left(\frac{2\pi}{n}\right)+\sin\left(\frac{4\pi}{n}\right)+\ldots+\sin\left(\frac{2(n-1)\pi}{n}\right)=0$$

Update: we couldn't get any relation formula on $\cos(\frac{2\pi}{n})$ and $\cos(\frac{2\pi}{n+1})$ if consider induction on $n$, even if we separate the question into even/odd parts. But the even case is easy since $$\cos\frac{2j\pi}{n}+\cos\frac{(n+2j)\pi}{n}=0,$$ hence the summation only left $\cos\pi=-1$.
A: Complement to @Solitary's answer:
Let $n\in\mathbb{N}_{\geqslant 2}$ and let define $\mu_n$ the set of the $n$th roots of unity in $\mathbb{C}$. One has: $$X^n-1=\prod_{\eta\in\mu_n}(X-\eta).$$
Examining the terms of degree $n-1$, one gets: $$-\sum_{\eta\in\mu_n}\eta=0.$$
