induction - Prove that $cosleft(frac{2pi}{n}right)+cosleft(frac{4pi}{n}right)+ldots+cosleft(frac{2(n-1)pi}{n}right)=-1$

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.

Answer

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.")