Extrema of Chebyshev polynomials (of the first kind) I can hardly find a proof why the extrema of the Chebyshev polynomials are 
$$
x_k=\cos(\frac{k}{n}\pi), k=1,...n
$$
and also why there are $n+1$ of them. The Chebyshev polynomials are here defined as $$T_0=1, T_1=x, T_{n+1}=2xT_n(x)-T_{n-1}(x)$$ with $$T_n(x)=\cos(n\arccos (x))$$ for $x\in [-1,1]$ and $n=0,1,...$
My idea:
$$
T_n'(x_i)=0 \Leftrightarrow sin(n t_i)=0 \Leftrightarrow nt_i=\pi i
$$
for $i=1,...,n$? Where do I go from here?
I would appreciate your help!

You're pretty much on the right track.
$$\frac d{dx}T_n(x)=\frac d{dx}\cos\left(n\cos^{-1}x\right)=-\sin\left(n\cos^{-1}x\right)\left(\frac{-n}{\sqrt{1-x^2}}\right)=0$$
So
$$\sin\left(n\cos^{-1}x\right)=0=\sin(k\pi)$$
Then
$$\cos^{-1}x=\frac{k\pi}n$$
And so
$$x=\cos\left(\frac{k\pi}n\right)$$
For $0\le k\le n$
If $k<0$ recall that $\cos(-x)=\cos(x)$ so that would be a duplicate solution. If $k>n$ then
$$\cos\left(\frac{k\pi}n\right)=\cos\left(\frac{(k-2n)\pi}n\right)$$
Where $|k-2n|<|k|$ so again a duplicate solution. Therefore there are only the $n+1$ solutions we found in the first place. Oh yeah, the solutions we found for $k\in\{0,n\}$ weren't necessarily valid because of the division by $\sqrt{1-x^2}=0$ but we can check that $T_n(1)=\cos\left(n\cos^{-1}(1)\right)=\cos(0)=1$ and $T_n(-1)=\cos\left(n\cos^{-1}(-1)\right)=\cos(n\pi)=(-1)^n$ are also extreme values.
