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!
On
We have $\left|T_n(x)\right|\leq 1$ for $|x|\leq 1$. So, if $a\in[-1,1]$ is such that $T_n(a)=\pm 1$, we have an extremum at $x=a$. You can check that $T_n(x_k)=\cos(k\pi)=\pm 1$ which means $x_k$ for $k=0,1,\ldots,n$ are extremum points. To show we don't have other extrema, we can use the derivative $$\left(T_n(x)\right)'=n\frac{\sin(n\arccos(x))}{\sqrt{1-x^2}} $$ Setting $\left(T_n(x)\right)'=0$, we find the critical points in $(-1,1)$ are precisely $x_k\; (k=1,\ldots,n-1)$, and we already know those are extremum points.
On
Let's examine $$ \frac{\mathrm{d}}{\mathrm{d}x} \cos(n \arccos x) = \frac{n}{\sqrt{1-x^2}} \sin(n \arccos x) \text{.} $$ At extrema, this expression is zero or undefined. The first factor is never zero, but it is undefined at $x = \pm 1$ (which are extrema of $T_n$). The second is zero precisely when $n \arccos x$ is an integer multiple of $\pi$, or \begin{align*} \sin(n \arccos x) &= 0 \\ n \arccos x &= \{\arcsin(0), \pi - \arcsin(0)\} + 2\pi \ell, \ell \in \Bbb{Z} \\ &= \{0, \pi\} + 2\pi \ell, \ell \in \Bbb{Z} \\ &= k \pi, k \in \Bbb{Z} \\ \arccos x &= \frac{k \pi}{n}, k \in \Bbb{Z} \text{,} \end{align*} but the range of arccosine is $[0,\pi]$, so $0 \leq k \leq n$. Then, $$ x = \cos\left( \frac{k \pi}{n} \right), k \in [0,n] \cap \Bbb{Z} \text{.} $$ When $k = 0$, $x = 1$ and when $k = n$, $x = -1$, so the two critical points which were points where the derivative was undefined are duplicated in this list of $n+1$ points.
On
We have
$$T_n(\cos x) = \cos(nx)$$ and hence
$$-T_n'(\cos x)\cdot\sin x = -n \sin (nx) $$
$$T_n'(\cos x) = n\frac{\sin(nx)}{\sin x} $$
The right side is zero if $nx$ is an integral multiple of $\pi$ but $x$ is not: $$x=\frac{k}{n}\pi \quad\text{ with }\quad 0<k<n, k/n\notin \Bbb Z$$
Thus the extrema of $T_n$ are at $\cos(x_k)$ with $0 < k<n$. Values outside that interval do not add more zeros due to periodicity and symmetry of (co)sine. This applies also in the special case $n=0$ as then there is no $k$ satisfying the constraint. .
Often $T_n$ are just considered over the interval $[-1,1]$ in which case one has to add the extrema at the interval boundaries $x_0$ and $x_n$ provided $n>0$.
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.