When does $|P_n(x)|=1$ hold?

Question

Let $P_n = \sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} (X^2-1)^k X^{n-2k}$ be the $n$-th Tchebyshev polynomial.

I want to solve $|P_n(x)|=1$ for $x\in [-1,1]$

I checked that $\large\cos\frac{k \pi}{n}$ with $k=0,\ldots,n$ fit, but why are they the only ones ?

I'm quite reluctant to use derivatives (I'd like a more simple argument).

Answer 1

It is not to hard to prove Valeriy's alternate form. This proof holds only for $x \in [-1,1]$ because we need $\cos^{-1}(x)$ to exist. \begin{align*} \cos(n \cos^{-1}(x)) &= \Re \left[ e^{i n \cos^{-1}(x)} \right] \\ &= \Re \left[ \left[ \cos (\cos^{-1}(x)) + i \sin(\cos^{-1}(x)) \right]^n \right] \\ &= \Re \left[ \left[ x \pm i \sqrt{1 - x^2} \right]^n \right] \\ &= \sum_{k \ge 0} {n \choose 2k} \left(\pm i \sqrt{1 - x^2}\right)^{2k} x^{n- 2k} \\ &= \sum_{k \ge 0} {n \choose 2k} (-1)^k (1 - x^2)^k x^{n- 2k}. \\ \end{align*}

So then \begin{align*} |P_n(x)| &= 1 \\ \implies n \cos^{-1}(x) &= k \pi \\ \implies \cos^{-1}(x) &= \frac{k \pi}{n} \\ \implies x &= \cos\left(\frac{k \pi}{n}\right). \\ \end{align*}

Answer 2

You can represent $P_n$ in alternate form: $P_n(x) = \cos(n\cos^{-1}(x))$. Now the roots of two equations for $|P_n(x)| = 1$ could be easily derived.