I am trying to prove that the zeros of $T_n(x)$, also called the chebyshev points are,
$x_i = \cos ((2i + 1)\frac{π}{2n}) \in (−1, 1), i = 0, 1, . . . , n − 1.$
I believe I have to use the fact that $T_n(x)=\cos(n \arccos(x))$ coupled with the fact that $\cos(x)=0$ when $x=(2k+1)(π/2)$ but I'm not sure where to go from there.