Given Chebyshev nodes on interval [a,b], show that we cannot find set of equispaced points ${y}_j$ in [a,b] st for all i there exist some j, ${y}_i={x}_j$
where Chebyshev nodes are defined
${x}_i = \frac{1}{2} (a+b) - \frac{1}{2} (b-a) \cos\left(\frac{2i-1}{2n}\pi\right)$
$i = 1, ...,n$
Idea: is enough to prove said property for the points $z_i=\cos\left(\frac{2i-1}{2n}\pi\right)$. Take the first points $z_1=\cos\left(\frac{\pi}{2n}\right)$, $z_2=\cos\left(\frac{3\pi}{2n}\right)$, $z_3=\cos\left(\frac{5\pi}{2n}\right)$. Using trigonometry, express $z_2$ and $z_3$ in function of $z_1$ and prove that $z_3-z_1$ isn't a rational multiple of $z_2-z_1$.
BTW, obviously $n>2$.