 Solution

For $x=\cos \theta$ with $0<\theta <\pi$, differentiation of the Chebyshev polynomial givesBoth sides are polynomials in $x$, so the identity holds at the endpoints and, in fact, for every real $x$.

Order the roots of a polynomial here as $x_j=\cos {\theta }_j$, ${\theta }_j=(2j-1)\pi /(2n)$, so $x_1>\cdots >x_n$. At these simple roots,For a polynomial $q$ of degree at most $n-1$, the Lagrange interpolation polynomial isFor $x>x_1$, the Chebyshev polynomial is positive: all its roots are at most $x_1$ and its leading coefficient is positive. Hence ${\ell }_j(x)T_n^{\prime }(x_j)=T_n(x)/(x-x_j)>0$. The assumed nodal bounds and interpolation of the degree-$n-1$ polynomial $T_n^{\prime }$ now giveContinuity includes $x=x_1$. Apply this argument to $q(-x)$ and use the parity of the Chebyshev polynomial to obtain the negative side. Thus the Chebyshev nodal derivative comparison proves