I'm trying to show that the derivative of Chebyshev polynomials at $x = 1$ satisfy $$T_k'(1) = k^2$$ for each $k ≥ 0$.
I can get the derivative to come out as
$$T'_k(x) = \frac{k \sin(k\theta)}{\sin(\theta)} $$
but after that it always ends up as just zero and not $k^2$....
What am I missing?