The Chebyshev polynomials of the first kind satisfy the recurrence relation $$ \begin{cases} T_{n}(x)=2xT_{n-1}(x)-T_{n-2}(x) \qquad n \geq 2 \\ T_{0}(x)=1, \ \ T_{1}(x)=x \\ \end{cases} $$ The polynomials $T_n(x)$ are orthogonal with respect to the weight function $\frac{1}{\sqrt{1-x^2}}$ defined on $x\in[-1,1]$. This means that $$\int_{-1}^1 T_n(x)T_m(x)\,\frac{dx}{\sqrt{1-x^2}}= \begin{cases} 0 &: n\ne m \\ l\neq 0 &: n=m \end{cases} $$ Prove that $T_n$ also satisfy a discrete orthogonality condition $$ \sum_{k=0}^{N-1}{T_i(x_k)T_j(x_k)} = \begin{cases} 0 &: i\ne j \\ l\neq 0 &: i=j \end{cases} \,\! $$ where the $x_k$ are the $N$ Chebyshev nodes of $T_N(x)$: $$x_k=\cos\left(\pi \frac{2k+1}{2N}\right), \ \ k=0,...,N-1$$ For continuous case it should be very simple, by letting $x = \cos (\theta)$ and using the defining identity $T_n(\cos (\theta)) = \cos (n\theta)$. But I can't imagine a way to prove the statement in the discrete case.
Any suggestions please?
Hint: use the formula: $$ \cos(ix)\cos(jx) = \frac{1}{2}\left(\cos((i+j)x)+\cos((i-j)x)\right)$$ and consider that: $$ \sum_{k=0}^{n}\cos(kx) = \text{Re} \sum_{k=0}^{n}\left(e^{ix}\right)^k. $$