 Solution

With $x=\cos \theta$, the weighted inner product for the Chebyshev polynomials becomesThe identity $2\cos (n\theta )\cos (m\theta )=\cos ((n-m)\theta )+\cos ((n+m)\theta )$ and the vanishing of ${\int }_0^{\pi }\cos (k\theta )d\theta$ for positive integers $k$ give zero when $n\ne m$, $\pi /2$ when $n=m>0$, and $\pi$ when $n=m=0$. ThusFor Gaussian quadrature, choose the $n$ distinct nodes as the zeros of the degree-$n$ orthogonal polynomial for the weight, here $T_n$, and choose the weights as the weighted integrals of their Lagrange interpolation basis functions. If a polynomial $p$ has degree at most $2n-1$, divide it as $p=q{T}_n+r$, where both $q$ and $r$ have degree at most $n-1$. Orthogonality makes the integral of $q{T}_n$ zero, while its quadrature sum also vanishes; interpolation integrates $r$ exactly. This proves degree $2n-1$ exactness. It is optimal because the squared nodal polynomial, of degree $2n$, has zero quadrature sum and strictly positive integral.

For $n=3$, $T_3(x)=4x^3-3x$, so the nodes are $-\sqrt{3}/2,0,\sqrt{3}/2$. Their interpolation basis functions areThe weighted zeroth and second moments are $\pi$ and $\pi /2$, and the first moment is zero. Integrating these three basis functions gives equal weights $\pi /3$, hence the Chebyshev–Gauss quadrature rulewhich is exact for every polynomial of degree at most $5$.