Interpolation- Barycentric coefficients for nodes that are Chebyshev points of the second kind. So I came across the following theorem:
If the interpolation node are Chebyshev points of the second kind given by :
$$ x_k=\cos \left( \frac{j\pi}{n}\right) \qquad ( 0 \leq j \leq n) $$
Then the barycetric coefficients are :
$\beta_j=(-1)^j\theta_j $
Where $\theta_j=\frac{1}{2}$ when $j=0$ or $j=n$ and $\theta_j=1$ in other cases.
But how come? I can't seem to come across the proof of this, and I can't seem to get this coefficients by calculating $\beta_i=1/\prod\limits_{j=0,j\neq i}^{j=n} \left( \cos \left( \frac{i\pi}{n}\right) - \cos \left( \frac{j\pi}{n}\right)\right)$
 A: The formula for $\beta_i$ is $(X - x_i)/P_n(X) = 1/P_n'(x)$ where $P_n$ is the monic polynomial with roots $x_i = \cos \frac{i\pi}{n}$.  The polynomial is $P_n(x) = (1 + \cos (n \arccos x))/2^n$, which can be differentiated and the derivative evaluated at interpolation points. 
A: Let $\pi(x)=\prod_{j=0}^n(x-x_j)$ be the node polynomial. Then $\beta_i = 1/\pi'(x_i)$. Denote by $T_n$ the Chebyshev polynomial; $T_n(\cos(\theta))=\cos(n\theta)$. Differentiating twice we get the identity
$(1-x^2)T_n''(x) -xT_n'(x)= -n^2 T_n(x)$. 
Now, $T_n'$ has a leading coefficient $n2^{n-1}X^{n-1}$ and has zeros at the $n-1$ points $(x_j)_{0<j<n}$. Therefore $\pi(x)=\frac{1}{n2^{n-1}}(x-1)(x+1)T_n'(x)$, so 
\begin{align*} \pi'(x_i) &= \frac{1}{n2^{n-1}} (2x_iT_n'(x_i) + (x_i^2-1) T_n''(x_i)) \\
&= \frac{1}{n2^{n-1}} (n^2 T_n(x_i) + x_iT_n'(x_i)).
\end{align*}
Evaluating this at the $x_i$ gives $\beta_i = (-1)^i 2^{n-1}/n$ for 
$0<i<n$, and half this value for $i=0$ or $n$. The common factor $2^{n-1}/n$ 
simplifies in the barycentric formula.  
This proof (in a slightly different presentation) may be found in L. Trefethen's Approximation Theory and Approximation Practice, Theorem 5.2.
