how do you find the weight of barycentric lagrange interpolation of chebyshev point? The barycentric lagrange formula
$$p(x)=\frac{\sum_{j=0}^{n}\frac{w_j}{x-x_j}f_j}{\sum_{j=0}^{n}\frac{w_j}{x-x_j}}, \qquad w_j=\frac{1}{\prod_{k\neq j}(x_j-x_k)}, \qquad j=0,\ldots, n$$
This is the chebyshev point, 
$$x_j=\cos\frac{(2j+1)\pi}{2n+2}, \qquad j=0,\ldots,n.$$
How $w_j$ become $w_j^*=(-1)^j\sin{\frac{(2j+1)\pi}{2n+2}}$ ?
 A: Let ${\theta} = \frac{{\pi}}{2 n+2}$.
Using the formula $\cos  \left(a\right)-\cos  \left(b\right) =-2 \sin  \frac{a+b}{2} \sin  \frac{a-b}{2}$, we get
$$\renewcommand{\arraystretch}{2}  \begin{array}{rcl}\displaystyle  \prod _{k \neq  j} \left({x}_{j}-{x}_{k}\right)&=&\displaystyle  {\left({-2}\right)}^{n} \prod _{k = 0 , k \neq  j}^{n} \left(\sin  \left(\left(j+k+1\right) {\theta}\right) \sin  \left(\left(j-k\right) {\theta}\right)\right)\\
&=&\displaystyle  {\left({-2}\right)}^{n} \prod _{k = 0 , k \neq  j}^{n} \sin  \left(\left(j+k+1\right) {\theta}\right) \prod _{k = 0}^{j-1} \sin  \left(\left(j-k\right) {\theta}\right) \prod _{k = j+1}^{n} \sin  \left(\left(j-k\right) {\theta}\right)\\
&=&\displaystyle  {\left({-2}\right)}^{n} {\left({-1}\right)}^{n-j} \prod _{k = 0}^{j-1} \sin  \left(\left(j-k\right) {\theta}\right) \prod _{k = 0 , k \neq  j}^{n} \sin  \left(\left(j+k+1\right) {\theta}\right) \prod _{k = j+1}^{n} \sin  \left(\left(2 n+2+j-k\right) {\theta}\right)
\end{array}$$
In the first product, we define $\ell  = j-k$. In the second product, we define $\ell  = j+k+1$ and in the third product, we
define $\ell  = 2 n+2+j-k$. It remains
$$\renewcommand{\arraystretch}{2}  \begin{array}{rcl}\displaystyle  \prod _{k \neq  j} \left({x}_{j}-{x}_{k}\right)&=&\displaystyle  {\left({-2}\right)}^{n} {\left({-1}\right)}^{n-j} \prod _{\ell  = 1}^{j} \sin  \left(\ell  {\theta}\right) \prod _{\ell  = j+1 , \ell  \neq  2 j+1}^{j+n+1} \sin  \left(\ell  {\theta}\right) \prod _{\ell  = j+n+2}^{2 n+1} \sin  \left(\ell  {\theta}\right)\\
&=&\displaystyle  \frac{{2}^{n} \prod _{\ell  = 1}^{2 n+1} \sin  \left(\ell  {\theta}\right)}{{\left({-1}\right)}^{j} \sin  \left(\left(2 j+1\right) {\theta}\right)}
\end{array}$$
Finally
$${w}_{j} = \frac{{\left({-1}\right)}^{j} \sin  \left(\displaystyle  \frac{2 j+1}{2 n+2} {\pi}\right)}{\displaystyle  {2}^{n} \prod _{\ell  = 1}^{2 n+1} \sin  \left(\frac{\ell}{2n+2}\pi\right)} = \frac{w_j^*}{d_n}$$
where the denominator $d_n$ does not depend on $j$. Hence one has
$$p(x) = \frac{\displaystyle\frac{1}{d_n}\sum \frac{w_j^*}{x-x_j}f_j}{\displaystyle\frac{1}{d_n}\sum \frac{w_j^*}{x-x_j}} = \frac{\displaystyle\sum \frac{w_j^*}{x-x_j}f_j}{\displaystyle\sum \frac{w_j^*}{x-x_j}}$$
