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 $w_j^*=(-1)^j\sin{\frac{(2j+1)\pi}{2n+2}}$ ?
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 $$p(x) = \frac{\displaystyle\frac{1}{d_n}\sum \frac{w_j^*}{x-x_j}f_j}{\displaystyle\frac{1}{d_n}\sum \frac{w_j^*}{x-x_j}} = \frac{\displaystyle\sum \frac{w_j^*}{x-x_j}f_j}{\displaystyle\sum \frac{w_j^*}{x-x_j}}$$