Alternate sum of Chebyshev polynomials

Question

The problem is

For all integer $n\ge1$, \begin{align}\frac{(-1)^n}{2^{n-1}}\left(\frac12+\sum _{k=1}^n (-1)^{k} T_k(x)\right)&=\prod _{j=0}^{n-1} \left(x-\cos \left(\frac{\pi  (2 j+1)}{2 n+1}\right)\right)\\&=\pm2^{-n}\sqrt{\frac{T_{2n+1}(x)+1}{x+1}}\end{align}

My attempt:

They are monic polynomials of same degree, I only need verify they have same roots.

The roots of right side are $x=\cos\frac{(2j+1)π}{2n+1}\;,j=0,1,⋯,n-1$.

So I need prove $$\sum_{k=1}^n(-1)^{k+1}\cos\left(k\frac{(2j+1)π}{2n+1}\right)=\frac12\quad,j=0,1,⋯,n-1$$

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$x=\cos\frac{(2j+1)π}{2n+1}\;,j=0,1,⋯,n-1$ are roots of $T_{2n+1}(x)=-1$.
Since $\cos\frac{(2j+1)π}{2n+1}$ are extrema of $T_{2n+1}(x)$, they are roots of its derivative: $$T_{2n+1}'(x)=2^{2n}(2n+1)\prod _{j=0}^{n-1} \left(x-\cos \left(\frac{\pi  (2 j+1)}{2 n+1}\right)\right)\prod _{j=0}^{n-1} \left(x+\cos \left(\frac{\pi  (2 j+1)}{2 n+1}\right)\right)$$

Answer 1

Replacing $j$ with $n-j$, $$\sum_{k=1}^n(-1)^{k+1}\cos\left(k\frac{(2(n-j)+1)π}{2n+1}\right)=\frac12\quad,j=1,⋯,n$$ Since $k\frac{(2(n-j)+1)π}{2n+1}=k\pi-k\frac{2jπ}{2n+1}$, it becomes $$\sum_{k=1}^n-\cos\left(k\frac{2jπ}{2n+1}\right)=\frac12\quad,j=1,⋯,n$$ Writing $\cos(x)=\frac{e^{ix}+e^{-ix}}2$, $$-\sum_{k=1}^n\exp\left(ik\frac{2jπ}{2n+1}\right)-\sum_{k=-n}^{-1}\exp\left(ik\frac{2jπ}{2n+1}\right)=1\quad,j=1,⋯,n$$ Writing $1=\exp\left(i0\frac{2jπ}{2n+1}\right)$, it becomes $$\sum_{k=-n}^n\exp\left(ik\frac{2jπ}{2n+1}\right)=0\quad,j=1,⋯,n$$ This is a geometric series with common ratio $\exp(i\frac{2jπ}{2n+1})\ne1$. $$\sum_{k=-n}^n\exp\left(ik\frac{2jπ}{2n+1}\right)=\frac{\exp\left(i(n+1)\frac{2jπ}{2n+1}\right)-\exp\left(-in\frac{2jπ}{2n+1}\right)}{\exp(i\frac{2jπ}{2n+1})-1}=0$$