I want to show that $\sum_{|j| < n} (n-|j|) \exp(ij\lambda)= \dfrac{\sin^2(\frac 1 2 n\lambda)}{\sin^2(\frac 1 2 \lambda)}$.
I know from Proving $\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$ \begin{equation} (1)\sum_{j=1}^{n-1} cos(j\lambda) = -\frac 1 2 + \frac{\sin(\frac{2n-1} 2 \lambda)}{2\sin(\frac \lambda 2)}. \end{equation}
and from the Hint in the first answer in How to show that $\frac{1}{\tan(x/2)}=2 \sum_{j=1}^{\infty}\sin(jx)$ in Cesàro way/sense? $ (2)\frac d {dx} \sum_{j=1}^{n-1} sin(j\lambda) =\frac d {d\lambda} \frac{\cos({\frac \lambda 2})-\cos(\frac{n+1}2 \lambda)}{2\sin(\frac\lambda 2)} = \frac{n\sin(\frac{n+1} 2 \lambda)\sin(\frac \lambda 2)+\cos(\frac{n\lambda}2) - 1}{4\sin^2(\frac \lambda 2)}. $
This is what i did so far: \begin{align} \sum_{|j| < n} (n-|j|) \exp(ij\lambda) &= n + 2\sum_{j=1}^{n-1}(n-j) cos(j\lambda) \\ &=n + 2n \sum_{j=1}^{n-1} cos(j\lambda) - \frac d {d\lambda}\sum_{j=1}^{n-1} sin(j\lambda)\\ &= \frac{n\sin(\frac{2n-1} 2 \lambda)\sin(\frac \lambda 2) + \frac 1 2 + \frac 1 2 n \sin(\frac {n+1} 2 \lambda)\sin(\frac \lambda 2) + \frac 1 2 \cos( \frac{n\lambda}2)}{\sin^2(\frac \lambda 2)}. \end{align}
If this and the proposition is true, it should be true that $ n\sin(\frac{2n-1} 2 \lambda)\sin(\frac \lambda 2) + \frac 1 2 + \frac 1 2 n \sin(\frac {n+1} 2 \lambda)\sin(\frac \lambda 2) + \frac 1 2 \cos( \frac{n\lambda}2) - \sin^2(\frac {n\lambda} 2) = 0. $
But for $n=2$, i get $ 3\sin(\frac {3\lambda} 2) \sin(\frac \lambda 2) + \frac 1 2 + \frac 1 2 \cos(\lambda) - \sin^2 \lambda \ne 0. $ (checked with Wolframalpha)
So there is probably something wrong with my calculations. But i checked everything twice and don't see my mistake. Do you see it?
EDIT
Trying to prove it as proposed in the second answer.
Let $z:=\exp(i\lambda)$ and $p(z):=\sum_{j=0}^{n-1} z^j$. First i want to show that $p(z)p(z^{-1})=\sum_{|j|
It is \begin{align} \sum_{j=1}^{n-1} z^{j}&= p(z) - 1 = \frac{1-z^n}{1-z} - 1.\\ \sum_{j=1}^{n-1} z^{-j}&= p(z^{-1})-1 = \frac{1-z^{-n}}{1-z^{-1}} - 1.\\ \sum_{j=1}^{n-1} jz^{-j} &= i \frac d {d\lambda} p(z^{-1}) = \frac{-nz^{-n}+(n-1)z^{-n-1}+z^{-1}}{(1-z^{-1})^2}.\\ \sum_{j=1}^{n-1} jz^{j} &= -i \frac d {d\lambda} p(z) = \frac{-nz^{n}+(n-1)z^{n+1}+z}{(1-z)^2}.\\ \end{align} So i have \begin{align} \sum_{|j|
The last step $p(z)p(z^{-1})= \frac{(z^{n/2}-z^{-n/2})^2}{(z^{1/2}-z^{-1/2})^2}$ was easy to show.
EDIT2
I did it now. It was a lot easier to show directly $p(z)p(z^{-1})=\sum_{|j|