complex analysis - Proving that $sum_{|j| < n} (n-|j|) exp(ijlambda)= frac{sin^2(frac 1 2 nlambda)}{sin^2(frac 1 2 lambda)}$

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|\end{align}
If i now put in the values from above and substract $p(z)p(z^{-1})$ i should get $0$. But Wolfram Alpha doesn't agree, so i am again on the wrong track. What did i wrong?

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|

Answer

Letting $\alpha=\frac{\lambda}{2}$ to make the expressions cleaner, we get:
$$\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 - 2\frac d {d\lambda}\sum_{j=1}^{n-1} \sin j\lambda\\
&= \frac{n\sin(2n-1)\alpha\sin\alpha + \frac 1 2 - \frac 1 2 n \sin (n+1)\alpha \sin \alpha - \frac 1 2 \cos n\alpha}{\sin^2\alpha}.
\end{align}$$
Note the minus signs in the numerator. I'm not sure if that solves the problem - there might be an error elsewhere. Wolfram Alpha says when $n=2$ the numerator is $\sin^2 2\alpha$, but $n=1$ doesn't agree.

The easier way to do this sort of problem is via the comment I gave above, using $z=e^{i\lambda}$.

Let $p(z)=\sum_{j=0}^{n-1} z^j$. Show that:$$p(z)p(z^{-1})=\sum_{|j|

Now, $p(z)=\frac{z^n-1}{z-1}$, and $p(z^{-1})=z^{-(n-1)}p(z)$. Now show: $$p(z)p(z^{-1})= \frac{(z^{n/2}-z^{-n/2})^2}{(z^{1/2}-z^{-1/2})^2}$$

Then let $z=e^{i\lambda}$.