A matrix w/integer eigenvalues and trigonometric identity Any intuition and/or rigorous arguments on the proofs of the following statements would be appreciated:
Let $n$ be a natural number. 
(a) Consider the following Toeplitz/circulant symmetric matrix:
$$
\Lambda_{kl} = \begin{cases}\frac{2n(n+1)}{3}, & k = l, \\[6pt]
\frac{-1}{1-\cos\frac{2\pi(k-l)}{2n+1}}, & k \ne l,\end{cases}
$$
where $k,l = 1,2, \ldots, 2n+1$.
Prove that its eigenvalues are natural numbers(!) 
$$2n, 4n-2, 6n-6, \ldots, n(n+1)$$
with multiplicity $2$ and $0$ w/multiplicity $1$.
(b) Find a formula for the right signs in the following trigonometric identity
$$
\tan\left(\frac{l\pi}{2n+1}\right) = 2\sum_{k=1}^n (\pm)\sin\left(\frac{k\pi}{2n+1}\right),\qquad l = 1,2,\ldots, n.
$$
and show that the signs are uniquely determined in the case of prime $2n+1$.
The problems (that are formulated to be self-contained) arise from discrete approximation of continuous inverse boundary problem (see http://en.wikibooks.org/wiki/On_2D_Inverse_Problems for background) and the statements were formulated w/computer help.
 A: Since (b) has been largely answered in the comments, I'll deal with (a) here.
It is the case here that there is a very simple (and otherwise useful and well-known) basis of eigenvectors. Let $\zeta$ be a primitive $2n+1$-th root of unity. Our eigenvectors for $\Lambda$ will be the “cyclic” vectors, whose coordinates are successive powers of a root of unity :
$$
C_s=(\zeta^s,\zeta^{2s},\zeta^{3s}, \ldots ,\zeta^{(2n)s},1) \ (0 \leq s \leq 2n)
\tag{1}
$$
Let us compute all the coordinates of $\Lambda C_s$ ; in other words, we must 
compute the sum
$$
f(i,s)=\sum_{j=1}^{2n+1} \Lambda_{ij} \zeta^{sj} \tag{2}
$$  
We have $f(i,s)=\frac{2}{3}n(n+1)\zeta^{si}+g(i,s)$, where
$$
g(i,s)=\sum_{j\neq i} \frac{\zeta^{sj}}{\frac{\zeta^{i-j}+\zeta^{j-i}}{2}-1}=
\sum_{t\neq 0} \frac{\zeta^{s(i+t)}}{\frac{\zeta^{-t}+\zeta^{t}}{2}-1}=
2\zeta^{si}\sum_{t=1}^{2n} \frac{(\zeta^t)^{s+1}}{(\zeta^{t}-1)^2}=
2\zeta^{si}\sum_{\eta\in U, \eta\neq 1} \frac{\eta^{s+1}}{(\eta-1)^2}
\tag{3}
$$
where $U$ is the set of all $(2n+1)$-th roots of unity. There are classical techniques to compute the RHS of (3) above ; see the answer to this question, where the final
formula obtained is
$$
\sum_{\eta\in U, \eta\neq 1} \frac{\eta^{s+1}}{(\eta-1)^2}
=\frac{6(2n+1-s)s-((2n+1)^2-1)}{12}=
\frac{s(2n+1-s)}{2}-\frac{n^2+n}{3}
 \tag{4}
$$
Combining (3) with (4), the terms in $\frac{n^2+n}{3}$ cancel out and we thus obtain
$$
\sum_{j=1}^{2n+1} \Lambda_{ij} \zeta^{sj}=s(2n+1-s)\zeta^{si} \tag{5}
$$
Or, to express things vectorially,
$$
\Lambda C_s=s(2n+1-s) C_s \tag{6}
$$
So $C_s$ is an eigenvector associated to the eigenvalue $s(2n+1-s)$ as wished.
