Trigonometric identity $\sum_k\frac1{\cos2\phi_k-\cos\Delta}=0$ Solving a problem I came across the following interesting identity ($\cos2\phi_k\ne\cos\Delta$ is assumed):
$$
\sum_{k=0}^{n-1}\frac1{\cos2\phi_k-\cos\Delta}=0,
$$
where $n\ge3$, $\Delta=\frac{2\pi}n$ and $\phi_k=\phi_0+k\Delta$.
Is there a simple proof of the identity?
 A: $\def\f{\phi}$
Since the question remains unanswered I will post here a solution which I have found in the meantime.
Consider a more general expression
$$
\sum_{k=0}^{n-1}\frac1{\cos l\Delta-\cos m\phi_k}=0,\tag1
$$
where $m,l$ are integers. What are the necessary and sufficient conditions on $n,l,m $ for (1) to hold?
The key observation is that the angles
$$
m\f_k=m\left(\f_0+k\Delta\right),\quad (k=0\dots n-1)
%\cos \f_k=\cos(\f_0+k\frac\Delta)
$$
are the solutions of the equation
$$
%\left[\cos\left(\frac nq\f\right)\right]^q=\left[\cos\left(\frac nq\f_0\right)\right]^q,
\cos(r\f)=\cos(r\f_0).
$$
where $r=\frac n{(n,m)}$, $(n,m)$ being the gcd of $m$ and $n$. Each solution enters the sum (1) $(n,m)$ times.
Let $\f=\arccos t$. Then
$$
P(t)=\cos(r\arccos t)-\cos(r\f_0)
$$
is a polynomial of $r$-th degree in $t$ (cf. Chebyshev polynomial), so that
$$
\sum_{k=0}^{r-1}\frac1{t-t_k}=\frac{P'(t)}{P(t)},
$$
$t_k=\cos m\f_k$ being the roots of $P(t)$.
Thus the identity (1) is equivalent to the statement:
$$
P'(t)\Big|_{t=\cos(l\Delta)}=0
$$
or
$$
\frac{r\sin(rl\Delta)}{\sin(l\Delta)}=0
$$
which implies:
$$
\frac{2l}{(m,n)}\in\mathbb Z,\quad \frac{2l}{n}\not\in\mathbb Z .\tag2
$$
One easily observes that for $m=2$, $l=1$, $n\ge3$ from the initial problem the condition (2) indeed holds.
