## Sums and products of tangents squared

Find the values of the following expressions ($n\geq 3$ odd):
\[
\begin{aligned}
S(n) &=
\sum_{k=0}^{\frac{n-3}{2}}\tan^2\left\{\tfrac{(2k+1)\pi}{2n}\right\}\,, \\
P(n) &=
\prod_{k=0}^{\frac{n-3}{2}}\tan^2\left\{\tfrac{(2k+1)\pi}{2n}\right\}\,, \\
\frac{S(n)}{P(n)} &=
\sum_{k=0}^{\tfrac{n-3}{2}}\,
\prod_{\substack{l=0 \\ l\ne k}}^{\tfrac{n-3}{2}}
\cot^2\left\{\tfrac{(2l+1)\pi}{2n}\right\}\,.
\end{aligned}
\]
Generalize the result to certain sums of products of tangents or
cotangents squared.

### Solution

The problem can be solved by looking at the
coefficients of the following polynomial (for the second equality use
$x^2-\cot^2{\alpha} = (x+\cot{\alpha})(x-\cot{\alpha})$ and recall
that $\cot(\pi-\alpha) = -\cot(\alpha)$):
\[
p_n(x) = x\prod_{k=0}^{\frac{n-3}{2}}\left(x^2 -
\cot^2\left\{\tfrac{(2k+1)\pi}{2n}\right\}\right) =
\prod_{k=0}^{n-1}\left(x - \cot\left\{\tfrac{(2k+1)\pi}{2n}\right\}\right)\,.
\]
Expanding the product we see that, except for the sign, $1/P(n)$ is
the coefficient of $x$ in $p_n(x)$, and $S(n)/P(n)$ is the coefficient
of $x^3$. The other coefficients yield various expressions involving
cotangents. So the problem amounts to finding coefficients of
$p_n(x)$. To that end we prove that $p_n(x)$ can be rewritten in the
following way:
\[
p_n(x) = \Re\{(x+i)^n\} \,,
\]
where $\Re(z)=$ real part of $z$ (applied to a polynomial we mean the
polynomial obtained by replacing the coefficients by their real parts;
note that if $p$ is a polynomial with complex coefficients and $a$ is
a real number then $\Re\{p\}(a)=\Re\{p(a)\}$). We note that the two
polynomials we are comparing have the same degree and same leading
coefficient, so we only need to show that they have exactly the same
roots, i.e., they are zero for exactly the same values of $x$. The
roots of $p_n(x)$ are $x_k=\cot\left\{\frac{(2k+1)\pi}{2n}\right\}$,
so we must prove that $\Re\{(x_k+i)^n\}=0$, or equivalently
$(x_k+i)^n$ is purely imaginary for $k=0,\dots, n-1$. In fact,
putting $\alpha_k = \frac{(2k+1)\pi}{2n}$ we get:
\[
\cot{\alpha_k}+i =
\frac{\cos{\alpha_k} + i \sin{\alpha_k}}{\sin{\alpha_k}}
=
\left\{\sin{\alpha_k}\right\}^{-1} e^{\alpha_k i} \,,
\]
so $\arg(\cot{\alpha_k}+i)=\alpha_k$, and
$\arg(\{\cot{\alpha_k}+i\}^n)=n\alpha_k = (k+1/2)\pi$, which is the
argument of a purely imaginary number.
So now we can expand $(x+i)^n$ using the binomial theorem and find its
real part:
\[
\Re\{(x_k+i)^n\} =
\sum_{l=0}^{\frac{n-1}{2}} (-1)^l \binom{n}{2l} x^{n-2l} \,.
\]
Finally, comparing coefficients we get:
\[
\begin{aligned}
\frac{1}{P(n)} &= \binom{n}{n-1} = n \,, \\
\frac{S(n)}{P(n)} &= \binom{n}{n-3} = \frac{n(n-1)(n-2)}{6} \,, \\
P(n) &= \frac{1}{n} \,, \\
S(n) &= \frac{(n-1)(n-2)}{6} \,,
\end{aligned}
\]
and much more:
\[
\begin{aligned}
\sum_{k=0}^{\frac{n-3}{2}}
\cot^2\left\{\tfrac{(2k+1)\pi}{2n}\right\}
&= \binom{n}{2} = \frac{n(n-1)}{2} \,, \\
\cot^2\left(\tfrac{\pi}{14}\right)\cot^2\left(\tfrac{3\pi}{14}\right)
&+
\cot^2\left(\tfrac{\pi}{14}\right)\cot^2\left(\tfrac{5\pi}{14}\right)
+
\cot^2\left(\tfrac{3\pi}{14}\right)\cot^2\left(\tfrac{5\pi}{14}\right)
= \binom{7}{4} = 35 \,,
\end{aligned}
\]
etc., etc., etc.
Miguel A. Lerma - 4/10/2003