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Very important sum of reciprocal cosine identity

Henrik

March 24, 2025

Abstract

We provide a proof for a very important sine identity.

Theorem 1. Let $n \in ℕ$ . Then

\sum_{k = 1}^{n} \frac{1}{1 - \cos (π \frac{k}{n})} = \frac{2 n^{2} + 1}{6} .

(1)

Proof. First, we derive a characterization of the left-hand side of (1). Let $x_{j} = \cos (π \frac{j}{n})$ for $j = 0, \dots, n$ and $p (x) = \prod_{j = 1}^{n} (x - x_{j})$ . Then we have the logarithmic derivative

\frac{p^{'} (x)}{p (x)} = \sum_{j = 1}^{n} \frac{1}{x - x_{j}} = \sum_{k = 1}^{n} \frac{1}{x - \cos (π \frac{k}{n})} .

(2)

Moreover, with $x = \cos 𝜃$ , let $U_{n - 1} (\cos 𝜃) \sin 𝜃 = \sin n𝜃$ be the Chebyshev polynomial of second kind. Then $p (x) = c (x + 1) U_{n - 1} (x)$ for some $c \neq 0$ . We thus get

\frac{p^{'} (x)}{p (x)} = \frac{U_{n - 1} (x) + (x + 1) U_{n - 1}^{'} (x)}{(x + 1) U_{n - 1} (x)} .

(3)

The derivative is characterized by $U_{n - 1}^{'} (x) = \frac{n T_{n} (x) - x U_{n - 1} (x)}{x^{2} - 1}$ where $T_{n} (\cos 𝜃) = \cos n𝜃$ is the Chebyshev polynomial of first kind. Substituting this into (3), we get

\begin{aligned} \frac{p^{'} (x)}{p (x)} & = \frac{1}{x + 1} + \frac{n T_{n} (x) - x U_{n - 1} (x)}{(x^{2} - 1) U_{n - 1} (x)} \\ = \frac{1}{\cos 𝜃 + 1} - \frac{n \cos n𝜃 - \cos 𝜃 \frac{\sin n𝜃}{\sin 𝜃}}{\sin 𝜃 \sin n𝜃} . \end{aligned}

We use the following expansion related to the Dirichlet kernel

\frac{\sin n𝜃}{\sin 𝜃} = \frac{e^{in𝜃} - e^{- in𝜃}}{e^{i𝜃} - e^{- i𝜃}} = \sum_{k = 0}^{n - 1} \cos ((n - 1 - 2 k) 𝜃) .

Then by applying L’Hôspital’s rule twice, we have

\begin{aligned} \frac{p^{'} (1)}{p (1)} & = \lim_{𝜃 \to 0} \frac{1}{\cos 𝜃 + 1} - \frac{n \cos n𝜃 - \cos 𝜃 \sum_{k = 0}^{n - 1} \cos ((n - 1 - 2 k) 𝜃)}{\sin 𝜃 \sin n𝜃} \\ = \frac{1}{2} + \frac{n^{3} - n - \sum_{k = 0}^{n - 1} {(n - 1 - 2 k)}^{2}}{2 n} \\ = \frac{2 n^{2} + 1}{6} \end{aligned}

which together with (2) implies (1). □

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