Product of Cotangents of Fractions of Pi

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Theorem

Let $m \in \Z$ such that $m > 1$.


Then:

$\ds \prod_{k \mathop = 1}^{m - 1} \cot \frac {k \pi} {2 m} = 1$


Proof

We have:

\(\ds \frac {k \pi} {2 m} + \frac {\paren {m - k} \pi} {2 m}\) \(=\) \(\ds \frac {\paren {k + m - k} \pi} {2 m}\)
\(\ds \) \(=\) \(\ds \frac \pi 2\)


That means:

$(1): \quad \dfrac \pi 2 - \dfrac {k \pi} {2 m} = \dfrac {\paren {m - k} \pi} {2 m}$


Let $m$ be odd.

Then $m - 1$ is even, and so:

\(\ds \prod_{k \mathop = 1}^{m - 1} \cot \frac {k \pi} {2 m}\) \(=\) \(\ds \prod_{k \mathop = 1}^{\frac {m - 1} 2} \cot \frac {k \pi} {2 m} \prod_{k \mathop = \frac {m - 1} 2 + 1}^{m - 1} \cot \frac {k \pi} {2 m}\)
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 1}^{\frac {m - 1} 2} \cot \frac {k \pi} {2 m} \cot \frac {\paren {m - k} \pi} {2 m}\)
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 1}^{\frac {m - 1} 2} \cot \frac {k \pi} {2 m} \map \cot {\frac \pi 2 - \frac {k \pi} {2 m} }\) from $(1)$
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 1}^{\frac {m - 1} 2} \cot \frac {k \pi} {2 m} \tan \frac {k \pi} {2 m}\) Cotangent of Complement equals Tangent
\(\ds \) \(=\) \(\ds 1\) Product of Tangent and Cotangent


Now suppose $m$ is even.

Then $m - 1$ is odd, and so:

\(\ds \prod_{k \mathop = 1}^{m - 1} \cot \frac {k \pi} {2 m}\) \(=\) \(\ds \paren {\prod_{k \mathop = 1}^{\frac m 2 - 1} \cot \frac {k \pi} {2 m} } \frac m 2 \frac \pi {2 m} \paren {\prod_{k \mathop = \frac m 2 + 1}^{m - 1} \cot \frac {k \pi} {2 m} }\)
\(\ds \) \(=\) \(\ds \paren {\prod_{k \mathop = 1}^{\frac m 2 - 1} \cot \frac {k \pi} {2 m} \cot \frac {\paren {m - k} \pi} {2 m} } \frac \pi 4\)
\(\ds \) \(=\) \(\ds \paren {\prod_{k \mathop = 1}^{\frac m 2 - 1} \cot \frac {k \pi} {2 m} \map \cot {\frac \pi 2 - \frac {k \pi} {2 m} } } \cot \frac \pi 4\) from $(1)$
\(\ds \) \(=\) \(\ds \paren {\prod_{k \mathop = 1}^{\frac m 2 - 1} \cot \frac {k \pi} {2 m} \tan \frac {k \pi} {2 m} } \cot \frac \pi 4\) Cotangent of Complement equals Tangent
\(\ds \) \(=\) \(\ds 1 \cot \frac \pi 4\) Product of Tangent and Cotangent
\(\ds \) \(=\) \(\ds 1\) Cotangent of $45 \degrees$

$\blacksquare$


Sources

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