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
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $148$