# Product of Cotangents of Fractions of Pi

## Theorem

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

Then:

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

## 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 $45 \degrees$

$\blacksquare$