Product of Sines of Fractions of Pi
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Theorem
Let $m \in \Z$ such that $m > 1$.
Then:
- $\ds \prod_{k \mathop = 1}^{m - 1} \sin \frac {k \pi} m = \frac m {2^{m - 1} }$
Proof
Consider the equation:
- $z^m - 1 = 0$
whose solutions are the complex roots of unity:
- $1, e^{2 \pi i / m}, e^{4 \pi i / m}, e^{6 \pi i / m}, \ldots, e^{2 \paren {m - 1} \pi i / m}$
Then:
\(\ds z^m - 1\) | \(=\) | \(\ds \paren {z - 1} \paren {z - e^{2 \pi i / m} } \paren {z - e^{4 \pi i / m} } \dotsm \paren {z - e^{2 \paren {m - 1} \pi i / m} }\) | product of all the roots | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {z^m - 1} {z - 1}\) | \(=\) | \(\ds \prod_{k \mathop = 1}^{m - 1} \paren {z - e^{2 k \pi i / m} }\) | dividing by $z - 1$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1 + z + \dotsb + z^{m-1}\) | \(=\) | \(\ds \prod_{k \mathop = 1}^{m - 1} \paren {z - e^{2 k \pi i / m} }\) | Sum of Geometric Sequence | ||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds m\) | \(=\) | \(\ds \prod_{k \mathop = 1}^{m - 1} \paren {1 - e^{2 k \pi i / m} }\) | setting $z = 1$ | |||||||||
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds m\) | \(=\) | \(\ds \prod_{k \mathop = 1}^{m - 1} \paren {1 - e^{-2 k \pi i / m} }\) | taking complex conjugate of both sides | |||||||||
\(\ds \leadsto \ \ \) | \(\ds m^2\) | \(=\) | \(\ds \prod_{k \mathop = 1}^{m - 1} \paren {\paren {1 - e^{2 k \pi i / m} } \paren {1 - e^{-2 k \pi i / m} } }\) | multiplying $(1)$ by $(2)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds m^2\) | \(=\) | \(\ds \prod_{k \mathop = 1}^{m - 1} \paren {1 - e^{2 k \pi i / m} - e^{-2 k \pi i / m} + 1}\) | multiplying out | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds m^2\) | \(=\) | \(\ds \prod_{k \mathop = 1}^{m - 1} \paren {2 - 2 \frac {e^{2 k \pi i / m} + e^{-2 k \pi i / m} } 2}\) | rearranging | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds m^2\) | \(=\) | \(\ds \prod_{k \mathop = 1}^{m-1} \paren {2 - 2 \cos \frac {2 k \pi} m}\) | Euler's Cosine Identity | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds m^2\) | \(=\) | \(\ds 2^{m - 1} \prod_{k \mathop = 1}^{m - 1} \paren {1 - \cos \frac {2 k \pi} m}\) | factoring out $m - 1$ instances of $2$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds m^2\) | \(=\) | \(\ds 2^{m - 1} \prod_{k \mathop = 1}^{m - 1} \paren {2 \sin^2 \frac {k \pi} m}\) | Double Angle Formula for Cosine: Corollary $2$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {m^2} {2^{2 m - 2} }\) | \(=\) | \(\ds \prod_{k \mathop = 1}^{m - 1} \paren {\sin^2 \frac {k \pi} m}\) | factoring out another $m - 1$ instances of $2$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac m {2^{m - 1} }\) | \(=\) | \(\ds \prod_{k \mathop = 1}^{m - 1} \paren {\sin \frac {k \pi} m}\) | taking square root of both sides |
$\blacksquare$
Also presented as
Some sources prefer to report this result as:
- $\ds \prod_{k \mathop = 1}^{m - 1} \sin \frac {k \pi} m = \frac {2 m} {2^m}$
on the grounds that it may be easier to remember.
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $52$