Product of Sines of Fractions of Pi

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: $52$