Product Formula for Sine

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Theorem

$\ds \map \sin {n z} = 2^{n - 1} \prod_{k \mathop = 0}^{n - 1} \map \sin {z + \frac {k \pi} n}$


Corollary

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

We have:

\(\ds 2 \sin \theta\) \(=\) \(\ds \frac {e^{i \theta} - e^{-i \theta} } i\) Sine Exponential Formulation
\(\ds \) \(=\) \(\ds \frac {e^{i \theta} e^{-i \theta} - e^{-i \theta} e^{-i \theta} } {i e^{-i \theta} }\)
\(\ds \) \(=\) \(\ds \paren {1 - e^{-2 i \theta} } \paren {-i e^{i \theta} }\)
\(\ds \) \(=\) \(\ds \paren {1 - e^{-2 i \theta} } \paren {e^{i \theta} e^{-i \pi / 2} }\) Euler's Formula/Examples/e^-i pi by 2
\(\ds \) \(=\) \(\ds \paren {1 - e^{-2 i \theta} } \paren {e^{i \theta - i \pi / 2} }\)
\(\ds \leadsto \ \ \) \(\ds \prod_{k \mathop = 0}^{n - 1} \paren {2 \sin \pi \paren {x + \frac k n} }\) \(=\) \(\ds \prod_{k \mathop = 0}^{n - 1} \paren {1 - e^{-2 i \pi \paren {x + k / n} } } \paren {e^{i \pi \paren {x - \paren {1 / 2} + \paren {k / n} } } }\)
\(\ds \) \(=\) \(\ds \paren {1 - e^{-2 i \pi n x} } \paren {e^{i \pi \paren {n x - 1 / 2} } }\)
\(\ds \) \(=\) \(\ds 2 \sin \pi n x\)