Prove-that-n-N-k-1-n-1-sin-kpi-2n-k-1-n-1-cos-kpi-2n-

Question Number 142698 by Willson last updated on 04/Jun/21

Prove that \forall n \in N^{*}

\underset{k = 1}{\prod^{n - 1}} \sin (\frac{k π}{2 n}) = \underset{k = 1}{\prod^{n - 1}} \cos (\frac{k π}{2 n})

Answered by Ar Brandon last updated on 04/Jun/21

\underset{k = 1}{\prod^{n - 1}} \sin (\frac{k π}{2 n}) = \underset{k = 1}{\prod^{n - 1}} \cos (\frac{π}{2} - \frac{k π}{2 n}) = \underset{k = 1}{\prod^{n - 1}} \cos (\frac{(n - k) π}{2 n})

= \cos (\frac{n - 1}{2 n} π) \cos (\frac{n - 2}{2 n} π) \cdot \cdot \cdot \cos (\frac{2}{2 n} π) \cos (\frac{1}{2 n} π)

= \cos (\frac{1}{2 n} π) \cos (\frac{2}{2 n} π) \cdot \cdot \cdot \cos (\frac{n - 2}{2 n} π) \cos (\frac{n - 1}{2 n} π)

= \underset{k = 1}{\prod^{n - 1}} \cos (\frac{k}{2 n} π)