we play with sin

Geometry Level 3

sin π 7 sin 2 π 7 sin 3 π 7 = ? \large \sin \frac \pi 7 \sin \frac {2\pi} 7 \sin \frac {3\pi}7 = \ ?


The answer is 0.3307.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Fletcher Mattox
May 16, 2020

An obscure (to me) identity from MathWorld simplifies things a bit.

k = 1 n 1 sin ( k π n ) = n 2 n 1 \displaystyle\prod_{k=1}^{n-1}\sin(\frac{k\pi}{n}) = \frac{n}{2^{n-1}}

By symmetry: sin ( k π n ) = sin ( n k π n ) \sin(\frac{{k\pi}}{n} ) = \sin(\frac{{n-k}\pi}{n})

k = 1 3 sin ( k π n ) = k = 4 6 sin ( k π n ) \displaystyle\prod_{k=1}^{3}\sin(\frac{k\pi}{n}) = \prod_{k=4}^{6}\sin(\frac{k\pi}{n})

k = 1 3 sin ( k π n ) = n 2 n 1 = 7 64 = 7 8 = 0.3307 \displaystyle{\prod_{k=1}^{3}\sin(\frac{k\pi}{n}) } = \sqrt{\frac{n}{2^{n-1}}} = \sqrt{\frac{7}{64}} = \frac{\sqrt{7}}{8} = 0.3307

k = 1 n 1 sin ( k π n ) = k = 1 n 1 ( e i π k n e i π k n 2 i ) sin ( θ ) = e i θ e i θ 2 i = k = 1 n 1 ( 1 2 i ) ( e i π k n ) ( e i 2 π k n + 1 ) = ( 1 2 i ) n 1 e i π n k = 1 n 1 k k = 1 n 1 ( 1 e 2 π i n k ) = 1 2 n 1 ( i ) n 1 e i π n ( n 1 ) n 2 k = 1 n 1 ( 1 w k ) w is nth root of unity = ( i ) n 1 2 n 1 ( i ) n 1 lim x 1 x n 1 x 1 = n 2 n 1 x n 1 = k = 0 n 1 ( x w k ) = ( x 1 ) k = 1 n 1 ( x w k ) k = 1 n 1 ( x w k ) = x n 1 x 1 \begin{aligned} \prod_{k=1}^{n-1} \sin(\frac{k \pi}{n}) &= \prod_{k=1}^{n-1} \left ( \frac{e^{\frac{i \pi k}{n}}-e^{-\frac{i \pi k}{n}}}{2i} \right ) & {\color{#D61F06} \sin(\theta) = \frac{e^{i \theta}-e^{-i\theta}}{2 i}} \\ &= \prod_{k=1}^{n-1} \left ( \frac{1}{2i} \right ) \cdot \left (- e^{-\frac{i \pi k}{n}} \right ) \cdot \left ( - e^{\frac{i 2\pi k}{n}} + 1 \right ) \\ &= \left ( \frac{-1}{2i} \right )^{n-1} \cdot e^{\frac{-i \pi}{n}\sum_{k=1}^{n-1} k} \cdot \prod_{k=1}^{n-1} \left ( 1 - e^{\frac{2\pi i}{n} k} \right ) \\ &= \frac{1}{2^{n-1} \cdot (-i)^{n-1}} \cdot e^{-\frac{i\pi}{n} \cdot \frac{(n-1)n}{2}} \cdot \prod_{k=1}^{n-1} \left ( 1 - w^k \right ) & \color{#D61F06} w \text{ is nth root of unity}\\ &= \frac{(-i)^{n-1}}{2^{n-1} \cdot (-i)^{n-1}} \cdot \lim_{x\rightarrow 1} \frac{x^n - 1}{x-1} = \frac{n}{2^{n-1}} \\ &\color{#D61F06} x^n - 1 = \prod_{k=0}^{n-1} \left ( x - w^k \right ) = (x - 1) \prod_{k=1}^{n-1} \left ( x - w^k \right ) \\ & \color{#D61F06} \Rightarrow \prod_{k=1}^{n-1} \left ( x - w^k \right ) = \frac{x^n - 1}{x-1} \end{aligned}

Hassan Abdulla - 1 year ago

Log in to reply

Aha! Very nice work.

Fletcher Mattox - 1 year ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...