Using complex numbers prove that $\sin(\frac {\pi}{m})\sin(\frac {2\pi}{m})\ldots\sin(\frac {(m-1)\pi}{m})= \frac{m}{2^{m -1}}$

Question

Prove that for $m=2,3,\dots$ $$ \sin\left(\frac{\vphantom{1}\pi}m\right)\sin\left(\frac{2\pi}m\right)\sin\left(\frac{3\pi}m\right)\cdots\sin\left(\frac{(m-1)\pi}m\right)=\frac{m}{2^{m-1}} $$

I have no idea how to begin at all, I'm trying to think of a way using de Moivre's theorem but I can't seem to figure it out.
I'm sorry for not showing effort but I'm completely stuck.

must be a mistake in the book then, but are you familiar with anything similar to this? — khaled014z, Sep 27 '18 at 17:35
oh i figured out the mistake, can i re-ask? correction is denominator of R.H.S is 2^(m-1) — khaled014z, Sep 27 '18 at 17:42
It should be $\frac{m}{2^{m-1}}$. The question looks as if it has a typo; it is missing a pair of braces. — robjohn, Sep 27 '18 at 17:54
This is closely related to Evaluation of $\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}$. — robjohn, Sep 27 '18 at 18:18
De Moivre's identity follows from Euler's formula, which can be used to get the representation $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$. Using this representation is the easiest way I know to show the (corrected) identity in the question. — robjohn, Sep 28 '18 at 15:26
Possible duplicate of More Questions from Mathematical Analysis by Apostol — rtybase, Sep 28 '18 at 17:46
@AhmadBazzi: you claim the left-hand side is $0$. However, none of the factors is $0$. I fixed the right-hand side, but the left-hand side is unchanged. — robjohn, Sep 28 '18 at 19:54
@AhmadBazzi: yes, but the product is up to $\frac{(m-1)\pi}n$ not $\frac{m\pi}n$. — robjohn, Sep 28 '18 at 20:09
@robjohn oh goshh.. i did not see that. Thanks for letting me know. — Ahmad Bazzi, Sep 28 '18 at 20:14

robjohn · Accepted Answer · 2018-09-27T18:25:06.863

Generalized from this answer: $$ \begin{align} \prod_{k=1}^{m-1}\left(\frac{e^{i\pi\frac km}-e^{-i\pi\frac km}}{2i}\right) &=\frac{e^{i\pi\frac{m(m-1)}{2m}}}{2^{m-1}i^{m-1}}\prod_{k=1}^{m-1}\left(1-e^{-i2\pi\frac km}\right)\\ &=\frac{e^{i\pi\frac{m-1}2}}{2^{m-1}i^{m-1}}\lim_{z\to1}\prod_{k=1}^{m-1}\left(z-e^{-i2\pi\frac km}\right)\\[3pt] &=\frac1{2^{m-1}}\lim_{z\to1}\frac{z^m-1}{z-1}\\[9pt] &=\frac{m}{2^{m-1}} \end{align} $$ The statement in the question seems to be missing a pair of braces: \frac{m}{2^m-1} rather than \frac{m}{2^{m-1}}.

score 1 · Answer 2 · edited Sep 27 '18 at 18:57

1

In this case left hand side of equation is not equal to right hand sides of equation .You can try it by putting random values of m.

Example ($m=3$): $$\sin \frac{\pi}{3} \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}\frac{\sqrt{3}}{2} = \frac{3}{4} \neq \frac{3}{2^3 - 1}$$

edited Sep 27 '18 at 18:57

robjohn

345,667

answered Sep 27 '18 at 17:25

Sourabh

171

1

this ought to be a comment rather than an answer. – Ahmad Bazzi Sep 27 '18 at 17:28
Well the equation is wrong – Sourabh Sep 27 '18 at 17:29
Well replace a value for $m$ and show your point. You could edit your answer. – Ahmad Bazzi Sep 27 '18 at 17:32
If m=3 then LHS of equation becomes sin(π/3)×sin(2π/3)=(√3/2)×(√3/2)=3/4 now RHS will be 3/(2^3-1)=3/7 hence LHS is not equal to RHS – Sourabh Sep 27 '18 at 17:36
You could edit your answer. I'll do it for you – Ahmad Bazzi Sep 27 '18 at 17:36

score 1 · Answer 3 · answered Sep 27 '18 at 18:11

$|\cos \theta + i\sin \theta - 1|\\ \sqrt {1 - 2\cos \theta} = 2\sin \frac {\theta}{2}\\ 2\sin\theta = |e^{2\theta i} - 1|$

$\prod_\limits{n=1}^{m-1} \sin \frac{n\pi}{m} = \frac {1}{2^{m-1}}\prod_\limits{n=1}^{m-1}|e^{\frac{2n\pi}{m} i} - 1|$

The set $\{e^{\frac{2n\pi}{m} i}\}$ make up the roots of $z^m - 1 = 0$ excluding the root at $z= 1$ or the roots of $(z^{m-1} + z^{m-2} + z^{m-3} + \cdots + 1)$

$(z - e^{\frac{2\pi}{m}i})(z - e^{\frac{4\pi}{m}i})(z - e^{\frac{6\pi}{m}i})\cdots(z - e^{\frac{(m-2)\pi}{m}i})= (z^{m-1} + z^{m-2} + z^{m-3} + \cdots + 1)\\ |z - e^{\frac{2\pi}{m}i}||z - e^{\frac{4\pi}{m}i}||z - e^{\frac{6\pi}{m}i}|\cdots|z - e^{\frac{(m-2)\pi}{m}i}|= |z^{m-1} + z^{m-2} + z^{m-3} + \cdots + 1| $

and now set z = 1

$|1 - e^{\frac{2\pi}{m}i}||1 - e^{\frac{4\pi}{m}i}||1 - e^{\frac{6\pi}{m}i}|\cdots|1 - e^{\frac{(m-2)\pi}{m}i}| = |1^{m-1} + 1^{m-2} + 1^{m-3} + \cdots + 1| = m$

$\prod_\limits{n=1}^{m-1} \sin \frac{n\pi}{m} = \frac {1}{2^{m-1}}\prod_\limits{n=1}^{m-1}|e^{\frac{2n\pi}{m} i} - 1| = \frac {m}{2^{m-1}}$

Using complex numbers prove that $\sin(\frac {\pi}{m})\sin(\frac {2\pi}{m})\ldots\sin(\frac {(m-1)\pi}{m})= \frac{m}{2^{m -1}}$

3 Answers3