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

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.

## closed as off-topic by Nosrati, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Shailesh, rtybaseSep 28 '18 at 17:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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
• yes.. my mistake i'll be cautious next time – khaled014z Sep 27 '18 at 17:46
• 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
• does it make a difference though? – khaled014z Sep 27 '18 at 17:55

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}}.

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

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

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

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

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

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

• 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