5
$\begingroup$

Prove that for any $n \in \mathbb{N}^*$ we have: $$\sin x \sin \left( x + \frac{\pi}{n} \right) \sin \left( x + \frac{2\pi}{n} \right)\cdots \sin \left( x + \frac{(n - 1)\pi}{n} \right) = \frac{\sin nx}{2^{n - 1}}$$

Here is what I did so far:

We know that $\sin x = \Im (e^{ix})$. So, we can rewrite our product the following way:

$\prod_{k = 1}^{n - 1} \sin \left( x + k\frac{\pi}{n} \right) = \prod_{k = 1}^{n - 1} \Im \left( e^{i \left( x + k\frac{\pi}{n} \right)} \right)$

At this point I got stuck, mainly because I'm not sure if I can take the $\Im$ of the whole product or not.

Thank you in advance!

$\endgroup$

marked as duplicate by Yiyuan Lee, Adam Hughes, Lucian, Arnaud D., hardmath Jan 14 '17 at 1:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ This might be a duplicate. $\endgroup$ – Daniel Fischer Nov 21 '16 at 21:23
0
$\begingroup$

This is a complex analysis problem. Let $z=cos(2x) + isin(2x)$, we have $$|z^n - 1 |=|2sin(nx)|$$

On the other hand, $$z^n - 1 = (z-1)(z-a)(z-a^2)...(z-a^{n-1}),$$ where $a=cos(2\pi/n) + i\sin(2\pi/n)$ is the $n$-th unit root. If we take norms on both sides, we get the other half of the formula.

There is a sign problem, which is easy to hand. ( for example, just look at value for $x\in (0, \pi/n)$; or instead of taking norms, write the terms as the product of a real number times a unit complex number)

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.