5
$\begingroup$

This question already has an answer here:

Prove that $$\sin\left(\frac{\pi}{n}\right)\sin\left(\frac{2\pi}{n}\right)\sin\left(\frac{3\pi}{n}\right).....\sin\left(\frac{(n-1)\pi}{n}\right)=\frac{n}{2^{n-1}}$$

Is there a proof without using complex numbers and $n-th$ roots of unity.

$\endgroup$

marked as duplicate by gen-z ready to perish, Nosrati, YuiTo Cheng, Lord Shark the Unknown, Shailesh Jun 22 at 5:38

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.

5
$\begingroup$

The following is the simplest proof I know. We have the identity \begin{equation} x^{2n} - 2x^n y^n \cos n\theta + y^{2n} = \bigg\{x^2 -2xy \cos \theta + y^2\bigg\}\bigg\{x^2-2xy \cos \bigg(\theta+\frac{2\pi}{n}\bigg)+y^2\bigg\}\cdots \end{equation} to $n$ factors adding $2\pi/n$ to each angle successively. This can be seen by noting the LHS and RHS share the same roots in $x$ using complex numbers, but given complex numbers are a trigonometric convenience I imagine there's a non-complex way to arrive at the identity. Let $x=y=1$, $\theta = 2\phi$, and apply $1-\cos\theta = 2\sin(\theta/2)$ \begin{equation} \sin n\phi = 2^{n-1} \sin \phi \sin\bigg(\phi + \frac{\pi}{n}\bigg)\sin\bigg(\phi + \frac{2\pi}{n}\bigg) \cdots \sin\bigg(\phi + \frac{(n-1)\pi}{n}\bigg) \end{equation} Divide by $\sin \phi$ and let $\phi \rightarrow 0$ to get the equation

$\endgroup$
0
$\begingroup$

Hint: Observe that we have $$ z^{n-1}+\cdots+z+1=\bigg(z-\exp(\frac{2\pi i}n)\bigg)\cdots\bigg(z-\exp(\frac{2\pi i(n-1)}n)\bigg) $$ Now let $z=1$ and use some trigonometry at the righthand side...

$\endgroup$

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