# $\sin(\frac{\pi}{n})\sin(\frac{2\pi}{n})…\sin(\frac{(n-1)\pi}{n})=\frac{n}{2^{n-1}}$ [duplicate]

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.

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

The following is the simplest proof I know. We have the identity $$$$x^{2n} - 2x^n y^n \cos n\theta + y^{2n} = \bigg\{x^2 -2xy \cos \theta + y^2\bigg\}\bigg\{x^2-2xy \cos \bigg(\theta+\frac{2\pi}{n}\bigg)+y^2\bigg\}\cdots$$$$ to $$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)$$ $$$$\sin n\phi = 2^{n-1} \sin \phi \sin\bigg(\phi + \frac{\pi}{n}\bigg)\sin\bigg(\phi + \frac{2\pi}{n}\bigg) \cdots \sin\bigg(\phi + \frac{(n-1)\pi}{n}\bigg)$$$$ Divide by $$\sin \phi$$ and let $$\phi \rightarrow 0$$ to get the equation
Hint: Observe that we have $$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...