Trigonometry prove

[man111]

Prove that $\displaystyle\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n} = \frac{\sqrt{n}}{2^{n-1}}$

[myra_panama]

We can prove, using the mathmatical induction .......

-- Чт мар 24, 2011 10:27:49 --

or we can do that:
We have :
$x^{2n-2}+x^{2n-4}+\cdots +x^2+1=\prod\limits_{k=1}^{n-1}(x^2-2x\cos {\frac{k\pi}{n}}+1)$

(Оффтоп)

can you prove this? if , not , i can help you//

for x=1 , we can easily prove that
$\displaystyle\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n} = \frac{\sqrt{n}}{2^{n-1}}$

[man111]

$x^{2n-2}+x^{2n-4}+\cdots +x^2+1=\prod\limits_{k=1}^{n-1}(x^2-2x\cos {\frac{k\pi}{n}}+1)$

How can I prove that....
please provide prove..
thanking you

[Руст]

Let $f(x)=\frac{x^{2n}-1}{x-1}$.
Then

$$2n=f(1)=\prod_{k=1}^{2n-1}(1-e^{\frac{2\pi i k}{2n}})=\prod_{k=1}^{2n-1}2\sin\frac{\pi k}{2n}(-ie^{\frac{\pi ik}{2n}})=2^{2n-1}(\prod_{k=1}^{n-1}\sin \frac{\pi k}{2n})^2.$$

[man111]

Thanks pyct and myra_panama