2022-01-17

Prove that
$\prod _{k=1}^{n-1}\mathrm{sin}\frac{k\pi }{n}=\frac{n}{{2}^{n-1}}$
Using ${n}^{th}$ root of unity
${\left({e}^{\frac{2ki\pi }{n}}\right)}^{n}=1$

nick1337

Step 1 $P=\prod _{k=1}^{n-1}\mathrm{sin}\left(\frac{k\pi }{n}\right)$ $=\left(2i{\right)}^{1-n}\prod _{k=1}^{n-1}\left({e}^{\frac{ik\pi }{n}}-{e}^{-\frac{ik\pi }{n}}\right)$ $=\left(2i{\right)}^{1-n}{e}^{-i\frac{n\left(n-1\right)}{2}\frac{\pi }{n}}\prod _{k=1}^{n-1}\left({e}^{\frac{2ik\pi }{n}}-1\right)$ $=\left(-2{\right)}^{1-n}\prod _{k=1}^{n-1}\left({\xi }^{k}-1\right)$ $={2}^{1-n}\prod _{k=1}^{n-1}\left(1-{\xi }^{k}\right)$ where $\xi ={e}^{\frac{2i\pi }{n}}$ Now note that ${x}^{n}-1=\left(x-1\right)\sum _{k=0}^{n-1}$ and ${x}^{n}-1=\prod _{k=0}^{n-1}\left(x-{\xi }^{k}\right)$ Cancelling $\left(x-1\right)$ we have $\prod _{k=1}^{n-1}\left(x-{\xi }^{k}\right)=\sum _{k=0}^{n-1}{x}^{k}$ Substituting $x=1$ we have $\prod _{k=1}^{n-1}\left(1-{\xi }^{k}\right)=n$ $\therefore P=n{2}^{1-n}$ Step 2 In order to note that ${x}^{n}-1=\prod _{k=0}^{n-1}\left(x-{\xi }^{k}\right)$, note that are roots of ${x}^{n}-1$. Therefore by polynomial reminder theorem we have ${x}^{n}-1=Q\left(x\right)\prod _{k=0}^{n-1}\left(x-{\xi }^{k}\right)$ Comparing degrees we find Q(x) has degree 0. Comparing highest coefficients we conclude $Q\left(x\right)=1$ Step 3 We may instead use the identity $|1-{e}^{\frac{2ik\pi }{n}}==2\mathrm{sin}\left(\frac{k\pi }{n}\right),$ to establish immediately that $P=\prod _{k=1}^{n-1}\mathrm{sin}\left(\frac{k\pi }{n}\right)={2}^{1-n}\prod _{k=1}^{n-1}|1-{e}^{\frac{2ik\pi }{n}}|$, and $={2}^{1-n}|\prod _{k=1}^{n-1}\left(1-{e}^{\frac{2ik\pi }{n}}\right)|$ continue by applying the foregoing logic to the product to obtain $P=n{2}^{1-n}$

Vasquez

Step 1 Consider ${z}^{n}=1$, each root is ${\xi }_{k}=\mathrm{cos}\frac{2k\pi }{n}+i\mathrm{sin}\frac{2k\pi }{n}={e}^{i\frac{2k\pi }{n}}$ So, we have ${z}^{n}-1=\prod _{k=0}^{n-1}\left(z-{\xi }_{k}\right)$ $⇒\left(z-1\right)\left({z}^{n-1}+\cdots +{z}^{2}+z+1\right)=\left(z-{\xi }_{0}\right)$ $\prod _{k=1}^{n-1}\left(z-{\xi }_{k}\right)$ $⇒\left(z-1\right)\left({z}^{n-1}+\cdots +{z}^{2}+z+1\right)=\left(z-1\right)$ $\prod _{k=1}^{n-1}\left(z-{\xi }_{k}\right)$ $⇒{z}^{n-1}+\cdots +{z}^{2}+z+1=\prod _{k=1}^{n-1}\left(z-{\xi }_{k}\right)$ By substituting $z=1$, $⇒n=\prod _{k=1}^{n-1}\left(1-{\xi }_{k}\right)$ Next, take the modulus on both sides, $|n|=n=|\prod _{k=1}^{n-1}\left(1-{\xi }_{k}\right)|=\prod _{k=1}^{n-1}|\left(1-{\xi }_{k}\right)|$ $1-{\xi }_{k}=1-\left(\mathrm{cos}\frac{2k\pi }{n}+i\mathrm{sin}\frac{2k\pi }{n}\right)=2\mathrm{sin}$ $\frac{k\pi }{n}\left(\mathrm{sin}\frac{k\pi }{n}-i\mathrm{cos}\frac{k\pi }{n}\right)$ $|1-{\xi }_{k}|=2\mathrm{sin}\frac{k\pi }{n}$ So, $n={2}^{n-1}\prod _{k=1}^{n-1}\mathrm{sin}\frac{k\pi }{n}$ $\prod _{k=1}^{n-1}\mathrm{sin}\frac{k\pi }{n}=\frac{n}{{2}^{n-1}}$

alenahelenash