Question Number 30743 by abdo imad last updated on 25/Feb/18
![decompose inside R[x] p(x)=x^(2n+1) −1 then find Π_(k=1) ^n sin( ((kπ)/(2n+1))) .](https://www.tinkutara.com/question/Q30743.png)
$${decompose}\:{inside}\:{R}\left[{x}\right]\:\:{p}\left({x}\right)={x}^{\mathrm{2}{n}+\mathrm{1}} \:−\mathrm{1}\:{then}\:{find} \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\:\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:. \\ $$
Commented by abdo imad last updated on 02/Mar/18
![the roots of p(x) ? z^(2n+1) −1=0 ⇔ z^(2n+1) =e^(i(2kπ)) so the roots are z_k = e^(i((2kπ)/(2n+1))) with k∈ [[0,2n]] ⇒ p(x)=Π_(k=0) ^(2n) (x−z_k )=(x−1) Π_(k=1) ^(2n) (x− e^(i((2kπ)/(2n+1))) ) ⇒ ((p(x))/(x−1)) =Π_(k=1) ^(2n) (x−e^(i((2kπ)/(2n+1))) ) ⇒lim_(x→1) ((p(x))/(x−1)) =Π_(k=1) ^(2n) (1−e^(i((2kπ)/(2n+1))) ) ⇒ (2n+1)=Π_(k=1) ^(2n) (1−cos(((2kπ)/(2n+1))) −isin(((2kπ)/(2n+1)))) =Π_(k=1) ^(2n) (2sin^2 ( ((kπ)/(2n+1))) −2i sin(((kπ)/(2n+1)))cos(((kπ)/(2n+1)))) =2^(2n) (−1)^n Π_(k=1) ^(2n) sin(((kπ)/(2n+1))) e^(i(((kπ)/(2n+1)))) =2^(2n) (−i)^n ( Π_(k=1) ^(2n) sin(((kπ)/(2n+1)))) e^(i(π/(2n+1))((2n(2n+1))/2)) =2^(2n) Π_(k=1) ^(2n) sin(((kπ)/(2n+1))) but Π_(k=1) ^(2n) sin(((kπ)/(2n+1)))=Π_(k=1) ^n sin(((kπ)/(2n+1))) Π_(k=n+1) ^(2n) sin(((kπ)/(2n+1)))the ch.k−n=j give Π_(k=n+1) ^(2n) sin(((kπ)/(2n+1))) =Π_(j=1) ^n sin( (((n+j)π)/(2n+1))) =Π_(j=1) ^n sin(π −((nπ +jπ)/(2n+1)))=Π_(j=1) ^n sin(((2nπ +π −nπ −jπ)/(2n+1))) =Π_(j=1) ^n sin ((((n+1−j)π)/(2n+1)))=Π_(k=1) ^n sin( ((kπ)/(2n+1)) ) by ch.n+1−j=k⇒ 2n+1=(2^n )^(2 ) (Π_(k=1) ^n sin(((kπ)/(2n+1))))^2 ⇒ Π_(k=1) ^n sin(((kπ)/(2n+1))) =((√(2n+1))/2^n ) .](https://www.tinkutara.com/question/Q31014.png)
$${the}\:{roots}\:{of}\:{p}\left({x}\right)\:?\:\:{z}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:{z}^{\mathrm{2}{n}+\mathrm{1}} ={e}^{{i}\left(\mathrm{2}{k}\pi\right)} \\ $$$${so}\:{the}\:{roots}\:{are}\:{z}_{{k}} =\:{e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{2}{n}+\mathrm{1}}} \:\:{with}\:{k}\in\:\left[\left[\mathrm{0},\mathrm{2}{n}\right]\right]\:\Rightarrow \\ $$$${p}\left({x}\right)=\prod_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:\left({x}−{z}_{{k}} \right)=\left({x}−\mathrm{1}\right)\:\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \left({x}−\:{e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{2}{n}+\mathrm{1}}} \right)\:\Rightarrow \\ $$$$\frac{{p}\left({x}\right)}{{x}−\mathrm{1}}\:=\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:\left({x}−{e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{2}{n}+\mathrm{1}}} \right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}} \:\:\frac{{p}\left({x}\right)}{{x}−\mathrm{1}}\:=\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \left(\mathrm{1}−{e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{2}{n}+\mathrm{1}}} \right) \\ $$$$\Rightarrow\:\left(\mathrm{2}{n}+\mathrm{1}\right)=\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \left(\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:−{isin}\left(\frac{\mathrm{2}{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\right) \\ $$$$=\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:\left(\mathrm{2}{sin}^{\mathrm{2}} \left(\:\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:−\mathrm{2}{i}\:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right){cos}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\right) \\ $$$$=\mathrm{2}^{\mathrm{2}{n}} \:\left(−\mathrm{1}\right)^{{n}} \:\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:{e}^{{i}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)} \\ $$$$=\mathrm{2}^{\mathrm{2}{n}} \left(−{i}\right)^{{n}} \left(\:\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\right)\:{e}^{{i}\frac{\pi}{\mathrm{2}{n}+\mathrm{1}}\frac{\mathrm{2}{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}}} \\ $$$$=\mathrm{2}^{\mathrm{2}{n}} \:\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:{but} \\ $$$$\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:\prod_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} {sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right){the} \\ $$$${ch}.{k}−{n}={j}\:{give} \\ $$$$\prod_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\prod_{{j}=\mathrm{1}} ^{{n}} \:{sin}\left(\:\frac{\left({n}+{j}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$=\prod_{{j}=\mathrm{1}} ^{{n}} \:{sin}\left(\pi\:−\frac{{n}\pi\:+{j}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\prod_{{j}=\mathrm{1}} ^{{n}} {sin}\left(\frac{\mathrm{2}{n}\pi\:+\pi\:−{n}\pi\:−{j}\pi}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$=\prod_{{j}=\mathrm{1}} ^{{n}} \:{sin}\:\left(\frac{\left({n}+\mathrm{1}−{j}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\:\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\:\right)\:{by}\:{ch}.{n}+\mathrm{1}−{j}={k}\Rightarrow \\ $$$$\mathrm{2}{n}+\mathrm{1}=\left(\mathrm{2}^{{n}} \right)^{\mathrm{2}\:} \:\:\left(\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \:\:\:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\frac{\sqrt{\mathrm{2}{n}+\mathrm{1}}}{\mathrm{2}^{{n}} }\:. \\ $$$$ \\ $$