Question Number 28370 by abdo imad last updated on 24/Jan/18
![1) factorizse p(x) =x^n −1 inside C[x] 2) find the value of Π_(k=1) ^(n−1) sin(((kπ)/n)) 3)find also the value of Π_(k=0) ^(n−1) sin(((kπ)/n) +θ).](https://www.tinkutara.com/question/Q28370.png)
$$\left.\mathrm{1}\right)\:{factorizse}\:{p}\left({x}\right)\:={x}^{{n}} \:−\mathrm{1}\:\:{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{{n}}\right) \\ $$$$\left.\mathrm{3}\right){find}\:{also}\:{the}\:{value}\:{of}\:\:\:\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{sin}\left(\frac{{k}\pi}{{n}}\:+\theta\right). \\ $$
Commented by abdo imad last updated on 26/Jan/18
![1)the roots of p(x) are the complex z_k = e^(i((2kπ)/n)) and k ∈[[0,n−1]] and p(x)=λ Π_(k=1) ^(n−1) (x−z_k ) .it s clear that λ=1 2)p(x)= (x−1) Π_(k=1) ^(n−1) ( x− e^(i((2kπ)/n)) ) for x≠1 ((p(x))/(x−1)) = Π_(k=1) ^(n−1) (x− e^(i ((2kπ)/n)) ) ⇒ lim_(x→1) ((x^n −1)/(x−1)) = Π_(k=1) ^(n−1) (1−cos(((2kπ)/n)) −isin(((2kπ)/n))) ⇒ n= Π_(k=1) ^(n−1) (2 sin^ (((kπ)/n)) −2isin(((kπ)/n))cos(((kπ)/n))) ⇒n= (−2i)^(n−1) Π_(k=1) ^(n−1) (sin(((kπ)/n))(e^(i((kπ)/n)) )) ⇒ n= (−2i)^(n−1) ( Π_(k=1) ^(n−1) sin(((kπ)/n))) e^(i(π/n)Σ_(k=1) ^(n−1) k) ⇒n= (−2i)^(n−1) e^(i(π/n)((n(n−1))/2)) Π_(k=1) ^(n−1) sin(((kπ)/n)) =(−i)^(n−1) i^(n−1) 2^(n−1) Π_(k=1) ^(n−1) sin(((kπ)/n))= 2^(n−1) Π_(k=1) ^(n−1) sin(((kπ)/n)) ⇒ Π_(k=1) ^(n−1) sin(((kπ)/n))= (n/2^(n−1) ) . (n≥2) ....be continued.....](https://www.tinkutara.com/question/Q28450.png)
$$\left.\mathrm{1}\right){the}\:{roots}\:{of}\:{p}\left({x}\right)\:{are}\:{the}\:{complex}\:{z}_{{k}} =\:{e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:\:{and}\:{k}\:\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$${and}\:{p}\left({x}\right)=\lambda\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left({x}−{z}_{{k}} \right)\:.{it}\:{s}\:{clear}\:{that}\:\lambda=\mathrm{1}\: \\ $$$$\left.\mathrm{2}\right){p}\left({x}\right)=\:\left({x}−\mathrm{1}\right)\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\:{x}−\:{e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:\right)\:\:\:{for}\:{x}\neq\mathrm{1} \\ $$$$\frac{{p}\left({x}\right)}{{x}−\mathrm{1}}\:=\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left({x}−\:{e}^{{i}\:\frac{\mathrm{2}{k}\pi}{{n}}} \right)\: \\ $$$$\Rightarrow\:\:{lim}_{{x}\rightarrow\mathrm{1}} \:\:\frac{{x}^{{n}} −\mathrm{1}}{{x}−\mathrm{1}}\:=\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left(\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)\:−{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)\right) \\ $$$$\Rightarrow\:{n}=\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{2}\:{sin}^{} \left(\frac{{k}\pi}{{n}}\right)\:−\mathrm{2}{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$$\Rightarrow{n}=\:\left(−\mathrm{2}{i}\right)^{{n}−\mathrm{1}} \:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({sin}\left(\frac{{k}\pi}{{n}}\right)\left({e}^{{i}\frac{{k}\pi}{{n}}} \right)\right) \\ $$$$\Rightarrow\:{n}=\:\left(−\mathrm{2}{i}\right)^{{n}−\mathrm{1}} \left(\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{{n}}\right)\right)\:{e}^{{i}\frac{\pi}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {k}} \\ $$$$\Rightarrow{n}=\:\left(−\mathrm{2}{i}\right)^{{n}−\mathrm{1}} \:{e}^{{i}\frac{\pi}{{n}}\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} \prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{{n}}\right) \\ $$$$=\left(−{i}\right)^{{n}−\mathrm{1}} \:{i}^{{n}−\mathrm{1}} \:\:\mathrm{2}^{{n}−\mathrm{1}} \prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{{n}}\right)=\:\mathrm{2}^{{n}−\mathrm{1}} \:\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{{n}}\right) \\ $$$$\Rightarrow\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{{n}}\right)=\:\frac{{n}}{\mathrm{2}^{{n}−\mathrm{1}} }\:\:.\:\:\:\:\left({n}\geqslant\mathrm{2}\right)\:\:\:….{be}\:{continued}….. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$