Question Number 27153 by abdo imad last updated on 02/Jan/18
$${find}\:{the}\:{value}\:{of}\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\:\right)\:. \\ $$
Commented by abdo imad last updated on 03/Jan/18
![let introduce the polynomial p(x)= x^(2n) −1 the roots of p(x) are the complex z_(k ) =e^(i((2k)/n)) and k∈[[0,2n−1]] and p(x)=λ Π_(k=0) ^(n−1) (x−z_k ) its clear that λ=1 and p(x)=Π_(k=0) ^(n−1) (x−z_k ) we have z_0 = 1 , z_1 = e^(i(π/n)) , z_2 = e^(i((2π)/n)) z_(n−1) = e^(i(((n−1)π)/n)) , z_n = −1 , z_(n+1) = e^(i(((n+1)π)/n)) , z_(2n−1) =e^(i(((2n−1)π)/(2n))) we see that z_(2n−1) =z_1 ^− , z_(2n−2) =z_2 ^− , z_(n+1) =z_(n−1) ^− ⇒ p(x)= (x^2 −1) Π_(k=1) ^(n−1) (x −z_k )(x−z_k ^− ) =(x^2 −1) Π_(k=1) ^(n−1) (x^2 −2cos(((kπ)/n))x +1) and for x^2 ≠1 ((p(x))/(x^2 −1)) = Π_(k=1) ^(n−1) ( x^2 −2cos(((kπ)/n))x +1) and by using hospital theoem lim_(x−>1) Π_(k=1) ^(n−1) ( x^2 −2cos(((kπ)/n))x +1) =lim_(x−>1) ((p^′ (x))/(2x)) Π_(k=1) ^(n−1) 2(1−cos(((kπ)/n)))= lim_(x−>1) ((2nx^(2n−1) )/(2x)) =n ⇒ 4^(n−1) Π_(k=1) ^(k=n−1) sin^2 (((kπ)/(2n)) )=n ⇒ Π_(k=1) ^(n−1) sin^2 (((kπ)/(2n)) ) = (n/4^(n−1) ) ⇒ Π_(k=1) ^(n−1) sin (((kπ)/(2n)) )= ((√n)/2^(n−1) ) ( n≥2)](Q27214.png)
$${let}\:{introduce}\:{the}\:{polynomial}\:{p}\left({x}\right)=\:{x}^{\mathrm{2}{n}} \:−\mathrm{1}\:\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$${are}\:{the}\:{complex}\:\:{z}_{{k}\:} ={e}^{{i}\frac{\mathrm{2}{k}}{{n}}} \:\:{and}\:{k}\in\left[\left[\mathrm{0},\mathrm{2}{n}−\mathrm{1}\right]\right]\:{and} \\ $$$${p}\left({x}\right)=\lambda\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{z}_{{k}} \:\right)\:{its}\:{clear}\:{that}\:\lambda=\mathrm{1}\:\:{and} \\ $$$${p}\left({x}\right)=\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{z}_{{k}} \right)\:{we}\:{have}\:{z}_{\mathrm{0}} =\:\mathrm{1}\:\:,\:{z}_{\mathrm{1}} =\:{e}^{{i}\frac{\pi}{{n}}} \:\:\:,\:{z}_{\mathrm{2}} =\:{e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \\ $$$${z}_{{n}−\mathrm{1}} =\:{e}^{{i}\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}} \:\:\:,\:\:{z}_{{n}} =\:−\mathrm{1}\:\:\:,\:{z}_{{n}+\mathrm{1}} =\:{e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}} \:\:,\:\:\:{z}_{\mathrm{2}{n}−\mathrm{1}} \:={e}^{{i}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}}} \\ $$$${we}\:{see}\:{that}\:\:{z}_{\mathrm{2}{n}−\mathrm{1}} \:={z}_{\mathrm{1}} ^{−} \:\:\:\:\:,\:\:{z}_{\mathrm{2}{n}−\mathrm{2}} ={z}_{\mathrm{2}} ^{−} \:\:\:,\:\:\:{z}_{{n}+\mathrm{1}} \:={z}_{{n}−\mathrm{1}} ^{−} \\ $$$$\Rightarrow\:{p}\left({x}\right)=\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({x}\:−{z}_{{k}} \:\right)\left({x}−{z}_{{k}} ^{−} \:\right) \\ $$$$=\left({x}^{\mathrm{2}} \:−\mathrm{1}\right)\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({x}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}\:+\mathrm{1}\right)\:{and}\:{for}\:{x}^{\mathrm{2}} \neq\mathrm{1} \\ $$$$\frac{{p}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:\:=\:\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\:{x}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}\:+\mathrm{1}\right)\:{and}\:{by}\:{using}\:{hospital}\:{theoem} \\ $$$${lim}_{{x}−>\mathrm{1}} \:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\:{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}\:+\mathrm{1}\right)\:={lim}_{{x}−>\mathrm{1}} \frac{{p}^{'} \left({x}\right)}{\mathrm{2}{x}} \\ $$$$\:\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \mathrm{2}\left(\mathrm{1}−{cos}\left(\frac{{k}\pi}{{n}}\right)\right)=\:{lim}_{{x}−>\mathrm{1}} \:\frac{\mathrm{2}{nx}^{\mathrm{2}{n}−\mathrm{1}} }{\mathrm{2}{x}}\:\:={n} \\ $$$$\Rightarrow\:\mathrm{4}^{{n}−\mathrm{1}} \:\prod_{{k}=\mathrm{1}} ^{{k}={n}−\mathrm{1}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}}\:\right)={n} \\ $$$$\Rightarrow\:\:\:\:\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}}\:\right)\:=\:\frac{{n}}{\mathrm{4}^{{n}−\mathrm{1}} } \\ $$$$\Rightarrow\:\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\:\left(\frac{{k}\pi}{\mathrm{2}{n}}\:\right)=\:\:\frac{\sqrt{{n}}}{\mathrm{2}^{{n}−\mathrm{1}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\:\:{n}\geqslant\mathrm{2}\right) \\ $$
Commented by Tinkutara last updated on 03/Jan/18
I have already asked this a long time back. See this at Q 19292.