Question Number 19292 by Tinkutara last updated on 08/Aug/17
$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{equality} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}}\:…\:\mathrm{sin}\:\frac{\left({n}\:−\:\mathrm{1}\right)\pi}{\mathrm{2}{n}}\:=\:\frac{\sqrt{{n}}}{\mathrm{2}^{{n}−\mathrm{1}} }\:. \\ $$
Commented by prakash jain last updated on 09/Aug/17
$$\mathrm{you}\:\mathrm{can}\:\mathrm{try}\:\mathrm{with}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{substitution} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}=\frac{{e}^{\frac{{i}\pi}{\mathrm{2}{n}}} −{e}^{−\frac{{i}\pi}{\mathrm{2}{n}}} }{\mathrm{2}{i}} \\ $$
Answered by Tinkutara last updated on 23/Aug/17
![x^(2n) −1=(x−1)(x−α)(x−α^2 )....(x−α^(2n−1) ) where α = e^(((2π)/(2n))i) = e^((π/n)i) On differentiating both sides, 2nx^(2n−1) =(x−α)(x−α^2 )....(x−α^(2n−1) ) +(x−1)(x−α^2 )(x−α^3 )....(x−α^(2n−1) ) +(x−1)(x−α)(x−α^3 )....+.... Put x = 1 2n=(1−α)(1−α^2 )(1−α^3 )....(1−α^(2n−1) ) Since 1 − e^(iθ) = 2 sin (θ/2) e^(((θ/2) − (π/2))i) 2n=(1−e^((π/n)i) )(1−e^(((2π)/n)i) )(1−e^(((3π)/n)i) )....(1−e^((π/n)(2n−1)i) ) =2^(2n−1) sin(π/(2n))sin((2π)/(2n))....sin(((2n−1)π)/(2n))e^(((π/(2n))−(π/2))i) e^((((2π)/(2n))−(π/2))i) ....e^((((π(2n−1))/(2n))−(π/2))i) =2^(2n−1) sin(π/(2n))sin((2π)/(2n))....sin(((2n−1)π)/(2n))e^(i[(π/(2n))(((2n(2n−1))/2))−(2n−1)(π/2)]) 2n=2^(2n−1) sin(π/(2n))sin((2π)/(2n))....sin(((2n−1)π)/(2n))e^0 ((2n)/2^(2n−1) )=sin(π/(2n))sin((2π)/(2n))....sin(((n−1)π)/(2n))sin((nπ)/(2n))....sin(π/(2n)) (n/2^(2n−2) ) = sin^2 (π/(2n)) sin^2 ((2π)/(2n)) .... sin^2 (((n−1)π)/(2n)) ((√n)/2^(n−1) ) = sin (π/(2n)) sin ((2π)/(2n)) .... sin (((n−1)π)/(2n))](https://www.tinkutara.com/question/Q20189.png)
$${x}^{\mathrm{2}{n}} −\mathrm{1}=\left({x}−\mathrm{1}\right)\left({x}−\alpha\right)\left({x}−\alpha^{\mathrm{2}} \right)….\left({x}−\alpha^{\mathrm{2}{n}−\mathrm{1}} \right)\:\mathrm{where}\:\alpha\:=\:{e}^{\frac{\mathrm{2}\pi}{\mathrm{2}{n}}{i}} \:=\:{e}^{\frac{\pi}{{n}}{i}} \\ $$$$\mathrm{On}\:\mathrm{differentiating}\:\mathrm{both}\:\mathrm{sides}, \\ $$$$\mathrm{2}{nx}^{\mathrm{2}{n}−\mathrm{1}} =\left({x}−\alpha\right)\left({x}−\alpha^{\mathrm{2}} \right)….\left({x}−\alpha^{\mathrm{2}{n}−\mathrm{1}} \right) \\ $$$$+\left({x}−\mathrm{1}\right)\left({x}−\alpha^{\mathrm{2}} \right)\left({x}−\alpha^{\mathrm{3}} \right)….\left({x}−\alpha^{\mathrm{2}{n}−\mathrm{1}} \right) \\ $$$$+\left({x}−\mathrm{1}\right)\left({x}−\alpha\right)\left({x}−\alpha^{\mathrm{3}} \right)….+…. \\ $$$$\mathrm{Put}\:{x}\:=\:\mathrm{1} \\ $$$$\mathrm{2}{n}=\left(\mathrm{1}−\alpha\right)\left(\mathrm{1}−\alpha^{\mathrm{2}} \right)\left(\mathrm{1}−\alpha^{\mathrm{3}} \right)….\left(\mathrm{1}−\alpha^{\mathrm{2}{n}−\mathrm{1}} \right) \\ $$$$\mathrm{Since}\:\mathrm{1}\:−\:{e}^{{i}\theta} \:=\:\mathrm{2}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:{e}^{\left(\frac{\theta}{\mathrm{2}}\:−\:\frac{\pi}{\mathrm{2}}\right){i}} \\ $$$$\mathrm{2}{n}=\left(\mathrm{1}−{e}^{\frac{\pi}{{n}}{i}} \right)\left(\mathrm{1}−{e}^{\frac{\mathrm{2}\pi}{{n}}{i}} \right)\left(\mathrm{1}−{e}^{\frac{\mathrm{3}\pi}{{n}}{i}} \right)….\left(\mathrm{1}−{e}^{\frac{\pi}{{n}}\left(\mathrm{2}{n}−\mathrm{1}\right){i}} \right) \\ $$$$=\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \mathrm{sin}\frac{\pi}{\mathrm{2}{n}}\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{2}{n}}….\mathrm{sin}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}}{e}^{\left(\frac{\pi}{\mathrm{2}{n}}−\frac{\pi}{\mathrm{2}}\right){i}} {e}^{\left(\frac{\mathrm{2}\pi}{\mathrm{2}{n}}−\frac{\pi}{\mathrm{2}}\right){i}} ….{e}^{\left(\frac{\pi\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}{n}}−\frac{\pi}{\mathrm{2}}\right){i}} \\ $$$$=\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \mathrm{sin}\frac{\pi}{\mathrm{2}{n}}\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{2}{n}}….\mathrm{sin}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}}{e}^{{i}\left[\frac{\pi}{\mathrm{2}{n}}\left(\frac{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}}\right)−\left(\mathrm{2}{n}−\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\right]} \\ $$$$\mathrm{2}{n}=\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \mathrm{sin}\frac{\pi}{\mathrm{2}{n}}\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{2}{n}}….\mathrm{sin}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}}{e}^{\mathrm{0}} \\ $$$$\frac{\mathrm{2}{n}}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }=\mathrm{sin}\frac{\pi}{\mathrm{2}{n}}\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{2}{n}}….\mathrm{sin}\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}}\mathrm{sin}\frac{{n}\pi}{\mathrm{2}{n}}….\mathrm{sin}\frac{\pi}{\mathrm{2}{n}} \\ $$$$\frac{{n}}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} }\:=\:\mathrm{sin}^{\mathrm{2}} \:\frac{\pi}{\mathrm{2}{n}}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\mathrm{2}\pi}{\mathrm{2}{n}}\:….\:\mathrm{sin}^{\mathrm{2}} \:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}} \\ $$$$\frac{\sqrt{{n}}}{\mathrm{2}^{{n}−\mathrm{1}} }\:=\:\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}}\:….\:\mathrm{sin}\:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}} \\ $$