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Question Number 37244 by subhendubera201314@gmail.com last updated on 11/Jun/18

sin^2 (π/11)+sin^2 (2π/11)+...+sin^2 (5π/11)=?

$$\mathrm{sin}^{\mathrm{2}} \left(\pi/\mathrm{11}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2}\pi/\mathrm{11}\right)+...+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{5}\pi/\mathrm{11}\right)=? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Jun/18

sin^2 a+sin^2 2a+sin^2 3a+sin^2 4a+sin^2 5a a=(Π/(11)) =((1−cos2a)/2)+((1−cos4a)/2)+((1−cos6a)/2)+((1−cos8a)/2)+ ((1−cos10a)/2) =5×(1/2)−(1/2)(c2a+c4a+c6a+c8a+c10a) when c2a=cos2a etc s=c2a+c4a+c6a+c8a+c10a s×2sina=2sina.(c2a+c4a+c6a+c8a+c10a) s×2sina=2sina.cos2a+2sina.cos4a+ 2sinacos6a+2sina.cos8a+2sina.cos10a now look 2sina.cos2a=sin3a−sina 2sina.cos4a=sin5a−sin3a 2sina.cos6a=sin7a−sin5a 2sina.cos8a=sin9a−sin7a 2sina.cos10a=sin11a−sin9a now add them...addition of right side is =sin11a−sina =2cos6a.sin5a addition of left is s×2sina so s×2sina=2cos6a.sin5a s=((cos6a.sin5a )/(sina)) (5/2)−(1/2)(((cos6a.sin5a)/(sina))) =(5/2)−(1/4)(((2cos6a.sin5a)/(sina))) =(5/2)−(1/4)(((sin11a−sina)/(sina))) a=(Π/(11)) so 11a=Π sin11a=sinΠ=0 =(5/2)−(1/4)(((0−sina)/(sina))) =(5/2)+(1/4) =((10+1)/4)=((11)/4)

$${sin}^{\mathrm{2}} {a}+{sin}^{\mathrm{2}} \mathrm{2}{a}+{sin}^{\mathrm{2}} \mathrm{3}{a}+{sin}^{\mathrm{2}} \mathrm{4}{a}+{sin}^{\mathrm{2}} \mathrm{5}{a} \\ $$$${a}=\frac{\Pi}{\mathrm{11}} \\ $$$$=\frac{\mathrm{1}−{cos}\mathrm{2}{a}}{\mathrm{2}}+\frac{\mathrm{1}−{cos}\mathrm{4}{a}}{\mathrm{2}}+\frac{\mathrm{1}−{cos}\mathrm{6}{a}}{\mathrm{2}}+\frac{\mathrm{1}−{cos}\mathrm{8}{a}}{\mathrm{2}}+ \\ $$$$\frac{\mathrm{1}−{cos}\mathrm{10}{a}}{\mathrm{2}} \\ $$$$=\mathrm{5}×\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left({c}\mathrm{2}{a}+{c}\mathrm{4}{a}+{c}\mathrm{6}{a}+{c}\mathrm{8}{a}+{c}\mathrm{10}{a}\right) \\ $$$${when}\:{c}\mathrm{2}{a}={cos}\mathrm{2}{a}\:{etc} \\ $$$${s}={c}\mathrm{2}{a}+{c}\mathrm{4}{a}+{c}\mathrm{6}{a}+{c}\mathrm{8}{a}+{c}\mathrm{10}{a} \\ $$$${s}×\mathrm{2}{sina}=\mathrm{2}{sina}.\left({c}\mathrm{2}{a}+{c}\mathrm{4}{a}+{c}\mathrm{6}{a}+{c}\mathrm{8}{a}+{c}\mathrm{10}{a}\right) \\ $$$${s}×\mathrm{2}{sina}=\mathrm{2}{sina}.{cos}\mathrm{2}{a}+\mathrm{2}{sina}.{cos}\mathrm{4}{a}+ \\ $$$$\:\:\mathrm{2}{sinacos}\mathrm{6}{a}+\mathrm{2}{sina}.{cos}\mathrm{8}{a}+\mathrm{2}{sina}.{cos}\mathrm{10}{a} \\ $$$${now}\:{look} \\ $$$$\mathrm{2}{sina}.{cos}\mathrm{2}{a}={sin}\mathrm{3}{a}−{sina} \\ $$$$\mathrm{2}{sina}.{cos}\mathrm{4}{a}={sin}\mathrm{5}{a}−{sin}\mathrm{3}{a} \\ $$$$\mathrm{2}{sina}.{cos}\mathrm{6}{a}={sin}\mathrm{7}{a}−{sin}\mathrm{5}{a} \\ $$$$\mathrm{2}{sina}.{cos}\mathrm{8}{a}={sin}\mathrm{9}{a}−{sin}\mathrm{7}{a} \\ $$$$\mathrm{2}{sina}.{cos}\mathrm{10}{a}={sin}\mathrm{11}{a}−{sin}\mathrm{9}{a} \\ $$$$ \\ $$$${now}\:{add}\:{them}...{addition}\:{of}\:{right}\:{side}\:{is} \\ $$$$={sin}\mathrm{11}{a}−{sina} \\ $$$$=\mathrm{2}{cos}\mathrm{6}{a}.{sin}\mathrm{5}{a} \\ $$$$ \\ $$$${addition}\:{of}\:{left}\:{is}\:{s}×\mathrm{2}{sina} \\ $$$${so} \\ $$$${s}×\mathrm{2}{sina}=\mathrm{2}{cos}\mathrm{6}{a}.{sin}\mathrm{5}{a} \\ $$$${s}=\frac{{cos}\mathrm{6}{a}.{sin}\mathrm{5}{a}\:}{{sina}} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{cos}\mathrm{6}{a}.{sin}\mathrm{5}{a}}{{sina}}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{2}{cos}\mathrm{6}{a}.{sin}\mathrm{5}{a}}{{sina}}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{sin}\mathrm{11}{a}−{sina}}{{sina}}\right) \\ $$$${a}=\frac{\Pi}{\mathrm{11}}\:\:{so}\:\mathrm{11}{a}=\Pi \\ $$$${sin}\mathrm{11}{a}={sin}\Pi=\mathrm{0} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{0}−{sina}}{{sina}}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{10}+\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{11}}{\mathrm{4}} \\ $$$$ \\ $$

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