sin-pi-14-sin-2pi-14-sin-3pi-14-sin-6pi-14-

Question Number 118899 by bobhans last updated on 20/Oct/20

sin ((π/(14)))sin (((2π)/(14)))sin (((3π)/(14)))...sin (((6π)/(14)))=?

$$\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{14}}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{14}}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{14}}\right)…\mathrm{sin}\:\left(\frac{\mathrm{6}\pi}{\mathrm{14}}\right)=? \\ $$

Answered by TANMAY PANACEA last updated on 20/Oct/20

taking help from S.L Loney Trigonometry 2^((n−1)/2) sin(((2π)/(2n)))sin(((4π)/(2n)))sin(((6π)/(2n)))...sin(((n−2)/(2n)))π=(√n) p=sin(((2π)/(28)))sin(((4π)/(28)))sin(((6π)/(28)))sin(((8π)/(28)))sin(((10π)/(28)))sin(((12π)/(28))) so comparing 2n=28→n=14 required answer=((√n)/2^((n−1)/2) ) =((√(14))/2^((13)/2) )=(((√7) ×(√2))/(2^6 ×(√2)))=((√7)/2^6 ) Answer

$${taking}\:{help}\:{from}\:{S}.{L}\:{Loney}\:{Trigonometry} \\ $$$$\mathrm{2}^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {sin}\left(\frac{\mathrm{2}\pi}{\mathrm{2}{n}}\right){sin}\left(\frac{\mathrm{4}\pi}{\mathrm{2}{n}}\right){sin}\left(\frac{\mathrm{6}\pi}{\mathrm{2}{n}}\right)…{sin}\left(\frac{{n}−\mathrm{2}}{\mathrm{2}{n}}\right)\pi=\sqrt{{n}}\: \\ $$$${p}={sin}\left(\frac{\mathrm{2}\pi}{\mathrm{28}}\right){sin}\left(\frac{\mathrm{4}\pi}{\mathrm{28}}\right){sin}\left(\frac{\mathrm{6}\pi}{\mathrm{28}}\right){sin}\left(\frac{\mathrm{8}\pi}{\mathrm{28}}\right){sin}\left(\frac{\mathrm{10}\pi}{\mathrm{28}}\right){sin}\left(\frac{\mathrm{12}\pi}{\mathrm{28}}\right) \\ $$$${so}\:{comparing} \\ $$$$\mathrm{2}{n}=\mathrm{28}\rightarrow{n}=\mathrm{14} \\ $$$${required}\:{answer}=\frac{\sqrt{{n}}}{\mathrm{2}^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} } \\ $$$$=\frac{\sqrt{\mathrm{14}}}{\mathrm{2}^{\frac{\mathrm{13}}{\mathrm{2}}} }=\frac{\sqrt{\mathrm{7}}\:×\sqrt{\mathrm{2}}}{\mathrm{2}^{\mathrm{6}} ×\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}^{\mathrm{6}} }\:{Answer} \\ $$

Commented by TANMAY PANACEA last updated on 20/Oct/20

Answered by mindispower last updated on 20/Oct/20

let z^(14) −1=0 ⇒z^(14) −1=Π_(k=0) ^(13) (X−e^((2ikπ)/(14)) )⇒Π_(k=1) ^(13) (1−e^((2ikπ)/(14)) )=14 (1−e^((2ikπ)/(14)) )=e^((ikπ)/(14)) (e^((−ikπ)/(14)) −e^((ikπ)/(14)) ) =−2isin(((kπ)/(14)))e^(ik(π/(14))) ⇒Π_(k=1) ^(13) (−2isin(((kπ)/(14)))).e^(i(π/(14))Σk) )=14 ⇒(−2i)^(13) Π_(k=1) ^6 sin(((kπ)/(14))).Π_(k=7) ^(13) sin(((kπ)/(14)))e^(i(π/(14))(7.13)) =14 ⇒(−2)^(13) .i.Π_(k=1) ^6 sin(((kπ)/(14))).Π_(k=1) ^7 sin((((14−k)π)/(14)))e^((1/2)13iπ) =14 sin(π−x)=sin(x),sin(((7π)/(14)))=sin((π/2))=1,e^((13iπ)/2) =i ⇒(2)^(13) [Π_(k=1) ^6 sin(((kπ)/(14)))]^2 =14 ⇒Π_(k=1) ^6 sin(((kπ)/(14)))=((√7)/2^6 )

$${let}\:{z}^{\mathrm{14}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{z}^{\mathrm{14}} −\mathrm{1}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{13}} {\prod}}\left({X}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{14}}} \right)\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\mathrm{13}} {\prod}}\left(\mathrm{1}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{14}}} \right)=\mathrm{14} \\ $$$$\left(\mathrm{1}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{14}}} \right)={e}^{\frac{{ik}\pi}{\mathrm{14}}} \left({e}^{\frac{−{ik}\pi}{\mathrm{14}}} −{e}^{\frac{{ik}\pi}{\mathrm{14}}} \right) \\ $$$$=−\mathrm{2}{isin}\left(\frac{{k}\pi}{\mathrm{14}}\right){e}^{{ik}\frac{\pi}{\mathrm{14}}} \\ $$$$\left.\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\mathrm{13}} {\prod}}\left(−\mathrm{2}{isin}\left(\frac{{k}\pi}{\mathrm{14}}\right)\right).{e}^{{i}\frac{\pi}{\mathrm{14}}\Sigma{k}} \right)=\mathrm{14} \\ $$$$\Rightarrow\left(−\mathrm{2}{i}\right)^{\mathrm{13}} \underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{14}}\right).\underset{{k}=\mathrm{7}} {\overset{\mathrm{13}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{14}}\right){e}^{{i}\frac{\pi}{\mathrm{14}}\left(\mathrm{7}.\mathrm{13}\right)} =\mathrm{14} \\ $$$$\Rightarrow\left(−\mathrm{2}\right)^{\mathrm{13}} .{i}.\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{14}}\right).\underset{{k}=\mathrm{1}} {\overset{\mathrm{7}} {\prod}}{sin}\left(\frac{\left(\mathrm{14}−{k}\right)\pi}{\mathrm{14}}\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{13}{i}\pi} =\mathrm{14} \\ $$$${sin}\left(\pi−{x}\right)={sin}\left({x}\right),{sin}\left(\frac{\mathrm{7}\pi}{\mathrm{14}}\right)={sin}\left(\frac{\pi}{\mathrm{2}}\right)=\mathrm{1},{e}^{\frac{\mathrm{13}{i}\pi}{\mathrm{2}}} ={i} \\ $$$$\Rightarrow\left(\mathrm{2}\right)^{\mathrm{13}} \left[\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{14}}\right)\right]^{\mathrm{2}} =\mathrm{14} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{14}}\right)=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}^{\mathrm{6}} } \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by TANMAY PANACEA last updated on 20/Oct/20