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Question Number 144947 by mathdanisur last updated on 30/Jun/21

Π_(k=1) ^(12) 2∙sin(((πk)/(24))) = ?

$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{12}} {\prod}}\mathrm{2}\centerdot{sin}\left(\frac{\pi{k}}{\mathrm{24}}\right)\:=\:? \\ $$

Answered by Olaf_Thorendsen last updated on 01/Jul/21

sin2θ = 2cosθsinθ 2cosθ = ((sin2θ)/(sinθ)) S = Π_(k=1) ^(12) 2sin(((kπ)/(24))) S = Π_(k=1) ^(12) 2cos((π/2)−((kπ)/(24))) S = Π_(k=1) ^(12) 2cos((((12−k)π)/(24))) S = Π_(k=0) ^(11) 2cos(((kπ)/(24))) S = 2Π_(k=1) ^(11) 2cos(((kπ)/(24))) S = 2Π_(k=1) ^(11) ((sin(((kπ)/(12))))/(sin(((kπ)/(24))))) S = 2((sin((π/(12))))/(sin((π/(24)))))×((sin((π/6)))/(sin((π/(12)))))×...×((sin(((11π)/(12))))/(sin(((11π)/(24))))) S = 2((sin(((11π)/(12))))/(sin((π/(24))))) S = 2((sin(π−((11π)/(12))))/(sin((π/(24))))) S = 2((sin((π/(12))))/(sin((π/(24))))) S = 4cos(π/(24))

$$\mathrm{sin2}\theta\:=\:\mathrm{2cos}\theta\mathrm{sin}\theta \\ $$$$\mathrm{2cos}\theta\:=\:\frac{\mathrm{sin2}\theta}{\mathrm{sin}\theta} \\ $$$$\mathrm{S}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{12}} {\prod}}\mathrm{2sin}\left(\frac{{k}\pi}{\mathrm{24}}\right) \\ $$$$\mathrm{S}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{12}} {\prod}}\mathrm{2cos}\left(\frac{\pi}{\mathrm{2}}−\frac{{k}\pi}{\mathrm{24}}\right) \\ $$$$\mathrm{S}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{12}} {\prod}}\mathrm{2cos}\left(\frac{\left(\mathrm{12}−{k}\right)\pi}{\mathrm{24}}\right) \\ $$$$\mathrm{S}\:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{11}} {\prod}}\mathrm{2cos}\left(\frac{{k}\pi}{\mathrm{24}}\right) \\ $$$$\mathrm{S}\:=\:\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\mathrm{11}} {\prod}}\mathrm{2cos}\left(\frac{{k}\pi}{\mathrm{24}}\right) \\ $$$$\mathrm{S}\:=\:\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\mathrm{11}} {\prod}}\frac{\mathrm{sin}\left(\frac{{k}\pi}{\mathrm{12}}\right)}{\mathrm{sin}\left(\frac{{k}\pi}{\mathrm{24}}\right)} \\ $$$$\mathrm{S}\:=\:\mathrm{2}\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{12}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{24}}\right)}×\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{12}}\right)}×...×\frac{\mathrm{sin}\left(\frac{\mathrm{11}\pi}{\mathrm{12}}\right)}{\mathrm{sin}\left(\frac{\mathrm{11}\pi}{\mathrm{24}}\right)} \\ $$$$\mathrm{S}\:=\:\mathrm{2}\frac{\mathrm{sin}\left(\frac{\mathrm{11}\pi}{\mathrm{12}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{24}}\right)} \\ $$$$\mathrm{S}\:=\:\mathrm{2}\frac{\mathrm{sin}\left(\pi−\frac{\mathrm{11}\pi}{\mathrm{12}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{24}}\right)} \\ $$$$\mathrm{S}\:=\:\mathrm{2}\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{12}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{24}}\right)} \\ $$$$\mathrm{S}\:=\:\mathrm{4cos}\frac{\pi}{\mathrm{24}} \\ $$

Commented by mathdanisur last updated on 01/Jul/21

cool Sir, thankyou, but answer 2(√3)

$${cool}\:{Sir},\:{thankyou},\:{but}\:{answer}\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$

Commented by Ar Brandon last updated on 01/Jul/21

S=4cos(π/(24))=4cos(((15)/2))° cos15°=(((√6)+(√2))/4)=2cos^2 (((15)/2))°−1 ⇒cos^2 (((15)/2))°=(((√6)+(√2)+4)/8) ⇒cos(((15)/2))°=((√((√6)+(√2)+4))/(2(√2)))

$$\mathrm{S}=\mathrm{4cos}\frac{\pi}{\mathrm{24}}=\mathrm{4cos}\left(\frac{\mathrm{15}}{\mathrm{2}}\right)° \\ $$$$\mathrm{cos15}°=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}=\mathrm{2cos}^{\mathrm{2}} \left(\frac{\mathrm{15}}{\mathrm{2}}\right)°−\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{15}}{\mathrm{2}}\right)°=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}+\mathrm{4}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{cos}\left(\frac{\mathrm{15}}{\mathrm{2}}\right)°=\frac{\sqrt{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}+\mathrm{4}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$

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