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Question Number 95703 by i jagooll last updated on 27/May/20
(1/(cos^2 10^o )) +(1/(sin^2 20^o )) + (1/(sin^2 40^o )) =?
$$\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}^{\mathrm{o}} }\:+\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{20}^{\mathrm{o}} }\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{40}^{\mathrm{o}} }\:=? \\ $$
Commented by john santu last updated on 27/May/20
this is equivalent to csc^2 ((π/9))+ csc^2 (((2π)/9))+ csc^2 (((4π)/9))
$$\mathrm{this}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\: \\ $$$$\mathrm{csc}^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{9}}\right)+\:\mathrm{csc}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)+\:\mathrm{csc}^{\mathrm{2}} \left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right) \\ $$
Answered by john santu last updated on 27/May/20
cos^2 10^o = sin^2 80^o sin 60^o =3sin 20^o −4sin^3 20^o =((√3)/2) sin 120^o =3sin 40^o −4sin^3 40^o =((√3)/2) sin (−240^o )=3sin (−80^o )−4sin^3 (−80^o )=((√3)/2) so sin 20^o ; sin 40^o & sin (−80^o ) are the roots of cubic equation of 3x−4x^3 =((√3)/2) or 4x^3 −3x=((√3)/2) ⇒x(4x^2 −3)=((√3)/2) ; squaring both sides x^2 (4x^2 −3)^2 =(3/4) , let x^2 = t ⇒t(4t−3)^2 = (3/4) ← are roots sin^2 20^o , sin^2 40^o , sin^2 80^o t(16t^2 −24t+9)−(3/4)=0 16t^3 −24t^2 +9t−(3/4)=0 64t^3 −96t^2 +36t−3=0 put t = (1/y) ⇒−3t^3 +36t^2 −96t+64=0 are roots (1/(sin^2 20^o )), (1/(sin^2 40^o )) and (1/(sin^2 80^o )) hence by vieta′s rule we get (1/(sin^2 20^o )) + (1/(sin^2 40^o )) + (1/(sin^2 80^o )) = ((−36)/(−3)) = 12
$$\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}^{\mathrm{o}} \:=\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{80}^{\mathrm{o}} \\ $$$$\mathrm{sin}\:\mathrm{60}^{\mathrm{o}} =\mathrm{3sin}\:\mathrm{20}^{\mathrm{o}} −\mathrm{4sin}\:^{\mathrm{3}} \mathrm{20}^{\mathrm{o}} \:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\mathrm{120}^{\mathrm{o}} =\mathrm{3sin}\:\mathrm{40}^{\mathrm{o}} −\mathrm{4sin}\:^{\mathrm{3}} \mathrm{40}^{\mathrm{o}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\left(−\mathrm{240}^{\mathrm{o}} \right)=\mathrm{3sin}\:\left(−\mathrm{80}^{\mathrm{o}} \right)−\mathrm{4sin}\:^{\mathrm{3}} \left(−\mathrm{80}^{\mathrm{o}} \right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{sin}\:\mathrm{20}^{\mathrm{o}} ;\:\mathrm{sin}\:\mathrm{40}^{\mathrm{o}} \:\&\:\mathrm{sin}\:\left(−\mathrm{80}^{\mathrm{o}} \right)\: \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{cubic}\:\mathrm{equation}\: \\ $$$$\mathrm{of}\:\mathrm{3x}−\mathrm{4x}^{\mathrm{3}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{or}\:\mathrm{4x}^{\mathrm{3}} −\mathrm{3x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}\left(\mathrm{4x}^{\mathrm{2}} −\mathrm{3}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:;\:\mathrm{squaring}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\mathrm{x}^{\mathrm{2}} \left(\mathrm{4x}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}}\:,\:\mathrm{let}\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{t}\: \\ $$$$\Rightarrow\mathrm{t}\left(\mathrm{4t}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{3}}{\mathrm{4}}\:\leftarrow\:\mathrm{are}\:\mathrm{roots}\: \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \mathrm{20}^{\mathrm{o}} ,\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{40}^{\mathrm{o}} ,\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{80}^{\mathrm{o}} \: \\ $$$$\mathrm{t}\left(\mathrm{16t}^{\mathrm{2}} −\mathrm{24t}+\mathrm{9}\right)−\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{16t}^{\mathrm{3}} −\mathrm{24t}^{\mathrm{2}} +\mathrm{9t}−\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{64t}^{\mathrm{3}} −\mathrm{96t}^{\mathrm{2}} +\mathrm{36t}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{put}\:\mathrm{t}\:=\:\frac{\mathrm{1}}{\mathrm{y}}\:\Rightarrow−\mathrm{3t}^{\mathrm{3}} +\mathrm{36t}^{\mathrm{2}} −\mathrm{96t}+\mathrm{64}=\mathrm{0} \\ $$$$\mathrm{are}\:\mathrm{roots}\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{20}^{\mathrm{o}} },\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{40}^{\mathrm{o}} }\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{80}^{\mathrm{o}} } \\ $$$$\mathrm{hence}\:\mathrm{by}\:\mathrm{vieta}'\mathrm{s}\:\mathrm{rule}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{20}^{\mathrm{o}} }\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{40}^{\mathrm{o}} }\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{80}^{\mathrm{o}} }\:=\:\frac{−\mathrm{36}}{−\mathrm{3}}\:=\:\mathrm{12} \\ $$
Commented by i jagooll last updated on 27/May/20
greatt
$$\mathrm{greatt} \\ $$

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