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Question Number 8289 by rhm last updated on 07/Oct/16

what is value sin 36° plese give me answer

$${what}\:{is}\:{value} \\ $$$${sin}\:\mathrm{36}° \\ $$$${plese}\:{give}\:{me}\:{answer} \\ $$

Commented by rhm last updated on 07/Oct/16

sir answer

$${sir}\:{answer} \\ $$

Commented by Rasheed Soomro last updated on 07/Oct/16

sin 36°=((√(2(√5) −10))/4)

$${sin}\:\mathrm{36}°=\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{5}}\:−\mathrm{10}}}{\mathrm{4}} \\ $$

Answered by ridwan balatif last updated on 15/Oct/16

let: sin18^o =sinθ sin36^o =cos54^o sin2θ =cos3θ 2sinθcosθ=cos2θcosθ−sinθsin2θ 2sinθcosθ=cos2θcosθ−sinθ.(2sinθcosθ) 2sinθ=cos2θ−2sin^2 θ 2sinθ=1−2sin^2 θ−2sin^2 θ 4sin^2 θ+2sinθ−1=0 sinθ_(1,2) =((−b±(√D))/(2a)) sinθ_(1,2) =((−2±(√(20)))/(2.4)) sinθ_(1,2) =((−1±(√5))/4) we take positive, because sin18^o has positive value sinθ=((−1+(√5))/4) so, sin18^o =((−1+(√5))/4) and cos18^o =((√(10+2(√5)))/4) sin36^o =2sin18^o cos18^o sin36^o =(((−1+(√5))/8))((√(10+2(√5))))

$$\mathrm{let}:\:\:\mathrm{sin18}^{\mathrm{o}} =\mathrm{sin}\theta \\ $$$$\mathrm{sin36}^{\mathrm{o}} =\mathrm{cos54}^{\mathrm{o}} \\ $$$$\mathrm{sin2}\theta\:\:=\mathrm{cos3}\theta \\ $$$$\mathrm{2sin}\theta\mathrm{cos}\theta=\mathrm{cos2}\theta\mathrm{cos}\theta−\mathrm{sin}\theta\mathrm{sin2}\theta \\ $$$$\mathrm{2sin}\theta\mathrm{cos}\theta=\mathrm{cos2}\theta\mathrm{cos}\theta−\mathrm{sin}\theta.\left(\mathrm{2sin}\theta\mathrm{cos}\theta\right) \\ $$$$\mathrm{2sin}\theta=\mathrm{cos2}\theta−\mathrm{2sin}^{\mathrm{2}} \theta \\ $$$$\mathrm{2sin}\theta=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta−\mathrm{2sin}^{\mathrm{2}} \theta \\ $$$$\mathrm{4sin}^{\mathrm{2}} \theta+\mathrm{2sin}\theta−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}\theta_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{b}\pm\sqrt{\mathrm{D}}}{\mathrm{2a}} \\ $$$$\mathrm{sin}\theta_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{2}\pm\sqrt{\mathrm{20}}}{\mathrm{2}.\mathrm{4}} \\ $$$$\mathrm{sin}\theta_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{we}\:\mathrm{take}\:\mathrm{positive},\:\mathrm{because}\:\mathrm{sin18}^{\mathrm{o}} \:\mathrm{has}\:\mathrm{positive}\:\mathrm{value} \\ $$$$\mathrm{sin}\theta=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{so},\:\mathrm{sin18}^{\mathrm{o}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:\:\:\mathrm{and}\:\mathrm{cos18}^{\mathrm{o}} =\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$$\mathrm{sin36}^{\mathrm{o}} =\mathrm{2sin18}^{\mathrm{o}} \mathrm{cos18}^{\mathrm{o}} \\ $$$$\mathrm{sin36}^{\mathrm{o}} =\left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{8}}\right)\left(\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}\right) \\ $$

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