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Question Number 139724 by mathlove last updated on 30/Apr/21

with out cientopic calculeter  sin9=?

$${with}\:{out}\:{cientopic}\:{calculeter} \\ $$$${sin}\mathrm{9}=? \\ $$

Commented by Dwaipayan Shikari last updated on 30/Apr/21

sin(9°) ?

$${sin}\left(\mathrm{9}°\right)\:? \\ $$

Commented by malwan last updated on 30/Apr/21

sin9 = (1/8)(√2)(1+(√5))−      (1/8)((√5)−1)(√(5+(√5)))  = (1/2)(√(2−(√((1/2)(5+(√5))))))

$${sin}\mathrm{9}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)− \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{8}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}+\sqrt{\mathrm{5}}\right)}} \\ $$

Answered by Dwaipayan Shikari last updated on 30/Apr/21

sin((π/(10)))=(((√5)−1)/4)  2sin((π/(20)))(√(1−sin^2 ((π/(20))))) =(((√5)−1)/4)  ⇒sin^2 ((π/(20)))−sin^4 ((π/(20)))=((((√5)−1)/8))^2   ⇒sin^4 ((π/(20)))−sin^2 ((π/(20)))+((3−(√5))/(32))=0  ⇒sin^2 ((π/(20)))=((1±(√(1−((3−(√5))/8))))/2)=((1±(√((5+(√5))/8)))/2)  sin((π/(20)))=(√((1−(1/2)(√((√5)φ)))/2))=(1/2)(√(2−(5)^(1/4) (√φ)))   φ=Golden Ratio

$${sin}\left(\frac{\pi}{\mathrm{10}}\right)=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{20}}\right)\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{20}}\right)}\:=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{20}}\right)−{sin}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{20}}\right)=\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{8}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{sin}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{20}}\right)−{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{20}}\right)+\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{32}}=\mathrm{0} \\ $$$$\Rightarrow{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{20}}\right)=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}}}}{\mathrm{2}}=\frac{\mathrm{1}\pm\sqrt{\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{8}}}}{\mathrm{2}} \\ $$$${sin}\left(\frac{\pi}{\mathrm{20}}\right)=\sqrt{\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\sqrt{\mathrm{5}}\phi}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\left(\mathrm{5}\right)^{\mathrm{1}/\mathrm{4}} \sqrt{\phi}}\:\:\:\phi={Golden}\:{Ratio} \\ $$

Commented by mathlove last updated on 01/May/21

what is ∅?

$${what}\:{is}\:\emptyset? \\ $$

Commented by Dwaipayan Shikari last updated on 01/May/21

Golden Ratio φ=((1+(√5))/2)=1.618..

$${Golden}\:{Ratio}\:\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\mathrm{1}.\mathrm{618}.. \\ $$

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