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Question Number 163171 by ZiYangLee last updated on 04/Jan/22

Prove that sin 36° = ((√(10−2(√5^ )))/4)

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

Answered by mr W last updated on 04/Jan/22

sin (3×18^° )=sin (54°)=cos (36°)=cos (2×18°) 3 sin 18°−4 sin^3 18°=1−2 sin^2 18° let t=sin 18° 3t−4t^3 =1−2t^2 4t^3 −2t^2 −3t+1=0 4t^3 −4t^2 +2t^2 −2t−t+1=0 (t−1)(4t^2 +2t−1)=0 t≠1 ⇒4t^2 +2t−1=0 ⇒t=((−1+(√5))/4)=sin 18° cos 18°=(√(1−((((√5)−1)/4))^2 ))=((√(10+2(√5)))/4) sin 36°=2 sin 18° cos 18° =2×((−1+(√5))/4)×((√(10+2(√5)))/4) =((√((10+2(√5))(6−2(√5))))/8) =((√(10−2(√5)))/4)

$$\mathrm{sin}\:\left(\mathrm{3}×\mathrm{18}^{°} \right)=\mathrm{sin}\:\left(\mathrm{54}°\right)=\mathrm{cos}\:\left(\mathrm{36}°\right)=\mathrm{cos}\:\left(\mathrm{2}×\mathrm{18}°\right) \\ $$$$\mathrm{3}\:\mathrm{sin}\:\mathrm{18}°−\mathrm{4}\:\mathrm{sin}^{\mathrm{3}} \:\mathrm{18}°=\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{18}° \\ $$$${let}\:{t}=\mathrm{sin}\:\mathrm{18}° \\ $$$$\mathrm{3}{t}−\mathrm{4}{t}^{\mathrm{3}} =\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \\ $$$$\mathrm{4}{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{4}{t}^{\mathrm{3}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}−{t}+\mathrm{1}=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}\right)=\mathrm{0} \\ $$$${t}\neq\mathrm{1} \\ $$$$\Rightarrow\mathrm{4}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}=\mathrm{sin}\:\mathrm{18}° \\ $$$$\mathrm{cos}\:\mathrm{18}°=\sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\mathrm{36}°=\mathrm{2}\:\mathrm{sin}\:\mathrm{18}°\:\mathrm{cos}\:\mathrm{18}° \\ $$$$\:\:\:\:\:\:=\mathrm{2}×\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}×\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:=\frac{\sqrt{\left(\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}\right)\left(\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}\right)}}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:=\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$

Commented by peter frank last updated on 05/Jan/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by henderson last updated on 07/Jan/22

please, sir W ! how you get the second line ... ?

$$\mathrm{please},\:\mathrm{sir}\:\:\mathrm{W}\:! \\ $$$$\mathrm{how}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{second}\:\mathrm{line}\:...\:?\: \\ $$

Commented by mr W last updated on 07/Jan/22

the formulas for sin 3α and cos 3α are very useful. you should memorise them. but if you can′t, you can also get them easily like in following way: sin 3α =sin (2α+α) =sin 2α cos α+cos 2α sin α =2sin α cos^2 α+(1−2 sin^2 α) sin α =2sin α (1−sin^2 α)+(1−2 sin^2 α) sin α =2sin α −2 sin^3 α+ sin α−2 sin^3 α =3sin α −4 sin^3 α =(3−4 sin^2 α)sin α

$${the}\:{formulas}\:{for}\:\mathrm{sin}\:\mathrm{3}\alpha\:{and}\:\mathrm{cos}\:\mathrm{3}\alpha \\ $$$${are}\:{very}\:{useful}.\:{you}\:{should}\:{memorise} \\ $$$${them}.\:{but}\:{if}\:{you}\:{can}'{t},\:{you}\:{can}\:{also} \\ $$$${get}\:{them}\:{easily}\:{like}\:{in}\:{following}\:{way}: \\ $$$$\mathrm{sin}\:\mathrm{3}\alpha \\ $$$$=\mathrm{sin}\:\left(\mathrm{2}\alpha+\alpha\right) \\ $$$$=\mathrm{sin}\:\mathrm{2}\alpha\:\mathrm{cos}\:\alpha+\mathrm{cos}\:\mathrm{2}\alpha\:\mathrm{sin}\:\alpha \\ $$$$=\mathrm{2sin}\:\alpha\:\mathrm{cos}^{\mathrm{2}} \:\alpha+\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\right)\:\mathrm{sin}\:\alpha \\ $$$$=\mathrm{2sin}\:\alpha\:\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\alpha\right)+\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\right)\:\mathrm{sin}\:\alpha \\ $$$$=\mathrm{2sin}\:\alpha\:−\mathrm{2}\:\mathrm{sin}^{\mathrm{3}} \:\alpha+\:\mathrm{sin}\:\alpha−\mathrm{2}\:\mathrm{sin}^{\mathrm{3}} \:\alpha \\ $$$$=\mathrm{3sin}\:\alpha\:−\mathrm{4}\:\mathrm{sin}^{\mathrm{3}} \:\alpha \\ $$$$=\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\right)\mathrm{sin}\:\alpha \\ $$

Commented by greg_ed last updated on 07/Jan/22

very good

$$\mathrm{very}\:\mathrm{good} \\ $$

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