Question

чему равен синус 36 градусов

Answer 1

$alpha =18^circ ; ,; ; 5 alpha =5cdot 18^circ =90^circ \\5 alpha =2 alpha +3alpha =90^circ ; ; to ; ; ; 2a=90^circ -3 alpha ; ; Rightarrow \\sin2 alpha =sin(90^circ -3 alpha )=cos3 alpha \\sin2 alpha =cos3alpha ; ; ; Rightarrow \\2sin alpha , cosalpha =4cos^3 alpha -3cos alpha \\2sin alpha , cosalpha -4cos^3alpha +3cosalpha =0\\cosalpha cdot (2sin alpha -4cos^2alpha +3)=0; ; Rightarrow \\cosalpha =cos18^circ textgreater 0; ; Rightarrow ; ; 2sinalpha -4cos^2alpha +3=0$

$2sinalpha-4(1-sin^2alpha )+3=0\\4sin^2 alpha +2sinalpha -1=0\\D=4-4cdot 4cdot (-1)=20; ; ,; ; sinalpha = frac{-2pm sqrt{20}}{8} = frac{-2pm 2sqrt5}{8}= frac{-1pm sqrt5}{4}\\sin alpha =sin18^circ textgreater 0; ; Rightarrow ; ; sin18^circ = frac{-1+sqrt5}{4} textgreater 0\\cos36^circ =cos(2cdot 18^circ )=cos^218^circ -sin^218^circ =(1-sin^218^circ )-sin^218^circ =\\=1-2sin^218^circ =1-2cdot Big (frac{-1+sqrt5}{4}Big )^2=1-2cdot frac{6+2sqrt5}{16}=1-frac{12+4sqrt5}{16}=$

$= frac{16-12+4sqrt5}{16}=frac{4+4sqrt5}{16}= frac{1+sqrt5}{4} \\sin^236^circ = 1-cos^236^circ ; ; Rightarrow ; ; sin36^circ =pm sqrt{1-cos^236^circ }\\Tak; kak; sin36^circ textgreater 0; ,; to; ; ; sin36^circ =+sqrt{1-cos^236^circ }=\\=sqrt{1-(frac{1+sqrt5}{4})^2}=sqrt{1- frac{6+2sqrt5}{16} }=\\= sqrt{ frac{10-2sqrt5}{16} } =sqrt{frac{5-sqrt5}{8}}= frac{sqrt{5-sqrt5}}{2sqrt2}$

Answer 2

спасибооо