Чему равен синус 36 градусов

Решите задачу:

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

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

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