Stim ca:[tex]sin(5a)=5sina-20sin^3a+16sin^5a[/tex] Pentru x=π/5 si notam x=sin(π/5) obtinem ecuatia [tex]0=5x-20x^3+16x^5|:x=>16x^4-20x^2+5=0\\
Notam \ x^2=t\\
16t^2-20t+5=0\\
t_{1/2}=\frac{1}{8}\cdot (5\pm\sqrt{5})[/tex] Vol alege valoare cu minus deoarece [tex]x=sin\frac{\pi}{5}<sin\frac{\pi}{4}[/tex] [tex]x=\sqrt{\frac{5-\sqrt{5}}{8}}[/tex] Dar stim ca [tex]sin^2 \frac{\pi}{5}+cos^2\frac{\pi}{5}=1\\
cos^2\frac{\pi}{5}=1-sin^2 \frac{\pi}{5}=1-\frac{5-\sqrt{5}}{8}[/tex] [tex]cos^2\frac{\pi}{5}=\frac{6+2\sqrt{5}}{16}=>cos\frac{\pi}{5}=\frac{1}{4}\sqrt{(1+\sqrt{5})^2}=\frac{1}{4}\sqrt{1+\sqrt{5}}[/tex]
