Răspunsuri

2014-07-12T15:58:35+03:00
Stim ca:sin(5a)=5sina-20sin^3a+16sin^5a
Pentru x=π/5 si notam x=sin(π/5) obtinem ecuatia
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})
Vol alege valoare cu minus deoarece
x=sin\frac{\pi}{5}<sin\frac{\pi}{4}
 x=\sqrt{\frac{5-\sqrt{5}}{8}}
Dar stim ca sin^2 \frac{\pi}{5}+cos^2\frac{\pi}{5}=1\\&#10;cos^2\frac{\pi}{5}=1-sin^2 \frac{\pi}{5}=1-\frac{5-\sqrt{5}}{8}
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}}