Question

sa se calculeze numarul   cos  ( $\pi /5$ ) 

Answer 1

Stim ca:$sin(5a)=5sina-20sin^3a+16sin^5a$
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
$x=sin\frac{\pi}{5}<sin\frac{\pi}{4}$
 $x=\sqrt{\frac{5-\sqrt{5}}{8}}$
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]
$cos^2\frac{\pi}{5}=\frac{6+2\sqrt{5}}{16}=>cos\frac{\pi}{5}=\frac{1}{4}\sqrt{(1+\sqrt{5})^2}=\frac{1}{4}\sqrt{1+\sqrt{5}}$