Bonjour :))
[tex]FORMULE\ DE\ MOIVRE\\(e^{i\theta})^{n}=e^{in\theta}\\\\w^{5}=(e^{i\frac{2\pi}{5}})^{5}=e^{i2\pi}=1[/tex]
[tex]w^{5}-1=(w-1)(aw^{4}+bw^{3}+cw^{2}+dw+e)\\w^{5}-1=aw^{5}+bw^{4}+cw^{3}+dw^{2}+ew-aw^{4}-bw^{3}-cw^{2}-dw-e\\w^{5}-1=aw^{5}+(b-a)w^{4}+(c-b)w^{3}+(d-c)w^{2}+(e-d)w-e\\\\Par\ identification,\ on\ a:\\a=1\\e=1\\\\b-a=0\ donc\ b=1\\c-b=0\ donc\ c=1\\d-c=0\ donc\ d=1\\e-d=0\ donc\ cela\ confirme\ e=1\\\\DONC\ \boxed{w^{5}-1=(w-1)(w^{4}+w^{3}+w^{2}+w+1)}[/tex]
[tex]w^{5}-1=0\ donc\ 1+w+w^{2}+w^{3}+w^{4}+0[/tex]
[tex]1+w+w^{2}+w^{3}+w^{4}=1+(e^{i\frac{2\pi}{5}})+(e^{i\frac{4\pi}{5}})+(e^{i\frac{6\pi}{5}})+(e^{i\frac{8\pi}{5}})\\\\\frac{6\pi}{5}=-\frac{4\pi}{5}\\\\\frac{8\pi}{5}=-\frac{2\pi}{5}\\\\Donc:\ 1+(e^{i\frac{2\pi}{5}})+(e^{-i\frac{2\pi}{5}})+(e^{i\frac{4\pi}{5}})+(e^{-i\frac{4\pi}{5}})\\\\FORMULE\ D'EULER\\e^{i\theta}+e^{-i\theta}=2cos(\theta)\\\\Donc: 1+2cos(\frac{2\pi}{5})+2cos(\frac{4\pi}{5})=0[/tex]
[tex]FORMULE\ COURS\ TRIGONOM\'ETRIE\\cos(2a)=2cos^{2}(a)-1\\\\cos(\frac{4\pi}{5})=cos(2*\frac{2\pi}{5})=2cos^{2}(\frac{2\pi}{5})-1[/tex]
[tex]1+w+w^{2}+w^{3}+w^{4}=1+2cos(\frac{2\pi}{5})+2cos(\frac{4\pi}{5})=0\\\\\Rightarrow 1+2cos(\frac{2\pi}{5})+2(2cos^{2}(\frac{2\pi}{5})-1)=0\\\\\Rightarrow 1+2cos(\frac{2\pi}{5})+4cos^{2}(\frac{2\pi}{5})-2=0\\\\\Rightarrow 4cos^{2}(\frac{2\pi}{5})+2cos(\frac{2\pi}{5})-1=0\\\\On\ pose\ x=cos(\frac{2\pi}{5})\\\\\Rightarrow 4x^{2}+2x-1=0[/tex]
[tex]\Delta=b^{2}-4ac\\\\\Delta=2^2-4*4*(-1)\\\\\Delta=4+16=20>0\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-2-2\sqrt{5}}{8}=\frac{-1-\sqrt{5}}{4}<0 \\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-2+2\sqrt{5}}{8}=\frac{-1+\sqrt{5}}{4}>0\\\\On\ sait\ d'apr\`es\ le\ cercle\ trigonom\'etrique\ que:\\\frac{2\pi}{5}<\frac{\pi}{2}\ \ donc\ cos(\frac{2\pi}{5})>0\\\\La\ valeur\ exacte\ de\ cos(\frac{2\pi}{5})\ est\ x_2=\frac{-1+\sqrt{5}}{4}[/tex]
N'hésite pas à revenir vers moi pour des questions. Bonne continuation :))
