SigMathS
Réponse 86:
Théorème des sinus
$\dfrac{a}{\sin \widehat{A}}=\dfrac{b}{\sin \widehat{B}}=\dfrac{c}{\sin \widehat{C}}=2R$
En utilisant le théorème des sinus on trouve
$a^2+b^2+c^2=4R^2(\sin^2 \widehat{A}+\sin^2\widehat{B}+\sin^2\widehat{C})$
$$\begin{align*}
&\sin^2 \widehat{A}+\sin^2\widehat{B}+\sin^2\widehat{C}\\
=&\sin^2 \dfrac{\pi}{7}+\sin^2\dfrac{2\pi}{7}+\sin^2\dfrac{4\pi}{7}\\
=&\dfrac{1-\cos \frac{2\pi}{7}}{2}+\dfrac{1-\cos \frac{4\pi}{7}}{2}+\dfrac{1-\cos \frac{8\pi}{7}}{2}\\
=&\dfrac{3-(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7})}{2}\\
=&\dfrac{3-(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7})}{2}\\
\end{align*}$$
les racines 7ème de l'unité sont de la forme $z_k=e^{i\frac{2k\pi}{7}}$ où $k\in\left\{{0,1,2,3,4,5,6}\right\}$ donc $\sum\limits_{k=0}^{6}{z_k}=0$ ce qui donne
$1+\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}\\
\qquad +\cos \frac{8\pi}{7}+\cos \frac{10\pi}{7}+\cos \frac{12\pi}{7}=0\\
\iff 1+\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}\\
\qquad +\cos \frac{6\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{2\pi}{7}=0\\
\iff 2(\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7})=-1\\
\iff \cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}=-\dfrac{1}{2}$
$a^2+b^2+c^2=4R^2\dfrac{3-\left({-\dfrac{1}{2}}\right)}{2}$ $\iff \boxed{a^2+b^2+c^2=7R^2}$