Exercice : Exercice basique 3
On considère un réel \(x\in [\frac{-\pi}{2} ;\frac{\pi}{2}]\) tels que \(sin(x)=\frac{\sqrt{2}-\sqrt{6}}{4}\)
Question
1.Déterminer la valeur exacte de \(cos(x)\)
Indice
Utilisez la relation \(cos^2x+sin^2x=1\)
Solution
\(sin(x)=\frac{\sqrt{2}-\sqrt{6}}{4}\)
\(cos^2 x+sin^2 x=cos^2 x +(\frac{\sqrt{2}-\sqrt{6}}{4})^2=1\)
\(\iff cos^2 x+sin^2 x=cos^2 x +\frac{(\sqrt{2}-\sqrt{6})^2}{16}=1\)
\(\iff cos^2 x +\frac{(\sqrt{2}-\sqrt{6})^2}{16}=1\)
\(\iff cos^2 x +\frac{\sqrt{2}^2-2\sqrt{2} \times \sqrt{6}+ \sqrt{6}^2}{16}=1\)
\(\iff cos^2 x +\frac{2-2\sqrt{12}+ 6}{16}=1\)
\(\iff cos^2 x +\frac{8-2\sqrt{12}}{16}=1\)
\(\iff cos^2 x =1-\frac{8-2\sqrt{12}}{16}\)
\(\iff cos^2 x =\frac{16}{16}-\frac{8-2\sqrt{12}}{16}\)
\(\iff cos^2 x =\frac{16-8+2\sqrt{12}}{16}\)
\(\iff cos^2 x =\frac{8+2\sqrt{12}}{16}\)
\(\iff cos x =\frac{\sqrt{8+2\sqrt{12}}}{4}\)
\(\iff cos x =\frac{\sqrt{\sqrt{2}^2+2\sqrt{2} \times \sqrt{6} +\sqrt{6}^2}}{4}\)
\(\iff cos x =\frac{\sqrt{(\sqrt{2}+\sqrt{6})^2}}{4}\)
\(\iff cos x =\frac{\sqrt{2}+\sqrt{6}}{4}\)
Question
2. On sait que \(x\in \{\frac{\pi}{12} ; \frac{5\pi}{12} ;\frac{-\pi}{12} ; \frac{-5\pi}{12}\}\)
Détrerminer la valeur exacte de \(x\)
Solution
\(sin(x)=\frac{\sqrt{2}-\sqrt{6}}{4}<0\)
donc \(x\notin \{\frac{\pi}{12} ; \frac{5\pi}{12}\}\)
\(cos x =\frac{\sqrt{2}+\sqrt{6}}{4}\)
\(|cos x[>|sin x|\) donc \(x=\frac{-\pi}{12}\)
