Chapitre VII Trigonométrie

\(sin(x)=\frac{\sqrt{2}-\sqrt{6}}{4}\)

\(cos^2(x)+sin^2(x)=1\)

\(\iff cos^2(x)+(\frac{\sqrt{2}-\sqrt{6}}{4})^2=1\)

\(\iff cos^2(x)+\frac{\sqrt{2}^2-2\sqrt{2} \times \sqrt{6}+\sqrt{6}^2}{16}=1\)

\(\iff cos^2(x)+\frac{2-2\sqrt{12}+6}{16}=1\)

\(\iff cos^2(x)+\frac{8-2\sqrt{4 \times 3}}{16}=1\)

\(\iff cos^2(x)+\frac{8-2 \times 2\sqrt{3}}{16}=1\)

\(\iff cos^2(x)+\frac{8-4\sqrt{3}}{16}=1\)

\(\iff cos^2(x)+\frac{2-\sqrt{3}}{4}=1\)

\(\iff cos^2(x)=1-\frac{2-\sqrt{3}}{4}\)

\(\iff cos^2(x)=\frac{4}{4}-\frac{2-\sqrt{3}}{8}\)

\(\iff cos^2(x)=\frac{4-(2-\sqrt{3})}{8}\)

\(\iff cos^2(x)=\frac{2+\sqrt{3}}{8}\)

\(\iff cos^2(x)=\frac{2+\sqrt{3}}{4}\)

\(\iff cos(x)=\sqrt{\frac{2+\sqrt{3}}{4}}\)

\(\iff cos(x)=\frac{\sqrt{2+\sqrt{3}}}{2}\) ou \(cos(x)=-\frac{\sqrt{2+\sqrt{3}}}{2}\)

or \(x\in [-\frac{\pi}{2} ;\frac{\pi}{2}]\) donc \(cos(x) \ge 0\)

Finalement :

\(cos(x)=\frac{\sqrt{2+\sqrt{3}}}{2}\)

Les angles \(\frac{\pi}{12}\)  et \(\frac{5\pi}{12}\) sont impossibles car \( sin(x)\) est négatif.

Or ces deux angles correspondent à des sinus positifs.

\(0 <-\frac{\pi}{12} < \frac{-\pi}{4}=\frac{-3\pi}{12}<\frac{-5\pi}{12}<\frac{-\pi}{2}=\frac{-6\pi}{12}\)

et donc

\(|sin(\frac{-5\pi}{12})|>cos(\frac{-5\pi}{12})\)

\(|sin(-\frac{\pi}{12})|<cos(\frac{-\pi}{12})\)

or

\(cos(x)=\frac{\sqrt{2+\sqrt{3}}}{2}\simeq 0,97\)

\(sin(x)=\frac{\sqrt{2}-\sqrt{6}}{4}\simeq -0,26\)

donc \(|sin(x)|<cos(x)\) et donc \(x=-\frac{\pi}{12}\)