Exercice : Exercice 14

Dans chacun des cas suivants, déterminer \(cos(x)\)

Question

1.\(x \in[\frac{\pi}{2} ;\pi]\) et \(sin(x)=\frac{1}{4}\)

Solution

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

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

\(\iff cos^2(x)+\frac{1}{16}=1\)

\(\iff cos^2(x)=1-\frac{1}{16}\)

\(\iff cos^2(x)=\frac{16}{16}-\frac{1}{16}\)

\(\iff cos^2(x)=\frac{15}{16}\)

\(\iff cos(x)=\sqrt{\frac{15}{16}}\) ou \(cos(x)=-\sqrt{\frac{15}{16}}\)

\(\iff cos(x)=\frac{\sqrt{15}}{4}\) ou \(cos(x)=-\frac{\sqrt{15}}{4}\)

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

donc \(cos(x)=-\frac{\sqrt{15}}{4}\)

Question

2.\(x \in[-\frac{\pi}{3} ;\frac{\pi}{3}]\) et \(sin(x)=-0,6\)

Solution

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

\(\iff cos^2(x)+(-0,6)^2=1\)

\(\iff cos^2(x)+0,36=1\)

\(\iff cos^2(x)=1-0,36\)

\(\iff cos^2(x)=0,64\)

\(\iff cos(x)=0,8\) ou \(cos(x)=-0,8\)

or \(x \in[-\frac{\pi}{3} ;\frac{\pi}{3}]\) donc \(\frac{1}{2}\le cos x\le 1\)

donc \(cos(x)=0,8\)

Question

2.\(x \in[-\frac{\pi}{2} ;0]\) et \(sin(x)=-\frac{2}{3}\)

Solution

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

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

\(\iff cos^2(x)+\frac{4}{9}=1\)

\(\iff cos^2(x)=1-\frac{4}{9}\)

\(\iff cos^2(x)=\frac{5}{9}\)

\(\iff cos(x)=\frac{\sqrt{5}}{3}\) ou \(cos(x)=-\frac{\sqrt{5}}{3}\)

or \(x \in[-\frac{\pi}{2} ;0]\) donc \(0\le cos x\le 1\)

donc \(cos(x)=\frac{\sqrt{5}}{3}\)