Exercice : Exercice 10
Déterminer la mesure principale des angles orientés de mesure suivante :
Question
a. \(\frac{9\pi}{4}\)
Solution
\(\frac{\frac{9\pi}{4}}{2\pi}=\frac{\frac{9\pi}{4}}{\frac{2\pi}{1}}\)
\(\iff \frac{\frac{9\pi}{4}}{2\pi}=\frac{9\pi}{4} \times \frac{1}{2\pi}\)
\(\if \frac{\frac{9\pi}{4}}{2\pi}=\frac{9}{8}=1,125\)
\(1<\frac{9}{8}<2\)
\(\frac{9\pi}{4}-1\times 2\pi=\frac{9\pi}{4}-\frac{8\pi}{4}=\frac{\pi}{4}\in ]-\pi ;\pi]\)
La mesure principale de \(\frac{9\pi}{4}\) est donc \(\frac{\pi}{4}\)
Question
b. \(\frac{192\pi}{6}\)
Solution
\(\frac{\frac{192\pi}{6}}{2\pi}=\frac{\frac{192\pi}{6}}{\frac{2\pi}{1}}\)
\(\iff \frac{\frac{192\pi}{6}}{2\pi}=\frac{192\pi}{6} \times \frac{1}{2\pi}\)
\(\if \frac{\frac{192\pi}{6}}{2\pi}=\frac{192}{12}=16\)
\(\frac{192\pi}{6}-16\times 2\pi=\frac{192\pi}{6}-32\pi=\frac{192\pi}{6}-\frac{192\pi}{6}=0\in ]-\pi ;\pi]\)
La mesure principale de \(\frac{192\pi}{6}\) est donc 0
Question
c. \(\frac{-5\pi}{4}\)
Solution
\(\frac{\frac{-5\pi}{4}}{2\pi}=\frac{\frac{-5\pi}{4}}{\frac{2\pi}{1}}\)
\(\iff \frac{\frac{-5\pi}{4}}{2\pi}=\frac{-5\pi}{4} \times \frac{1}{2\pi}\)
\(\if \frac{\frac{-5\pi}{4}}{2\pi}=\frac{-5}{8}=-0,625\)
\(-1<\frac{-5}{8}<0\)
\(\frac{-5\pi}{4}-0\times 2\pi=\frac{-5\pi}{4} \notin ]-\pi ;\pi]\)
\(\frac{-5\pi}{4}-(-1) \times 2\pi=\frac{-5\pi}{4}+2\pi=\frac{-5\pi}{4}+\frac{8\pi}{4}=\frac{3\pi}{4} \in ]-\pi ;\pi]\)
La mesure principale de \(\frac{-5\pi}{4}\) est donc \(\frac{3\pi}{4}\)
Question
d. \(\frac{-33\pi}{2}\)
Solution
\(\frac{\frac{-33\pi}{2}}{2\pi}=\frac{\frac{-33\pi}{2}}{\frac{2\pi}{1}}\)
\(\iff \frac{\frac{-33\pi}{2}}{2\pi}=\frac{-33\pi}{2} \times \frac{1}{2\pi}\)
\(\if \frac{\frac{-33\pi}{2}}{2\pi}=\frac{-33}{4}=-8,25\)
\(-9<\frac{-33}{4}<-8\)
\(\frac{-33\pi}{2}-(-8)\times 2\pi=\frac{-33\pi}{2}+16\pi=\frac{-33\pi}{2}+\frac{32\pi}{2}=\frac{-\pi}{2} \in ]-\pi ;\pi]\)
La mesure principale de \(\frac{-33\pi}{2}\) est donc \(\frac{-\pi}{2}\)
Question
e. \(\frac{16\pi}{7}\)
Solution
\(\frac{\frac{16\pi}{7}}{2\pi}=\frac{\frac{16\pi}{7}}{\frac{2\pi}{1}}\)
\(\iff \frac{\frac{16\pi}{7}}{2\pi}=\frac{16\pi}{7} \times \frac{1}{2\pi}\)
\(\if \frac{\frac{16\pi}{7}}{2\pi}=\frac{16}{14}=\frac{8}{7}\)
\(1<\frac{8}{7}<2\)
\(\frac{16\pi}{7}-1\times 2\pi=\frac{16\pi}{7}-\frac{14\pi}{7}=\frac{2\pi}{7} \in ]-\pi ;\pi]\)
La mesure principale de \(\frac{16\pi}{7}\) est donc \(\frac{2\pi}{7}\)
Question
f. \(\frac{52\pi}{3}\)
Solution
\(\frac{\frac{52\pi}{3}}{2\pi}=\frac{\frac{52\pi}{3}}{\frac{2\pi}{1}}\)
\(\iff \frac{\frac{52\pi}{3}}{2\pi}=\frac{52\pi}{3} \times \frac{1}{2\pi}\)
\(\if \frac{\frac{52\pi}{3}}{2\pi}=\frac{52}{6}=\frac{26}{3}\)
\(\frac{24}{3}<\frac{26}{3}<\frac{27}{3}\)
\(8<\frac{26}{3}<9\)
\(\frac{52\pi}{3}-8\times 2\pi=\frac{52\pi}{3}-16\pi=\frac{52\pi}{3}-\frac{48\pi}{3}=\frac{4\pi}{3} \notin ]-\pi ;\pi]\)
\(\frac{52\pi}{3}-9\times 2\pi=\frac{52\pi}{3}-18\pi=\frac{52\pi}{3}-\frac{54\pi}{3}=\frac{-2\pi}{3} \in ]-\pi ;\pi]\)
La mesure principale de \(\frac{52\pi}{3}\) est donc \(\frac{-2\pi}{3}\)