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Donner un réel dans l'intervalle $]-\pi;\pi]$ qui mesure, en radians, l'angle $(\vec u,\vec v)$ si:
a. $(\vec u,\vec v) = \dfrac{5\pi}3$;
Corrigé
\[\frac{5\pi}3 - 2\pi = \frac{5\pi}3 - \frac{6\pi}3 = -\frac{\pi}3
\implies (\vec u,\vec v) = -\frac{\pi}3.\]
b. $(\vec u,\vec v) = \dfrac{7\pi}5$;
Corrigé
\[\frac{7\pi}5 - 2\pi =\frac{7\pi}5 - \frac{10\pi}5 = -\frac{3\pi}5 \implies (\vec u,\vec v) = -\frac{3\pi}5.\]
c. $(\vec u,\vec v) = \dfrac{17\pi}5$;
Corrigé
\[\frac{17\pi}5 - 2\times 2\pi = \frac{17\pi} 5 - \frac{20\pi} 5 = -\frac{3\pi}5
\implies
(\vec u,\vec v) = -\frac{3\pi}5.\]
d. $(\vec u,\vec v) = \dfrac{50\pi}7$;
Corrigé
\[\frac{50\pi}7 - 4\times 2\pi = \frac{50\pi}7 - \frac{56\pi}7 = -\frac{6\pi}7
\implies
(\vec u,\vec v) = -\frac{6\pi}7.\]
e. $(\vec u,\vec v) = -\dfrac{55\pi}{8}$;
Corrigé
\[-\frac{55\pi}8 + 3\times 2\pi = -\frac{55\pi}8 + \frac{48\pi}8 = -\frac{7\pi}8
\implies
(\vec u,\vec v) = -\frac{7\pi}8.\]
f. $(\vec u,\vec v) = \dfrac{38\pi} 6$.
Corrigé
\[\frac{38\pi}6 -3\times 2\pi = \frac{38\pi}6 - \frac{36\pi}6 = \frac{2\pi}6=\frac{\pi}3
\implies
(\vec u,\vec v) = \frac{\pi}3.\]
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