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Résolvez l'équation \( \cos x = -{1\over 2} \).
\( S=\left\{\dfrac{2\pi}{3}\right\} \)
\( S=\left\{\dfrac{2\pi}{3},\, \dfrac{4\pi}{3}\right\} \)
\( S=\left\{\dfrac{2\pi}{3}+2k\pi,\, \dfrac{4\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{3}+2k\pi,\, -\dfrac{\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
Sans calculatrice, calculez \(tg\, \theta\) si \(\theta=\dfrac{5\pi}{6}\) .
\( -\dfrac{\sqrt{3}}{3} \)
\( -\sqrt{3} \)
\( \dfrac{\sqrt{3}}{3} \)
\( 150 \)
Si \(tg\,\theta=\dfrac{5}{12}\) alors \(cotg\, \theta=\)
\(\dfrac{12}{5} \)
\( \dfrac{5}{12} \)
\( \dfrac{7}{12} \)
n'existe pas
Convertissez en degrés l'angle \(-\pi \over 3\) .
\( \dfrac{1}{6} \mbox{ degrés}\)
\( 3 \mbox{ degrés}\)
\(60 \mbox{ degrés}\)
\( 300 \mbox{ degrés}\)
Convertissez en radians l'angle \(390^\circ \).
\(30\mbox{ radians}\)
\(\dfrac{\pi}{3} \mbox{ radians}\)
\( \dfrac{\pi}{6}\mbox{ radians}\)
\( 2\pi \mbox{ radians}\)
Sachant que \(ABCD\) est un carré inscrit dans un cercle de centre \(O \), comparez les angles \(\widehat{COD}\) et \(\widehat{CAD} \).
\( 2\widehat{COD}=\widehat{CAD} \)
\( \widehat{COD}=2\widehat{CAD} \)
\( \widehat{COD}=\widehat{CAD} \)
\( \widehat{COD}=\dfrac{1}{2}\widehat{CAD} \)
Convertissez en degrés l'angle \(5\pi \over 2 \).
\(90\mbox{ degrés}\)
\(\dfrac{5}{2} \mbox{ degrés}\)
\( \dfrac{1}{4}\mbox{ degrés}\)
\( \dfrac{\pi}{2}\mbox{ degrés}\)
Sans calculatrice, calculez \(\sin\theta\) si \(\theta=315^{\circ}\).
\( -\dfrac{1}{2} \)
\( -\dfrac{\sqrt{2}}{2} \)
\( \dfrac{\sqrt{2}}{2} \)
\( \dfrac{7\pi}{4} \)
\(\sin (3\pi +a)= \)
\( \sin a \)
\( -\sin a \)
\( \cos a \)
\( \pi+\sin a \)
\(\sin ({\pi \over 2}-a)= \)
\(\cos a \)
\( -\cos a \)
