Auto-Math
Sachant que \(ABCDEF\) est un hexagone régulier inscrit dans un cercle de centre \(O\) , comparez les angles \(\widehat{AOB} \) et \(\widehat{ADB}\).
\( \widehat{AOB}=\widehat{ADB} \)
\( 2\widehat{AOB}=\widehat{ADB} \)
\(\widehat{AOB}=2\widehat{ADB} \)
\( \widehat{AOB}=\dfrac{1}{2}\widehat{ADB} \)
Donnez la valeur de \( \cos {\pi \over 6} \).
\( 30 \)
\( \dfrac{1}{2} \)
\(\dfrac{\sqrt{3}}{2} \)
\( -\dfrac{\sqrt{3}}{2} \)
Déterminez à l'aide du cercle trigonométrique la valeur de \(\sin\dfrac{11\pi}{6} \).
\( -\dfrac{1}{2} \)
\( \dfrac{\sqrt{3}}{2} \)
Convertissez en radians l'angle \(-75^\circ \).
\( -75\mbox{ radians}\)
\( \dfrac{\pi}{36} \mbox{ radians}\)
\( \dfrac{5\pi}{12}\mbox{ radians}\)
\( \dfrac{19\pi}{12}\mbox{ radians}\)
Résolvez l'équation \(tg\, 3x = \dfrac{\sqrt{3}}{3}\) .
\( S=\left\{\dfrac{\pi}{18}\right\} \)
\( S=\left\{\dfrac{\pi}{18}+k\dfrac{\pi}{3};\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{18}+2k\dfrac{\pi}{3};\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{18}+2k\pi;\, k\in\mathbb{Z}\right\} \)
Convertissez en radians l'angle \(30^\circ \).
\(30\mbox{ radians}\)
\( \dfrac{\pi}{6} \mbox{ radians}\)
\( \dfrac{\pi}{3} \mbox{ radians}\)
\(\dfrac{\pi}{30}\mbox{ radians}\)
Donnez la valeur de \(\cos {3\pi \over 4}\) .
\( -\dfrac{\sqrt{2}}{2} \)
\( \dfrac{\sqrt{2}}{2} \)
\( 135 \)
Résolvez l'équation \(\cos 2x = \dfrac{\sqrt{2}}{2}\) .
\( S=\left\{\dfrac{\pi}{8}+k\pi,\, -\dfrac{\pi}{8}+k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{8}+2k\pi,\, -\dfrac{\pi}{8}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{4}\right\} \)
\( S=\left\{\dfrac{\pi}{8}\right\} \)
Si \(\sin\theta=\dfrac{3}{5} \) alors \(tg\, \theta=\)
\( \dfrac{4}{3} \)
\( \dfrac{3}{4} \)
\(\dfrac{2}{5} \)
n'existe pas
Si \(\alpha=53^{\circ}\) , alors le supplémentaire de \(\alpha\) vaut
\(37^{\circ} \)
\( 127^{\circ} \)
\( 233^{\circ} \)
\( 413^{\circ} \)
