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Si \(\alpha=53^{\circ}\) , alors l'opposé de \(\alpha\) vaut
\( 35^{\circ} \)
\( 233^{\circ} \)
\( 413^{\circ} \)
\( -53^{\circ} \)
Sans calculatrice, calculez \(\cos\theta\) si \( \theta=\dfrac{5\pi}{6}\) .
\(150 \)
\( -\dfrac{1}{2} \)
\( \dfrac{\sqrt{3}}{2} \)
\( -\dfrac{\sqrt{3}}{2} \)
\(\sin (3\pi +a)= \)
\( \sin a \)
\( -\sin a \)
\( \cos a \)
\( \pi+\sin a \)
Donnez la valeur de \( \sin {\pi \over 2}\) .
1
-1
0
90
Résolvez l'équation \(\cos(3x+\pi) = \cos x \).
\( S=\left\{-\dfrac{\pi}{2}+k\pi,\, -\dfrac{\pi}{4}+k\dfrac{\pi}{2};\, k\in\mathbb{Z}\right\} \)
\( S=\left\{-\dfrac{\pi}{2},\, -\dfrac{\pi}{4};\, k\in\mathbb{Z}\right\} \)
\( S=\left\{-\dfrac{\pi}{2}+k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{-\dfrac{\pi}{2}+2k\pi,\, -\dfrac{\pi}{4}+2k\dfrac{\pi}{2};\, k\in\mathbb{Z}\right\} \)
Si \(\alpha=53^{\circ}\) , alors le supplémentaire de \(\alpha\) vaut
\(37^{\circ} \)
\( 127^{\circ} \)
Convertissez en degrés l'angle \(5\pi \over 2 \).
\(90\mbox{ degrés}\)
\(\dfrac{5}{2} \mbox{ degrés}\)
\( \dfrac{1}{4}\mbox{ degrés}\)
\( \dfrac{\pi}{2}\mbox{ degrés}\)
Résolvez l'équation \(\sin 5x +1=0 \).
\( S=\left\{-\dfrac{1}{5}\right\} \)
\( S=\left\{\dfrac{3\pi}{10},\, -\dfrac{\pi}{10}\right\} \)
\( S=\left\{\dfrac{3\pi}{10}+2k\pi,\, -\dfrac{\pi}{10}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{3\pi}{10}+2k\dfrac{\pi}{5};\, k\in\mathbb{Z}\right\} \)
Convertissez en degrés l'angle\( \pi \over 12 \).
\(\dfrac{\pi}{15} \mbox{ degrés}\)
\(15 \mbox{ degrés}\)
\( 12 \mbox{ degrés}\)
\( 7,5 \mbox{ degrés}\)
Sachant que \(ABCDEF\) est un hexagone régulier inscrit dans un cercle de centre \(O\) , comparez les angles \(\widehat{DBA}\) et \(\widehat{DOA}\).
\( \widehat{DBA}=\widehat{DOA} \)
\( 2\widehat{DBA}=\widehat{DOA} \)
\(\widehat{DBA}=2\widehat{DOA} \)
\( \dfrac{1}{2}\widehat{DBA}=\widehat{DOA} \)
