Devoir 01 S02 (D) 2026, Trigonométrie
📅 March 31, 2026 | 👁️ Views: 1
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\begin{enumerate}
\item \note{1} Déterminer l’abscisse curviligne principale du point $A\left(\frac{2029\pi}{3}\right)$\\
\item \note{3} Placer les points $A$, $B\left(\frac{7\pi}{6}\right)$ et le point $C$ tel que
$\left(\overline{\overrightarrow{OB},\overrightarrow{OC}}\right)\equiv \frac{5\pi}{4}\,[2\pi]$
\item \note{1} Déterminer l’abscisse curviligne principale du point $C$
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\item \note{4} Simplifier les expressions suivantes.\\
\(A = 2\sin(x-5\pi) - 3\cos\left(4\pi+x\right) + \sin\left(\frac{7\pi}{2}+x\right),\)\\
\(B = \cos\left(\frac{\pi}{12}\right) + \cos\left(\frac{5\pi}{12}\right) + \cos\left(\frac{7\pi}{12}\right) + \cos\left(\frac{11\pi}{12}\right).\)
\item \note{1.5} Sachant que \(\sin \frac{7\pi}{8} = \frac{\sqrt{2-\sqrt2}}{2}\).
Calculer \(\cos \frac{7\pi}{8} \).
\end{enumerate}
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Soit l'intervalle \(I = \left]-\pi;-\frac{\pi}{2}\right[\cup\left]\frac{\pi}{2};\pi\right[\).
Pour tout ~$x\in I$~ On pose :
\[f(x) = 2\sin x - 2\cos x + \sqrt{3}\tan x - \sqrt3.\]
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\item
\begin{enumerate}[label=\alph*.]
\item \note{1} Vérifier que \(f(x) = (2\cos x + \sqrt3)(\tan x - 1)\).
\item \note{1.5} Calculer : \(f\left(\frac{\pi}{4}\right),\quad f(\pi)\quad\text{et}\quad f\left(\frac{5\pi}{6}\right)\).
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\item
\begin{enumerate}[label=\alph*.]
\item \note{3} Résoudre dans \(I\) les équations :
\(2\cos x + \sqrt3 = 0 \quad\text{et}\quad \tan x - 1 = 0.\)
\item \note{0.5} En déduire dans \(I\) les solutions de \(f(x)=0\).
\end{enumerate}
\item
\begin{enumerate}[label=\alph*.]
\item \note{3} Résoudre dans \(I\) les inéquations :
\(2\cos x + \sqrt3 > 0 \quad\text{et}\quad \tan x - 1 \le 0.\)
\item \note{1} Donner le tableau de signe de \(f(x)\) sur \(I\).
\item \note{0.5} En déduire dans \(I\) les solutions de l’inéquation \(f(x) < 0\).
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﴿فَلِلَّهِ ٱلْحَمْدُ رَبِّ ٱلسَّمَـٰوَٰتِ وَرَبِّ ٱلْأَرْضِ رَبِّ ٱلْعَـٰلَمِينَ (36)• وَلَهُ ٱلْكِبْرِيَآءُ فِى ٱلسَّمَـٰوَٰتِ وَٱلْأَرْضِ ۖ وَهُوَ ٱلْعَزِيزُ ٱلْحَكِيمُ (37)• ﴾ (الجاثية الآيات 36-37)
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