Control 01 Semester 02 en Trigonometrie (B)
📅 March 19, 2025 | 👁️ Views: 6

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%%%%%%%%%%%%%%%%%%%%%%%%%%%% Start of the exam %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\exo{Exercice 1: (\textit{14 pts}) }\\
\hspace*{0.3cm}Soit $(O; OI, OJ)$ un repère orthonormé lié à un cercle trigonométrique ~$(\mathscr C)$~
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Placer les deux points suivants sur le cercle trigonométrique: ~$A\left(\frac{-\pi}{3}\right)$~,
~$B\left(\frac{-137\pi}{6}\right)$~
\question[1]
Construire un triangle ~$ABC$~ réctangle en ~$A$~ tel que
~$\widehat{\left(\overrightarrow{AB},\overrightarrow{AC}\right)}\equiv \frac{\pi}{2} [2\pi]$
\question[3]
Calculer ~$\sin\left(\frac{-\pi}{3}\right)$~ , ~$\cos\left(\frac{5\pi}{3}\right)$~
et ~$\tan\left(\frac{78\pi}{3}\right)$~
\question[3]
Simpilfier ~~
$A(x)=\sin(x)\cdot\cos\left(\frac{21\pi}{2}-x\right)-\sin\left(\frac{\pi}{2}-x\right)\cdot\cos(17\pi-x)$
\question[3]
Sachant que $\sin \frac{7\pi}{8}=\frac{\sqrt{2-\sqrt2}}{2}\quad$ Calculer ~$\cos \frac{7\pi}{8}$~
et ~$\cos \frac{-7\pi}{8}$~ puis ~$\cos \frac{\pi}{8}$~
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Résoudre les équations suivantes:\\
$x\in \mathbb R \quad \cos\left(x\right)=\frac{\sqrt3}{2}\quad$; $\quad x \in \mathbb R \quad \cos\left(2x- \frac{\pi}{4}\right)=\frac{1}{2}\quad$ ;~~~
$x\in ]-2\pi,3\pi]\quad \sin 2x =-\frac{\sqrt3}{2}\quad$
\question[1]
\textbf{Bonus Question}: Résoudre
$\quad x\in ]-\pi,\pi]\quad 2\sin^2(7\pi+x)-3\sqrt3\cos\left(\frac{9\pi}{2}-x\right)+3=0$\\
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