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\begin{document}
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\noindent
\begin{center}
    \begin{tabular}{@{}p{0.22\textwidth}p{0.57\textwidth}p{0.17\textwidth}}
        %\toprule
            \multirow{2}{*}{\parbox{\linewidth}{Prof MOSAID \newline \mylink }}
            & \Centering {Série Trigonométrie} & \hfill  TCS \\
        \bottomrule
    \end{tabular}
\end{center}
\textbf{\underline{Exercice 1:}}\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.98\textwidth}}
    & 1. Déterminer les abscisses curvilines principaux de
    \(A(\frac{789\pi}{7})\) et \(B(\frac{-214\pi}{5})\)\\
    & 2. Représenter sur le cercle trigonométrique les points d'abscisses curvilines:
    \(\frac{-\pi}{6}\), \(\frac{2\pi}{3}\), \(\frac{23\pi}{2}\), \(\frac{-59\pi}{4}\)\\
    & 3. Représenter sur le cercle trigonométrique les points \(M_k\) d'abscisses curvilines
    \(\frac{\pi}{3}+k\frac{\pi}{4}, \quad k\in \mathbb{Z}\)\\
    & 4. Déterminer les abscisses curvilines \(x\) du point \(M\) dans les cas:\\
    & \hspace*{1cm}{\large\textbullet}
    \(M(\frac{\pi}{4})\), et \(x\in [ \frac{34\pi}{6}, \frac{43\pi}{3}] \)
    {\large\textbullet} \(M(\frac{-2\pi}{5})\), et \(x\in [ \frac{-33\pi}{5}, \frac{-13\pi}{5}] \) \\
\end{tabular}
\\
\textbf{\underline{Exercice 2:}}\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.98\textwidth}}
    & 1. Construir un triangle isocèle en \(A\) tel que
    \((\overline{\overrightarrow{AB},\overrightarrow{AC}})\equiv \frac{2\pi}{5}[2\pi]\)\\
    & 2. Déterminer en radian les mesures :
    \((\overline{\overrightarrow{BA},\overrightarrow{BC}})\),
    \((\overline{\overrightarrow{BA},\overrightarrow{AC}})\) et
    \((\overline{\overrightarrow{CB},\overrightarrow{AC}})\)\\
\end{tabular}
\\
\\
\textbf{\underline{Exercice 3:}}\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.98\textwidth}}
    & Soit \(A(\frac{-\pi}{3})\) un point du cercle trigonométrique,
    donner la mesure principale de l'angle
    \((\widehat{\overrightarrow{OA},\overrightarrow{OM}})\)
    dans les cas:
    \hspace*{1cm}{\large\textbullet} \(M(\frac{27\pi}{2})\)\hspace*{1cm}
    {\large\textbullet}
    \((\overline{\overrightarrow{OJ},\overrightarrow{OM}})\equiv \frac{23\pi}{8}[2\pi]\)\\
\end{tabular}
\\
\\
\textbf{\underline{Exercice 4:}}\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.98\textwidth}}
    & Déterminer les lines trigonométrique suivantes:
      \(\cos \frac{7\pi}{6}\), \(\tan -\frac{73\pi}{3}\),
    \(\sin \frac{15\pi}{4}\), \(\sin \frac{-23\pi}{3}\)\\
\end{tabular}
\textbf{\underline{Exercice 5:}}\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.50\textwidth}p{0.50\textwidth}}
    & Soit \(x\in ]\frac{\pi}{2},\pi[\) on pose \(A=\frac{\tan x -1}{\tan^2 x+1}\)
    & 2. Calculer \(A\) sachant que \(\sin x=\frac{4}{5}\)\\
    & 1. Montrer que \(A=\cos x\sin x -\cos^2x\)
    & 3. Calculer \(x\) sachant que \(A=0\)\\
\end{tabular}
\\
\textbf{\underline{Exercice 6:}}\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.98\textwidth}}
    & Sachant que \(\sin \frac{7\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}\), Calculer:
      \(\cos \frac{7\pi}{8}\), \(\tan \frac{7\pi}{8}\),
    \(\sin \frac{\pi}{8}\), \(\sin \frac{3\pi}{8}\)
    \(\sin \frac{-25\pi}{8}\), \(\tan \frac{-78\pi}{8}\), \(\cos \frac{327\pi}{8}\)\\
\end{tabular}
\\
\\
\textbf{\underline{Exercice 7:}} Simplifier:\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.50\textwidth}p{0.43\textwidth}}
    &\begin{itemize}[topsep=6pt, partopsep=0pt, parsep=0pt, itemsep=0pt, after=\vspace*{-1.5\baselineskip}, before=\vspace*{-\baselineskip}, leftmargin=*, labelsep=0pt]
        \item \(A = \cos^6 x+\sin^6 x+3\cos^2 x\cdot \sin^2 x\)
        \item \(B = (1+\sin x+\cos x)^2-2(1+\sin x)(1+\cos x)\)
        \item \(C=\sin(15\pi-x)\cdot\cos(\frac{5\pi}{2}-x)-\sin(\frac{5\pi}{2}-x)\cdot\cos(3\pi-x)\)
        \item \(D=\sin x \cdot \cos(\frac{21\pi}{2}-x)-\cos(17\pi-x)\cdot\sin(\frac{\pi}{2}-x)\)
    \end{itemize}
    &\begin{itemize}[topsep=3pt, partopsep=0pt, parsep=0pt, itemsep=0pt, after=\vspace*{-1.5\baselineskip}, before=\vspace*{-\baselineskip}, leftmargin=*, labelsep=0pt]
        \item \(E=2(\cos^6 x+\sin^6 x)-3(\cos^4 x+\sin^4 x)\)
        \item \(F=\cos^2 \frac{\pi}{8}+\cos^2 \frac{3\pi}{8}+
            \cos^2 \frac{5\pi}{8}+\cos^2 \frac{7\pi}{8}\)
        \item \(K=1+\sin \frac{\pi}{7}+\sin \frac{2\pi}{7}+\sin \frac{3\pi}{7}
        +\cdots+\sin \frac{13\pi}{7}\)
        \item \mylink \hspace*{0.5cm}\mylink
    \end{itemize}\\
\end{tabular}
\\
\\
\textbf{\underline{Exercice 8:}}\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.98\textwidth}}
    & Soit \(x\in\mathbb{R},\quad A=\cos^4 x+\sin^4 x -(\sin x\cos x)(\cos x-\sin x)^2\)\\
    & 1. Montrer que \(A = 1 -\sin x \cdot\cos x\) \hspace*{0.5cm}
    2. Sachant que \(\sin \frac{11\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4}\),
    calculer \(A\) pour \(x=\frac{11\pi}{12}\)\\
\end{tabular}
\\
\\
\textbf{\underline{Exercice 9:}} Résoudre les équations et inéquations
    et représenter sur le cercle trigonométrique\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.50\textwidth}p{0.43\textwidth}}
    &\begin{itemize}[topsep=6pt, partopsep=0pt, parsep=0pt, itemsep=0pt, after=\vspace*{-1.5\baselineskip}, before=\vspace*{-\baselineskip}, leftmargin=*, labelsep=0pt]
        \item \(x\in]-\pi,\pi[\quad \cos x = -\frac{1}{2}\), \(\quad \cos x < -\frac{1}{2}\)
        \item \(x\in]-\frac{\pi}{3},\frac{2\pi}{3}[\quad \cos(2x- \frac{\pi}{3}) < \frac{-1}{2}\),
            poser \((X=2x-\frac{\pi}{3})\)
        \item \(x\in]-2\pi,\pi[\quad \sin(x) \le \frac{-\sqrt{3}}{2}\)
        \item \(x\in[-\pi,2\pi]\quad \sin(x- \frac{\pi}{3}) = \frac{-\sqrt{3}}{2}\)
        \item \(x\in\mathbb{R}, \tan x = \sqrt{3}\) {\large\textbullet}
            \(x\in[0,\pi], \cos(2x- \frac{\pi}{4}) \ge -\frac{\sqrt{2}}{2}\)
    \end{itemize}
    &\begin{itemize}[topsep=3pt, partopsep=0pt, parsep=0pt, itemsep=0pt, after=\vspace*{-1.5\baselineskip}, before=\vspace*{-\baselineskip}, leftmargin=*, labelsep=0pt]
        \item \(x\in \mathbb{R}\quad \tan(2x- \frac{\pi}{6}) = \frac{-1}{2}\)
        \item \(x\in]-\pi,\pi[\quad \cos(2x- \frac{\pi}{4}) = \cos(x+ \frac{\pi}{3})\)
        \item \(x\in \mathbb{R}\quad \tan^2x+(\sqrt{3}-1)\tan x-\sqrt{3}=0\)
        \item \(x\in \mathbb{R}\quad 2\cos^2x-7\cos x+3=0\)
        \item \mylink \hspace*{0.5cm}\mylink
        \item \mylink \hspace*{0.5cm}\mylink
    \end{itemize}\\
\end{tabular}
\\
\\
\textbf{\underline{Exercice 10:}}\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.98\textwidth}}
    & 1. Etudier les signes de \(\tan x -\sqrt{3}\) et \(\tan x -1\) sur
        \(]\frac{-\pi}{2},\frac{\pi}{2}[\)\\
    & 2. En déduir les solution de l'inéquation: \hspace*{0.5cm}
    \(x\in]\frac{-\pi}{2},\frac{\pi}{2}[\quad \tan x +\frac{\sqrt{3}}{\tan x}<1+\sqrt{3}\)\\
\end{tabular}
\\
\\
\textbf{\underline{Exercice 11:}}\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.98\textwidth}}
    & Soit \(p(x)=4\sin^x-2(\sqrt{2}-\sqrt{3})\sin x -\sqrt{6}\)\\
    & 1.1 résoudre dans \(\mathbb{R}\) l'équation \(4t^2-2(\sqrt{2}-\sqrt{3})t-\sqrt{6}=0\)\\
    & 1.2 En déduir une factorisation du trinôme:  \(4t^2-2(\sqrt{2}-\sqrt{3})t-\sqrt{6}\)\\
    & 1.3 résoudre: \(x\in]0,2\pi[\quad p(x)=0\)\\
    & 2.1 Factoriser \(p(x)\)\\
    & 2.2 résoudre: \(x\in]0,2\pi[\quad p(x)<0\)\\
\end{tabular}
\vspace*{-0.2cm}
\begin{quote}
\textit{``Why was the math book sad? Because it had too many problems. But then it found its sine of relief after solving them all!''} \\
\end{quote}
\vspace*{-1.2cm}
\textcolor{white}{.}\hfill \underline{MOSAID le \today}\\
\textcolor{white}{.}\hfill \mylink
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