Devoir Libre 1 S02 Calcul Trigonométrique
📅 March 29, 2024 | 👁️ Views: 510

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\noindent\begin{tabularx}{\textwidth}{@{} lCr @{}}
Lycee Taghzirt\textbf{/}Prof MOSAID &
2023-2024\textbf{/Devoir Libre: Trigonométrie}&
TCSF\textbf{/}...h\\
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\textbf{\underline{Exercice 1:}}\\
\noindent
\begin{tabular}{@{}p{0.01\textwidth}|p{0.98\textwidth}}
& \mylabel[green]{1} Simplifier:\\
& \hspace*{0.5cm} $A(x)=\cos^6x+\sin^6x+3\cos^2x\cdot\sin^2x\quad$ et $\quad B(x)=2\left(\cos^6x+\sin^6x\right)-3\left(\cos^4x+\sin^4x\right)$ \\
&\mylabel[green]{2} Calculer $\tan \frac{-78\pi}{8}\quad$ et $\quad \cos \frac{327\pi}{8}$\\
& \mylabel[green]{3} Placer le point $A\left(\frac{-78\pi}{8}\right)$ sur le cercle trigonométrique\\
& \mylabel[green]{4} Construir un triangle $ABC$ réctangle et isocèle en $B$ tel que $\left(\widehat{\overrightarrow{AC},\overrightarrow{AB}}\right)$ est négative
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\textbf{\underline{Exercice 2:}}\\
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& \mylabel[green]{1} Soit $x \in \mathbb{R}\quad$. Simplifier: $A(x)=\sin(15\pi-x)\cdot\cos\left(\frac{5\pi}{2}-x\right)-\sin\left(\frac{5\pi}{2}-x\right)\cdot\cos(3\pi-x)$\\
& \mylabel[green]{1} Calculer:\\
& \hspace*{0.5cm} $B=\tan \frac{\pi}{5}+\tan \frac{2\pi}{5}+\tan \frac{3\pi}{5}+\tan \frac{4\pi}{5}\quad$ ; $\quad C=\cos^2 \frac{\pi}{8}+\cos^2 \frac{3\pi}{8}+\cos^2 \frac{5\pi}{8}+\cos^2 \frac{7\pi}{8}$\\
& \hspace*{0.5cm} $D=1+\sin \frac{\pi}{7}+\sin \frac{2\pi}{7}+\sin \frac{3\pi}{7}+\cdots+\sin \frac{13\pi}{7}$
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\textbf{\underline{Exercice 3:}}\\
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& Soit $x \in \mathbb{R}$. Soit $A=\cos^4x+\sin^4x-(\sin x\cdot\cos x)(\cos x-\sin x)^2$ \\
& \mylabel[green]{1} Montrer que $A=1-\sin x\cdot\cos x$\\
& \mylabel[green]{2} Sachant que $\sin \frac{11\pi}{12}=\frac{\sqrt6-\sqrt2}{2}\quad$ Calculer $A$ pour $x= \frac{11\pi}{12}$\\
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\textbf{\underline{Exercice 4:}}\\
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& \mylabel[green]{1} Résoudre dans $\mathbb{R}$ les équations suivantes:\\
& \hspace*{1cm} $\quad \sqrt2-2\sin\left(x+ \frac{\pi}{4}\right)=0\quad$; $\quad 2\cos\left(2x- \frac{\pi}{3}\right)=0\quad$; $\quad \sqrt3\tan(2x)-1=0$\\
& \mylabel[green]{2} Résoudre:\\
& \hspace*{0.5cm}$x\in ]-2\pi,3\pi]\quad \sqrt3+2\sin x =0\quad$ ; $\quad x\in ]-\pi,\pi]\quad 2\sin^2(7\pi+x)-3\sqrt3\cos\left(\frac{9\pi}{2}-x\right)+3=0$\\
& \hspace*{1cm} $x\in \mathbb{R}: \quad 2\cos^2x+\cos x-1=0\quad$ ; $\quad x \in [-\pi,\pi]\quad 2\cos^2x+\cos x-1\le0$
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