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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Définitions de la feuille d'exercices

\newcommand{\ch}{Chapitre 4}
\newcommand{\Ch}{Vecteurs et droites}
\newcommand{\Cl}{1SSI}
\newcommand{\Annee}{2011-2012}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%		Packages
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	Début
\begin{document}
 
\debutSsCadreCha

\begin{multicols}{2}
 
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\Rqs{\item On a évidemment : $$(\vec u ; \vec v)=(\overrightarrow{OA} ; \overrightarrow{OB}) = 
(\overrightarrow{OA'} ; \overrightarrow{OB'})= (\overrightarrow{OA} ; \overrightarrow{OB'}) \dots$$
\item Les mesures de $(\overrightarrow{OA} ; \overrightarrow{OB}) $ sont en fait les mesures en radians de l'angle orient\'e $\widehat{AOB}$, 
mais cette nouvelle définition nous permet de savoir que l'unité est le radian, et que l'on tient compte de l'orientation 
du plan, sans le préciser à chaque fois. De plus, elle va s'avérer très pratique pour les manipulations algébriques.
}

\end{multicols}

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\begin{tabular}{*{7}{|>{\centering}m{1.5cm}} |}
\hline 
Angle $\theta$  & $0$ & $\dfrac{\pi}{6}$ & $\dfrac{\pi}{4}$ &  $\dfrac{\pi}{3}$ & $\dfrac{\pi}{2}$ & $\pi$  \tabularnewline\hline
$\cos \theta$ & $\dots$ & $\dots $ & $\dots$ & $\dots$ & $\dots$  & $\dots$ \tabularnewline\hline
$\sin \theta$ & $\dots$ & $\dots$ & $\dots$ & $\dots$ & $\dots$ & $\dots$  \tabularnewline\hline
%$\tan \theta $ & 0 & $ \dfrac{\sqrt{3}}{3}$ & 1 & $\sqrt{3}$ & n'existe pas & 0 \tabularnewline \hline
\end{tabular}
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\Dem{
\begin{multicols}{2}
Pour calculer les valeurs de $\sin \dfrac{\pi}{4}$ et de $\cos \dfrac{\pi}{4}$ on 
se place dans le quadrilatère $OHMK$ ci-contre, de diagonale $OM=1$.\\

On sait que $OHMK$ est un carré, car il possède 4 angles droits et que l'angle $(\overrightarrow{OH} ; \overrightarrow{OM})$ mesure $\dfrac{\pi}{4}$.\\
On appelle $a$ la mesure de son côté.\\

Par définition du cosinus et sinus d'un nombre réel, on a $a=\cos \dfrac{\pi}4=\sin\dfrac{\pi}4$. Cherchons cette valeur.\\
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\end{multicols}
Dans le triangle $OHM$ rectangle en $H$, d'après le théorème de Pythagore, on a :\\
\vspace{2cm}

%$$OM^2=OH^2 + MH^2 \iff 1^2 =2 a^2 + a^2 \iff a^2 = \dfrac{1}{2} \stackrel{a>0}{\iff} a = \dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2} $$
Donc \fbox{$\cos \dfrac{\pi}{4} =\sin \dfrac{\pi}{4}= \dots\dots $}.\\

\begin{multicols}{2}
Pour calculer les valeurs du cosinus et du sinus de $\dfrac{\pi}{3}$ et $\dfrac{\pi}{6}$,
 on se place dans le triangle $OIM$ ci-contre, de côté $OM=1$. \\

On cherche $OH=\cos\dfrac{\pi}3$ et $OK=\sin\dfrac{\pi}3$.\\

On sait que le triangle $OIM$ est équilatéral car\\
 $OM=OI$ et que $(\overrightarrow{OH} ; \overrightarrow{OM})$ mesure $\dfrac{\pi}{3}$.
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\end{multicols}
Donc sa hauteur $[MH]$ est aussi sa médiane. D'où $H$ est le milieu de $[OI]$ et\quad  $OH=$\fbox{$\cos\dfrac{\pi}3=\dots\dots$}.\\ 
De plus, d'après le théorème de Pythagore dans le triangle $OHM$ rectangle en $H$ on a :\\
\vspace{2cm}
%$$OM^2 = OH^2 + MH^2 \iff 1^2 = \left(\dfrac12\right)^2 + OK^2 \iff OK^2 = 1-\dfrac14 = \dfrac34 \stackrel{OK>0}{\iff} OK = \dfrac{\sqrt{3}}{2}$$

Donc \fbox{$\sin\dfrac{\pi}3 = \dots \dots$}.\\

De plus la hauteur $[MH]$ du triangle $OIM$ est aussi sa bissectrice donc $(\overrightarrow{MO} ; \overrightarrow{MI})=\dfrac{\pi}6 \quad [2\pi]$.\\
On en déduit dans le triangle $OHM$ rectangle  en $H$ que \quad \fbox{ $\cos\dfrac{\pi}6 =\dots\dots= \dots\dots$ }\quad  et \quad 
\fbox{$\sin\dfrac{\pi}6 = \dots\dots = \dots\dots$ }
}


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\begin{multicols}{2}
\Expl{
Grâce aux angles associés, trouver les valeurs des cosinus et sinus des angles suivants :
$$\dfrac{3\pi}4 \qquad \qquad \dfrac{5\pi}4 \qquad \qquad -\dfrac{\pi}4 \qquad \qquad \dfrac{7\pi}4 $$
$$-\dfrac{\pi}3 \qquad \qquad \dfrac{2\pi}3 \qquad \qquad -\dfrac{2\pi}3 \qquad \qquad \dfrac{5\pi}3$$
$$\dfrac{5\pi}6 \qquad \qquad \dfrac{11\pi}6 \qquad \qquad -\dfrac{\pi}6 \qquad \qquad -\dfrac{5\pi}6 $$
}

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\end{multicols}




\end{document}

 