%!TEX encoding = UTF-8 Unicode

\documentclass[a4paper, reqno, 12pt]{article} %[a4paper, reqno]{amsart}

%\newcommand{\version}{\iftrue} % true : sans commentaires
%\newcommand{\affichecode}{\iffalse}

\newcommand{\version}{\iffalse} %       avec commentaires
\newcommand{\affichecode}{\iftrue}

\input{entete-section}
\usepackage[top=2cm, bottom=2cm, left=1.5cm, right=1.5cm]{geometry}
\usepackage{minitoc}
\definecolor{shadecolor}{rgb}{0.82,0.94,0.75}

\pagestyle{fancy}
\fancyhead[L]{Olivier Lader}	
\fancyhead[CE]{Exemple}
\fancyhead[CO]{Exemple}
\fancyhead[R]{\annee}
\fancyfoot[C]{\thepage}

\begin{document}

\title{$\LaTeX$}

\tableofcontents

\section{Informatique}

\subsection{Instructions conditionnelles}

\begin{cadre} \begin{defi}
Dans un {\bf algorithme}, on est parfois amené à exécuter une ou plusieurs instructions uniquement si une certaine condition est vérifiée, c'est ce qu'on appelle les {\bf instructions conditionnelles}. 

Si la condition n'est pas vérifiée, on peut soit exécuter un autre bloc d'instructions, soit ne rien faire. 

Dans ces deux cas, on exécute ensuite la suite de l'algorithme.
\end{defi} \end{cadre}

\begin{multicols}{2}
\begin{algorithm}[H] \begin{algorithmic}[1]
\State x $\leftarrow$ entier aléatoire entre 1 et 6 
\If{ x = 6 }
\State Afficher "Gagné !"
\Else
\State Afficher "Perdu..."
\EndIf
\State Afficher "À bientôt !"
\end{algorithmic} \end{algorithm}
\columnbreak
\begin{lstlisting}
import random

x = random.randint(1,6)
if x == 6:
    print("gagné!")
else:
    print("Perdu...")
print("A bientôt!")
\end{lstlisting}
\end{multicols}


\subsection{Boucles bornées}

\begin{cadre} \begin{defi}
Lorsqu'on veut exécuter un nombre déterminé de fois un même bloc d'instructions, on utilise une {\bf boucle bornée}, aussi appelée boucle {\bf Pour}.

Ces boucles sont munies d'une variable compteur que l'on peut utiliser dans la boucle.
\end{defi} \end{cadre}

\begin{exemple} \qquad
\begin{multicols}{2}
\begin{algorithmic}[1]
\For{i allant de 1 à 10}
\State a $\leftarrow$ 2 $\times$ i
\State Afficher a
\EndFor
\State Afficher "Terminé"
\end{algorithmic}
\columnbreak
\begin{lstlisting}
for i in range(1,11): # Attention la seconde borne est 11 et non pas 10!
    a = 2*i
    print(a)
print("Terminé")
\end{lstlisting}
\end{multicols}
Dans la boucle précédente, la variable {\tt i} est le compteur. À chaque passage dans la boucle, {\tt i} est incrémenté de 1 c'est-à-dire, augmente de $1$.
\end{exemple}

\begin{cadre} \begin{defi}
La boucle bornée:

\begin{minipage}{0.5\linewidth}
\begin{algorithm}[H]
\begin{algorithmic}[1]
\For{ $i$ allant de $a$ à $b$ avec un pas de $1$}
\State \ldots
\EndFor
\end{algorithmic}
\end{algorithm}
\end{minipage}

s'écrira:
\begin{lstlisting}
For i in range(a,b+1):      # attention au +1 !
    ...
\end{lstlisting}
\end{defi} \end{cadre}



\subsection{Boucles non bornées}

\begin{cadre} \begin{defi}
Lorsqu'on veut répéter un même bloc d'instructions tant qu'une certaine condition est vérifiée, on utilise une {\bf boucle non bornée}, aussi appelée {\bf Tant que}.
\end{defi} \end{cadre}

\begin{exemple} \qquad
\begin{multicols}{2}
\begin{algorithmic}[1]
\State p $\leftarrow$ 1
\While{ p $\leq$ 100 }
\State p $\leftarrow$ p $\times  2$
\EndWhile
\State Afficher p
\end{algorithmic}
\columnbreak
\begin{lstlisting}
p = 1
while p <= 100:
    p = p*2
print(p)
\end{lstlisting}
\end{multicols}
Comme la condition de la boucle {\tt Tant que} de l'exemple précédent porte sur la variable {\tt p}, cette variable doit être initialisée préalablement (c'est-à-dire qu'il faut lui donner une valeur au départ): ici, on l'a initialisé à $1 = 2^0$, la première puissance de 2.

Il faudra aussi veiller à ce que la condition dans la boucle {\tt Tant que} finisse par être remplie (sinon le programme ne s'arrêtera jamais).
\end{exemple}

\subsection{Tri par fusion (merge sort)}

L'algorithme est basé sur le principe fondamentale en informatique: \og diviser pour régner\fg{}. Nous verrons en fin de cette section à quel point se principe est efficace.

\medskip

L'idée est de casser (de manière récursive) la liste en deux et de trier les deux sous-listes puis de les fusionner. Voici un exemple:

\begin{center} \begin{tikzpicture}[x=1cm, y=1cm]
\foreach \x/\n in {0/1,2/3,4/4,6/2,8/7,10/6,12/5}
    \draw[shift={(\x,0)}] (0,0) rectangle (1,1) ++(-.5,-.5) node{$\n$};
\foreach \x/\n in {0.5/1,2.5/3,3.5/4,6.5/2,7.5/7,10.5/6,11.5/5}
    \draw[shift={(\x,2)}] (0,0) rectangle (1,1) ++(-.5,-.5) node{$\n$};
\foreach \x in{3.5,7.5,11.5}{
    \draw[->, shift={(\x,0)}] (0,1.8) -- (-1,1.2);
    \draw[->, shift={(\x,0)}] (0,1.8) -- (1,1.2);
    \draw[shift={(\x,0)}, line width=2pt, color=rouge2] (0,1.9) -- (0,3.1);
}
\foreach \x/\n in {1/1,2/3,3/4,7.5/2,8.5/7,9.5/6,10.5/5}
    \draw[shift={(\x,4)}] (0,0) rectangle (1,1) ++(-.5,-.5) node{$\n$};
\draw[->, shift={(2,2)}] (0,1.8) -- (-1,1.2);
\draw[->, shift={(2,2)}] (0,1.8) -- (1.5,1.2);
\draw[->, shift={(9.5,2)}] (0,1.8) -- (-2,1.2);
\draw[->, shift={(9.5,2)}] (0,1.8) -- (2,1.2);
\draw[shift={(2,2)}, line width=2pt, color=rouge2] (0,1.9) -- (0,3.1);
\draw[shift={(9.5,2)}, line width=2pt, color=rouge2] (0,1.9) -- (0,3.1);
\foreach \x/\n in {2.5/1,3.5/3,4.5/4,5.5/2,6.5/7,7.5/6,8.5/5}
    \draw[shift={(\x,6)}] (0,0) rectangle (1,1) ++(-.5,-.5) node{$\n$};
\draw[->, shift={(5.5,4)}] (0,1.8) -- (-3,1.2);
\draw[->, shift={(5.5,4)}] (0,1.8) -- (4,1.2);
\draw[shift={(5.5,4)}, line width=2pt, color=rouge2] (0,1.9) -- (0,3.1);
\foreach \y in {1.5, 3.5, 5.5}
    \draw[color=bleu2] (-.4,\y) node[left]{division};
\foreach \y in {-.5, -2.5, -4.5}
    \draw[color=bleu2] (-.4,\y) node[left]{fusion};
% fusions:
\foreach \x/\n in {0/1,2.5/3,3.5/4,6.5/2,7.5/7,10.5/5,11.5/6}
    \draw[shift={(\x,-2)}] (0,0) rectangle (1,1) ++(-.5,-.5) node{$\n$};
\foreach \x in{3.5,7.5,11.5}{
    \draw[<-, shift={(\x,-2)}] (0,1.2) ++(-2pt,0) -- (-1,1.8);
    \draw[<-, shift={(\x,-2)}] (0,1.2) ++(2pt, 0) -- (1,1.8);
}
\foreach \x/\n in {.5/1,1.5/3,2.5/4,7.5/2,8.5/5,9.5/6,10.5/7}
    \draw[shift={(\x,-4)}] (0,0) rectangle (1,1) ++(-.5,-.5) node{$\n$};
\draw[<-, shift={(2,-4)}] (0,1.2) ++(-2pt,0) -- (-1.5,1.8);
\draw[<-, shift={(2,-4)}] (0,1.2) ++(2pt, 0) -- (1.5,1.8);
\draw[<-, shift={(9.5,-4)}] (0,1.2) ++(-2pt,0) -- (-2,1.8);
\draw[<-, shift={(9.5,-4)}] (0,1.2) ++(2pt, 0) -- (2,1.8);
\foreach \x/\n in {1.5/1,2.5/2,3.5/3,4.5/4,5.5/5,6.5/6,7.5/7}
    \draw[shift={(\x,-6)}] (0,0) rectangle (1,1) ++(-.5,-.5) node{$\n$};
\draw[<-, shift={(5,-6)}] (0,1.2) ++(-2pt,0) -- (-3,1.8);
\draw[<-, shift={(5,-6)}] (0,1.2) ++(2pt, 0) -- (4.5,1.8);
\end{tikzpicture} \end{center}

Une vidéo illustrant cet algorithme: \url{https://youtu.be/JSceec-wEyw}

\begin{exo} \qquad
\begin{enumerate}
\item On considère les listes: {\tt T1 = [1,4,5,7,9]} et {\tt T2 = [2,3,6]}, toutes les deux triées. Écrire un programme qui fusionne les deux listes en une liste triée.
\item Écrire une fonction {\tt fusion(T1,T2)} qui permet de fusionner les listes triées {\tt T1} et {\tt T2} en une liste triée {\tt T} et retourne {\tt T}.
\affichecode
\begin{lstlisting}
def fusion(T1,T2):
    T = []
    i = 0
    j = 0
    while i < len(T1) and j < len(T2):
        x = T1[i]
        y = T2[j]
        if x < y: 
            T.append(x)
            i += 1
        else:
            T.append(y)
            j += 1 
    while i < len(T1): # il reste encore des éléments de T1 à ajouter
        T.append(T1[i])
        i += 1
    while j < len(T2): # sinon, il reste encore des éléments de T2 à ajouter
        T.append(T2[j])
        j += 1
    return T
\end{lstlisting}
\fi
\item Écrire une fonction {\tt fusion\_rec(T1,T2)} qui permet de faire la fusion mais de manière récursive.
\affichecode
\begin{lstlisting}
def fusion_rec(T1,T2):
    if T1 == []:
        return T2
    elif T2 == []:
        return T1
    elif T1[0] < T2[0]:
        return [T1[0]] + fusion(T1[1:], T2)
    else:
        return [T2[0]] + fusion(T1, T2[1:])
\end{lstlisting}
\fi
\end{enumerate}
\end{exo}

\espace{80}

\begin{mybox2} \begin{thm}[tri fusion]
Le tri fusion peut être codé en python de la manière suivante:
\begin{lstlisting}
def tri_fusion(L):
    if len(L) <= 1:
        return L
    else:
        n = len(L) // 2
        return fusion( tri_fusion(L[:n]), tri_fusion(L[n:]) )
\end{lstlisting}
Sa complexité temporelle dans le pire des cas est
\[ \mathcal O (n \log_2(n)) \]
et dans le meilleur des cas est 
\[ \mathcal O (n \log_2(n)) \]
\end{thm} \end{mybox2}


\subsection{Exercices}

\begin{exo}
On considère l'algorithme suivant:
\begin{center} \begin{tikzpicture}[x=.75cm, y=.9cm] \begin{scriptsize}
\node[draw, rectangle] (L1) at (0,0) {Entrée};
\node[draw, diamond, aspect=2.5] (B1) at (0,-2) {Est-ce un multiple de 11?};
\node[draw, rectangle] (L2) at (0,-4) {ajoute 8};
\draw[->] (L1) -- (B1);
\draw[->, rounded corners=8pt] (L2.west) -- (-3.5,-4) -- (-3.5,-2) -- (B1.west);
\draw[->] (B1) -- (L2) node[pos=.5, right]{non};
\node[draw,rectangle] (L3) at (6,-2) {soustrais 3};
\draw[->] (B1) -- (L3) node[pos=.5, above]{oui};
\node[draw, diamond, aspect=2.5] (B2) at (6,-4) {Est-ce un multiple de 3?};
\draw[->] (L3) -- (B2);
\node[draw] (L4) at (6,-6) {ajoute 88};
;
\draw[->, rounded corners=8pt] (L4.west) -- (6-3.5,-6) -- (6-3.5,-4) -- (B2.west);
\draw[->] (B2) -- (L4) node[pos=.5, right]{non};
\node[draw] (L5) at (12,-4) {ajoute 5};
\draw[->] (B2) -- (L5) node[pos=.5, above]{oui};
\node[draw, diamond, aspect=2.5] (B3) at (12,-6) {Est-ce un multiple de 5?};
\draw[->] (L5) -- (B3);
\node[draw] (L6) at (12,-8) {ajoute 264};
\draw[->, rounded corners=8pt] (L6.west) -- (12-3.5,-8) -- (12-3.5,-6) -- (B3.west);
\draw[->] (B3) -- (L6) node[pos=.5, right]{non};
\node[draw] (L7) at (18,-6) {sortie};
\draw[->] (B3) -- (L7) node[pos=.5, above]{oui};
\end{scriptsize} \end{tikzpicture} \end{center}
\begin{enumerate}
\item En entrant le nombre 437 dans l'algorithme précédent, quel nombre obtient-on en sortie? Justifier.
\item On admet qu'en Python, la commande {\tt x \% a == 0} permet de tester si {\tt a} divise {\tt x}. Compléter le programme suivant pour qu'il corresponde à l'algorithme:
\begin{lstlisting}
x = int(input("Entrez un nombre"))
while x % 11 != 0:
   x = x + ...
x = x - ...
while ...
    x = ...
x = ...
while ...
    x = 
print("sortie : {}".format(x)) 
\end{lstlisting}
\affichecode
\begin{lstlisting}
x = int(input("Entrez un nombre entier"))
while x % 11 != 0:
    x += 8
x = x-3
while x % 3 != 0:
    x += 88
x += 5
while x % 5 != 0:
    x += 264
print("sortie: {}".format(x))
\end{lstlisting}
\fi
\end{enumerate}
\end{exo}

\begin{exo}
Pour les 20 ans de Mathématiques Sans Frontières Gérard a inventé un algorithme pour arriver de 1989 à 2009 en quelques étapes.
\begin{center} \begin{tikzpicture}[x=1.3cm, y=.8cm]
\node[draw, ellipse] (L1) at (0,10) {N = 1989};
\node[draw, text width=50ex] (L2) at (0,8) {Choisir un des chiffres du nombre N et lui 

ajouter l'un des trois autres. 

$S$ est la somme résultant de cette addition.};
\draw[->] (L1) -- (L2);
\node[draw, diamond, aspect=2.5] (L3) at (0,6) {$S < 10$};
\node[draw, text width=25ex] (L4g) at (-2,4) {Remplacer le chiffre

choisi par $S$ dans

l'écriture de $N$.};
\node[draw, text width=25ex] (L4d) at (2,4) {Remplacer le chiffre

choisi par $S-10$ dans

l'écriture de $N$.};
\node[draw, diamond, aspect=2.5] (L5) at (0,1) {N = 2009};
\node[draw, ellipse] (L6) at (4,1) {STOP};
\draw (L2) -- (L3);
\draw[->] (L3.west) -- (-2,6) -- (L4g) node[pos=.5, left]{Oui};
\draw[->] (L3.east) -- (2,6) -- (L4d) node[pos=.5, right]{Non};;
\draw[->] (L4g) -- (-2,2.5) -- (0,2.5) -- (L5);
\draw[->] (L4d) -- (2,2.5) -- (0,2.5) -- (L5);
\draw[->] (L5.west) -- (-4,1) node[pos=.5, below]{Non} -- (-4,8) -- (L2);
\draw[->] (L5.east) -- (L6) node[pos=.5, below]{Oui};
\draw (4.5,8) node[right, text width=30ex]{Par exemple: 

je choisis 8 et j'ajoute 9};
\draw (4.5,6) node[right]{$S=17$};
\draw (4.5,4) node[right]{$N=1979$};
\end{tikzpicture} \end{center}
Donner une suite, la plus courte possible, de nombres allant de 1989 à 2009 suivant l'algorithme de Gérard.
\comment{En 5 étapes: 1989-1089-1189-2189-2109-2009}
\end{exo}

\section{Mathématiques}

\begin{mybox} \begin{propriété}
Soit $a$ un nombre réel et $r >0$. Pour tout nombre réel $x$, on a 
{\small \begin{align*}
|x-a| \leq r \quad&\iff\quad  a-r \leq x \leq a + r \\
\quad&\iff\quad x \in [a-r;a+r] 
\end{align*} }
\begin{center} \begin{tikzpicture}[x=1.3cm, y=1cm] \begin{scriptsize}
\draw[->] (-1,0) -- (7,0) node[right]{$\R$};
\draw[line width=1.5pt, color=rouge] (0,0) -- (4,0);
\draw[line width=1.5pt, color=rouge] (0,0) ++(2pt,4pt) -- ++(-2pt,0) -- ++(0,-8pt) -- ++(2pt,0);
\draw[line width=1.5pt, color=rouge] (4,0) ++(-2pt,4pt) -- ++(2pt,0) -- ++(0,-8pt) -- ++(-2pt,0);
\draw (2,2pt)  -- ++(0,-4pt) node[below]{$a$};
\draw (0,2pt)  -- ++(0,-4pt) node[below]{$a-r$};
\draw (4,2pt)  -- ++(0,-4pt) node[below]{$a+r$};
\draw[->] (2,0.5) ++(-2pt,0)  -- (0,.5) node[pos=.5, above]{$-r$};
\draw[->] (2,0.5) ++(2pt,0)  -- (4,.5) node[pos=.5, above]{$+r$};
\end{scriptsize} \end{tikzpicture} \end{center}
Dans ce cas, le nombre $a$ est appelé {\bf centre de l'intervalle} et le nombre $r$ {\bf rayon} de l'intervalle.
\end{propriété} \end{mybox}

\begin{exemple}
Voici une représentation graphique d'une fonction $f: [-5,5] \to \R$.
\begin{center} \begin{tikzpicture}[only marks, x = 0.7cm, y=0.7cm]
% Cadrillage
\draw[color = gray] (-5,-3) grid[step=.5, very thin] (5,4);
% Axes
\draw[->] (-5.5,0) -- coordinate (x axis mid) (5.5, 0) node[right]{$x$};
\draw[->] (0,-3.5) -- coordinate (y axis mid) (0, 4.5) node[above]{$y$};
\foreach \x in {-5,-4,...,5}
	\draw (\x, 2pt) -- (\x, -3pt) node[below] {\small $\x$\;};
\foreach \y in {-3,-2,...,-1,1,2,...,4}
	\draw (2pt, \y) -- (-3pt, \y) node[left] {\small $\y$};

% Courbe
\draw[thick] (-5,-3) .. controls +(30:1) and +(-100:.5) .. (-4,0);
\draw[thick] (-4,0) .. controls +(80:.5) and +(180:1) .. (-3,1);
\draw[thick] (-3,1) .. controls  +(0:.5) and +(180:1) .. (-1,0);
%\draw[thick] () .. controls +() and +() .. ();
\draw[thick] (-1,0) .. controls +(0:1) and +(-110:.3) .. (0,1);
\draw[thick] (0,1) .. controls +(70:.3) and +(180:1.5) .. (2,3.5);
\draw[thick] (2,3.5) .. controls +(0:1) and +(150:2) .. (5,-1);
\draw (3.5,2) node{$\mathcal C_f$};
\draw[thick, color=rouge] (-5.5,-3) -- (6,-3) node[right]{minimum = -3};
\draw[thick, color=rouge] (-5,0) -- (-5,-3);
\draw[thick, color=rouge] (-5.5,3.5) -- (6,3.5) node[right]{maximum = 3.5};
\draw[thick, color=rouge] (2,0) -- (2,3.5);
\end{tikzpicture}
\end{center}
Le tableau de variations de la fonction $f$:
\begin{center} \begin{tikzpicture}[scale=.8, y=.9cm, x=1cm]  \begin{scriptsize}
\foreach \x in {0,1.5,12.5}
	\draw (\x,1) -- (\x,-2);
\foreach \y in {1,0,-2}
	\draw (0,\y) -- (12.5,\y);
\draw (.7,.5) node{$x$}; 
\draw (.7,-1) node{$f(x)$}; 
\draw (2, .5) node{$-5$};
\draw (4.5, .5) node{$-3$};
\draw[->] (2.2, -1.5) -- (4.3,-.5) node[pos=0,left]{-3} node[pos=1, right]{1};
\draw (7, .5) node{$-1$};
\draw (9.5, .5) node{$2$};
\draw (12, .5) node{$5$};		
\draw[->] (4.7,-.5) -- (6.8,-1.5) node[pos=1, right]{0};
\draw[->] (7.2,-1.5) -- (9.3,-.5) node[pos=1, right]{3.5};
\draw[->] (10,-.5) -- (11.8,-1.5) node[pos=1, right]{-1};
\end{scriptsize} \end{tikzpicture} \end{center}
Le maximum de $f$ sur l'intervalle $[-5;5]$ est 3.5 et il est atteint en 2. Le minimum de $f$ sur l'intervalle $[-5;5]$ est -3 et il est atteint en -5. 
\end{exemple}

Considérons deux évolutions successives:
\begin{center}
\begin{tikzpicture}
\node[draw] (V0) at (0,0) {$v_1$};
\node[draw] (V1) at (3,0) {$v_2$};
\node[draw] (V2) at (6,0) {$v_3$};
\draw[->] (V0) .. controls +(60:2) and +(120:2) .. (V2) node[pos=.5, above]{\small $\times (1 + t)$};
\draw[->] (V0) .. controls +(-60:1) and +(-120:1) .. (V1) node[pos=.5, below]{\small $\times (1 + t_1)$};
\draw[->] (V1) .. controls +(-60:1) and +(-120:1) .. (V2) node[pos=.5, below]{\small $\times (1 + t_2)$};
\end{tikzpicture}
\end{center}
D'après le propriété précédente, on a les trois relations suivantes:
\[ v_2 = (1 + t_1) v_1 \qquad\text{,}\qquad v_3 = (1 + t_2) v_2 \qquad\text{et}\qquad v_3 = (1 + t) v_1 \]
D'où

\begin{mybox} \begin{propriété} 
Le coefficient multiplicateur de $v_0$ à $v_2$ est égal au produit des coefficients multiplicateurs successifs et le taux d'évolution globale est égal à
\[ t = (1+t_1)\times(1+t_2) - 1 \]
\end{propriété} \end{mybox}


\begin{exemple}
Considérons l'application $f\,:\llbracket 1; 5\rrbracket \to \{A,B,C,D,E\}$ définie ainsi:
\begin{center} \begin{tikzpicture}[x=.6cm, y=.4cm]
\draw[->] (0.5,7) node[left]{$\llbracket 1; 5\rrbracket$} -- (4.5,7) node[right]{$\{A,B,C,D,E\}$} node[pos=.5, above]{$f$};
\foreach \x in {1,2,...,5}{
	\node (A\x) at (0,6-\x) {$\x$}; 
	\node (B\x) at (7,6-\x) {\AlphAlph{\x}}; }
\draw (7,3) ellipse (1 and 3);
\draw (0,3) ellipse (1 and 3);
\draw[->, color=rouge] (A1) -- (B2);
\draw[->] (A2) -- (B3);
\draw[->] (A3) -- (B1);
\draw[->] (A4) -- (B4);
\draw[->] (A5) -- (B5);
\draw[dash pattern=on 5pt off 2pt on 2pt off 2pt, color=gris] (10,-1) -- (10,8);
\draw[shift={(13,0)}, ->] (2.5,7) node[left]{$\{A,B,C,D,E\}$} -- (6.5,7) node[right]{$\llbracket 1; 5\rrbracket$} node[pos=.5, above]{$f^{-1}$};
\foreach \x in {1,2,...,5}{
	\node (A\x) at (20,6-\x) {$\x$}; 
	\node (B\x) at (13,6-\x) {\AlphAlph{\x}}; }
\draw[shift={(13,0)}] (7,3) ellipse (1 and 3);
\draw[shift={(13,0)}] (0,3) ellipse (1 and 3);
\draw[shift={(13,0)},<-, color=rouge] (A1) -- (B2);
\draw[shift={(13,0)},<-] (A2) -- (B3);
\draw[shift={(13,0)},<-] (A3) -- (B1);
\draw[shift={(13,0)},<-] (A4) -- (B4);
\draw[shift={(13,0)}, <-] (A5) -- (B5);
\end{tikzpicture} 
\end{center}
On observe que $f$ est une bijection et son application réciproque $f^{-1}$ est représenté ci-dessus (on observe qu'on retourne les flèches dans l'autre sens pour obtenir la réciproque).
\end{exemple}

\subsection{Fonctions}

\begin{exemples} \qquad \\
\begin{center} \begin{tabular}{c|c}
Résoudre graphiquement $f(x) > 3$ & Résoudre graphiquement $g(x) < 1$ \\
sur l'intervalle $[0;4]$  & sur l'intervalle $[0;5]$  \\ \hline
\begin{tikzpicture}[x=1cm, y=.8cm] \begin{scriptsize}
\draw[->] (-1,0) -- (6,0) node[right]{$x$};
\draw[->] (0,-1) -- (0,4.2) node[above]{$y$};
\foreach \x in {1,2,3,4,5}
	\draw (\x,2pt) -- ++(0,-5pt) node[below]{$\x$};
\foreach \y in {1,2,3,4}
	\draw (2pt,\y) -- ++(-5pt,0) node[left]{$\y$};
\draw (0,0) node[below left]{$0$};
\draw[domain=-0:4, thick] plot(\x,{4 - (\x-2)*(\x-2)});
\draw[domain=1:3, thick, color=rouge] plot(\x,{4 - (\x-2)*(\x-2)});
\draw[dashed, thick,  color=bleu] (-3pt,3) node[left]{$3$} -- (5,3);
\draw[thick, color=rouge] (1,0) node{]} -- (3,0) node{[};
\draw (3.8,1) node[right]{$\mathcal C_f$};
\end{scriptsize} \end{tikzpicture}
& 
\begin{tikzpicture}[x=1cm, y=.8cm] \begin{scriptsize}
\draw[->] (-1,0) -- (6,0) node[right]{$x$};
\draw[->] (0,-2) -- (0,3) node[above]{$y$};
\foreach \x in {1,2,3,4,5}
	\draw (\x,2pt) -- ++(0,-5pt) node[below]{$\x$};
\foreach \y in {-1,1,2}
	\draw (2pt,\y) -- ++(-5pt,0) node[left]{$\y$};
\draw (0,0) node[below left]{$0$};
\draw[domain=0:5, thick] plot(\x,{-0.1 * (\x + 1)*(\x-2)*(\x-4)+1});
\draw[domain=0:2, color=rouge, thick] plot(\x,{-0.1 * (\x + 1)*(\x-2)*(\x-4)+1});
\draw[domain=4:5, color=rouge, thick] plot(\x,{-0.1 * (\x + 1)*(\x-2)*(\x-4)+1});
\draw[dashed, thick,  color=bleu] (-3pt,1) node[left]{$1$} -- (5,1);
\draw[thick, color=rouge] (0,0) node{[} -- (2,0) node{[};
\draw[thick, color=rouge] (4,0) node{]} -- (5,0) node{]};
\draw (3,1.6) node{$\mathcal C_g$};
\end{scriptsize} \end{tikzpicture}
\\ \hline
L'ensemble des solutions est & L'ensemble des solutions est \\
\hide{$]1;\,3[$} &\hide{$[0;\,2[ \,\cup\, ]4;\,5]$} 
\end{tabular} \end{center} 
\end{exemples}

\begin{cadre} \begin{defi}
Soit $f: I \to \R$ une fonction dérivable en un point $a$ de $I$ et $A$ le point d'abscisse $a$ de la courbe représentative de la fonction $f$. La {\red tangente à la courbe} $\mathcal C_f$ au point d'abscisse $a$, notée $T_a(f)$, est la droite passant par le point $A(a;f(a))$ et dont le coefficient directeur est $f'(a)$.
\end{defi} \end{cadre}

\begin{mybox} \begin{propriété}
Soit $f: I \to \R$ une fonction dérivable en un point $a$ de $I$, alors l'équation de la tangente à la courbe $\mathcal C_f$ au point d'abscisse $a$ est:
\[  y = f'(a) ( x - a) + f(a) \]
\end{propriété} \end{mybox}


Graphiquement: 

\begin{center} \centering \begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=.5cm,y=.5cm]
\draw[->,color=black] (-3.34,0) -- (18,0);
\draw[->,color=black] (0,-2) -- (0,13.89);
\draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$};
\draw[color=rouge,fill=rouge,fill opacity=0.1] (7,3.6) -- (8,3.6) -- (8,4.8);
% Courbe:
\draw[color = blue, smooth,samples=100,domain=-3:13] plot(\x,{1/10*((\x)-1)^2});
% Tangente:
\draw [color=rouge,domain=3:14] plot(\x,{(-4.8--1.2*\x)/1});
\begin{scriptsize}
\fill [color=gray] (7,3.6) circle (1.5pt);
\draw (6.6,4) node {$A$};
\draw[color=rouge] (8,3.6) node[anchor=north east] {1};
\draw[color=rouge] (8,4.23) node[anchor=west] {$f'(a)$};
\draw (7, .1) -- (7,-.2) node[below] {$a$};
\draw (.1, 3.6) -- (-.2, 3.6) node[anchor = east]{$f(a)$};
\draw[dashed] (7,0) -- (7,3.6);
\draw[dashed] (0,3.6) -- (7,3.6);
\draw (10,7) node[right]{$T_a(f): y = f'(a) ( x - a) + f(a)$};
\draw (10.2,10) node{$\mathcal C_f$};
\end{scriptsize}
\end{tikzpicture} 
\end{center}

\begin{figure}[h!] \begin{center} \begin{tikzpicture}[x=.8cm, y=.8cm] \begin{scriptsize}
\draw[->] (-5,0) -- (10,0) node[right]{$x$};
\draw[->] (0,-5) -- (0,10) node[above]{$y$};
\draw [color=gris!70,dash pattern=on 3pt off 3pt, xstep=1,ystep=1] (-5,-5) grid (10,10);
\draw[color=white] (7,1.946) -- (9,2.197) node[pos=.5, sloped, above, color=black]{$y = \ln(x)$};
\draw[domain=0.011:10, samples=200, thick] plot(\x,{ln(\x)});
\draw[domain=-5:2.3, samples=200, thick] plot(\x,{exp(\x)});
\foreach \x in {1}
	\draw (\x,2pt) -- ++(0,-4pt) node[below]{$\x$};
\foreach \y in {1}
	\draw (2pt,\y) -- ++(-4pt,0) node[left]{$\y$};
\draw (0,0) node[below left]{$0$};
\draw[fill=rouge] (1,0) circle (1.5pt);
\draw[fill=rouge] (0,1) circle (1.5pt);
\draw[dashed] (0,1) -- (2.718,1) -- (2.718,-2pt) node[below]{$e$};
\draw[dashed] (-2pt,2.718) node[left]{$e$} -- (1,2.718) -- (1,-2pt);
\draw (2,7) node[right]{$y = e^x$};
\draw[fill=rouge] (2.718,1) circle (1.5pt);
\draw[fill=rouge] (1,2.718) circle (1.5pt);
\draw [color=gris,dash pattern=on 4pt off 2pt on 2pt off 2pt] (-5,-5) -- (10,10) node[pos=.8, sloped, above, color=black]{$y = x$};
\def \x {1.6}
\def \y {5}
\draw[dashed, color=bleu] (\x,-2pt) node[below]{$y$} -- (\x,\y) -- (-2pt,\y) node[left]{$x = e^y$};
\draw[dashed, color=bleu] (\y,-2pt) node[below]{$x$} -- (\y,\x) -- (-2pt,\x) node[left]{$y = \ln(x)$};
\begin{scope}[shift={(3.3,3.3)}]
\draw[color=vert, rotate=45] (3pt,0) -- (3pt,3pt) -- (0,3pt);
\end{scope}
\draw[color=vert] (\x,\y) -- (\y,\x);
\draw[fill=rouge] (\x,\y) circle (1.5pt);
\draw[fill=rouge] (\y,\x) circle (1.5pt);
\end{scriptsize} \end{tikzpicture}  \end{center}
\caption{Le logarithme népérien et l'exponentielle}
\label{symlogexpo}
\end{figure}

\begin{mybox} \begin{propriété}[relation de Chasles]
Soit $f:[a;\,b] \to \R$ continue. Soit $c$ un nombre de l'inervalle $[a;b]$, alors
\begin{center} \begin{tikzpicture}[x=1.5cm, y=.8cm] 
\begin{scriptsize}
\def \a {3}
\def \b {7}
\def \fa {2}
\def \fb {2.2}
\draw[fill=rouge!20, color=rouge!10] (\b,0) -- (\a,0) -- (\a,\fa) .. controls +(-10:2) and +(-150:3) .. (\b, \fb) -- (\b, 0);
\draw[fill=rouge!20, color=rouge!10] (\a,0) -- (1,0) -- (1,.5) .. controls +(30:2) and +(170:1) .. (\a, \fa) -- (\a, 0);
\draw[pattern=north west lines, pattern color=rouge!90] (\a,0) -- (1,0) -- (1,.5) .. controls +(30:2) and +(170:1) .. (\a, \fa) -- (\a, 0);
\draw[thick, color=black] (0,.5) .. controls +(-20:.8) and +(-150:1) .. (1, .5);
\draw[thick, color=rouge] (1,.5) .. controls +(30:2) and +(170:1) .. (\a, \fa);
\draw[thick, color=rouge] (\a,\fa) .. controls +(-10:2) and +(-150:3) .. (\b, \fb);
\draw[thick, color=black] (\b,\fb) .. controls +(30:1) and +(-150:.5) .. (8, 3) node[pos=.9, above]{$\mathcal C_f$};
\draw[->] (-.5,0) -- (8,0) node[right]{$x$};
\draw[->] (0,-8pt) -- (0,4) node[right]{$y$};
\draw (0,0) node[below left]{$O$};
\draw (1,2pt) -- ++(0,-4pt) node[below]{$a$};
\draw (\a,2pt) -- ++(0,-4pt) node[below]{\textcolor{bleu}{$c$}};
\draw (\b,2pt) -- ++(0,-4pt) node[below]{$b$} -- (\b,\fb);
\end{scriptsize} 
\draw (2,.1) node[above, color=black]{$\int_a^{\textcolor{bleu}{c}} f(x) \dx$};
\draw (5,.1) node[above, color=black]{$\int_{\textcolor{bleu}{c}}^b f(x) \dx$};
\draw[thick, color=rouge!90] (1,.5) -- (1,0) -- (7,0) -- (7,2.2);
\draw[<-, color=rouge!90] (3.7,1.87) -- (4,2.5) node[right, color=black]{$\int_a^b f(x)\dx$};
\end{tikzpicture} \end{center}
\[ \int_a^{\textcolor{bleu}{c}} f(x) \dx + \int_{\textcolor{bleu}{c}}^b f(x) \dx = \int_a^b f(x) \dx \]
\end{propriété} \end{mybox}

\begin{mybox2} \begin{thm}[Sommes de Riemann]
Soit $f:[a;b] \to \R$ une fonction continue sur $[a;b]$, alors
\[ \limite \frac{b-a} n \sum_{k=0}^{n-1} f\left( a + k \frac {b-a} n \right) = \int_a^b f(x) \dx \]
\end{thm} \end{mybox2}

\begin{rmq}
Ce théorème est encore vrai pour une fonction à valeurs complexes.
\end{rmq}


\begin{rmq}
Ce procédé a été utilisé par Bernard Riemann dans \og über die Darstellbarkeit einer Funktion durch eine trigonometrische Reihe \fg{} en 1854 pour définir plus rigoureusement l'intégrale d'une fonction continue sur un intervalle. D'ailleurs, en sa mémoire, on parle aussi d'{\bf intégrale de Riemann} pour la notion d'intégrale que nous étudions dans ce chapitre.
\end{rmq}

En reprenant la démonstration, on montre de même que:

\begin{corollaire}
Soit $f:[a;b] \to \R$ une fonction continue sur $[a;b]$, alors
\[ \limite \frac{b-a} n \sum_{k=1}^{n} f\left( a + k \frac {b-a} n \right) = \int_a^b f(x) \dx \]
\end{corollaire}

En effectuant la demi-somme des deux suites précédente, on déduit immédiatement:

\begin{mybox2} \begin{corollaire}[méthode des trapèzes]
Soit $f:[a;b] \to \R$ une fonction continue sur $[a;b]$. Pour tout $0 \leq k \leq n$, posons 
\[ a_k = a + k \frac {b-a} n \]
alors
\[ \limite \frac{b-a} {n} \sum_{k=0}^{n-1}\frac{f(a_k) + f(a_{k+1})} 2  = \int_a^b f(x) \dx \]
\end{corollaire} \end{mybox2}

\begin{center} \begin{tikzpicture}[x=3cm, y=2.5cm] \begin{scriptsize}
\draw[fill=rouge!10, color=rouge!10] (2.5,1.5) -- (2.5,0) -- (-2,0) -- (-2,2) .. controls +(30:2) and +(-180:1) .. (1.5,1) .. controls +(0:.3) and +(-180:.2) .. (1.9,1.3) .. controls +(0:.5) and +(120:.5) .. (2.5,1.6);
\draw[->] (-2.5,0) -- (3,0) node[right]{$x$};
\draw[->] (0,-8pt) -- (0,2.3) node[right]{$y$};
\draw (1,2pt) -- ++(0,-4pt) node[below]{$I$};
\draw (2pt,1) -- ++(-4pt,0) node[left]{$J$};
\draw (0,0) node[below left]{$O$};
\end{scriptsize} 
\def \a {-2}
\def \b {2.5}
\draw (\a,2pt) -- ++(0,-4pt) node[below]{$a$};
\draw (\b,2pt) -- ++(0,-4pt) node[below]{$b$};
\draw[thick] (2.5,1.5) -- (2.5,0) -- (-2,0) -- (-2,2) .. controls +(30:2) and +(-180:1) .. (1.5,1) .. controls +(0:.3) and +(-180:.2) .. (1.9,1.3) .. controls +(0:.5) and +(120:.5) .. (2.5,1.6) -- (2.5,0);
\draw (-1.5,2.2) ++(0,3pt) node[above]{$\mathcal C_f$};
\draw[pattern=north west lines, pattern color=bleu2!60] (1.5,0) -- (1.9,0) -- (1.9,1.3) -- (1.5,1) -- (1.5,0);
\draw[color=bleu2] (1.5,-2pt) node[below, color=black]{$a_k$} -- (1.5,1);
\draw[color=bleu2] (1.9,-2pt) node[below, color=black]{$a_{k+1}$} -- (1.9,1.3);
\draw[color=bleu2] (1.5,1) -- (1.9,1.3);
\draw[<-] (1.7, 0.7) -- (1,2.5) node[above]{Aire(trapèze) $= \text{hauteur} \times \frac{(\text{petite base + grande base}) } 2 = \frac {b-a} n \times  \frac{f(a_k) + f(a_{k+1})} 2$};
\end{tikzpicture} \end{center}



\begin{exemple}
Considérons l'équation différentielle d'ordre 1:
\begin{equation} \tag{E}
y' + t y = t
\end{equation}
Posons $f: \R^2 \to \R$ l'application définie par $f(t,y) = t - ty$, alors $(E)$ équivaut à 
\[ \forall \, t \in \R: y'(t) = f(t,y(t))  \]
Soit $y:\R \to \R$ une solution de $(E)$, alors pour déterminer le coefficient directeur $y'(x)$ de la tangente au point $M(x;y)$ sur la courbe représentative de la solutio $y$, il suffit de calculer $f(x,y)$. Dans la figure suivante, nous avons représenté pour différents points $M(x;y)$ un segment de droite de pente $f(x;y)$. Nous obtenons ainsi le {\red champs de directions} de l'équation différentielle:
%{{{
\begin{center} \begin{tikzpicture}[x=1.35cm, y=1.35cm] \begin{scriptsize}
\draw[->] (-6,0) -- (6,0) node[right]{$t$};
\draw[->] (0,-4pt) -- (0,4.5) node[right]{$y$};
\draw (-6.0,0.0) -- ++(0.05,-0.3);
\draw (-6.0,0.2) -- ++(0.05,-0.24);
\draw (-6.0,0.4) -- ++(0.05,-0.18);
\draw (-6.0,0.6) -- ++(0.05,-0.12);
\draw (-6.0,0.8) -- ++(0.05,-0.06);
\draw (-6.0,1.0) -- ++(0.05,0.0);
\draw (-6.0,1.2) -- ++(0.05,0.06);
\draw (-6.0,1.4) -- ++(0.05,0.12);
\draw (-6.0,1.6) -- ++(0.05,0.18);
\draw (-6.0,1.8) -- ++(0.05,0.24);
\draw (-6.0,2.0) -- ++(0.05,0.3);
\draw (-6.0,2.2) -- ++(0.05,0.36);
\draw (-6.0,2.4) -- ++(0.05,0.42);
\draw (-6.0,2.6) -- ++(0.05,0.48);
\draw (-6.0,2.8) -- ++(0.05,0.54);
\draw (-6.0,3.0) -- ++(0.05,0.6);
\draw (-6.0,3.2) -- ++(0.05,0.66);
\draw (-6.0,3.4) -- ++(0.05,0.72);
\draw (-6.0,3.6) -- ++(0.05,0.78);
\draw (-6.0,3.8) -- ++(0.05,0.84);
\draw (-5.8,0.0) -- ++(0.05,-0.29);
\draw (-5.8,0.2) -- ++(0.05,-0.232);
\draw (-5.8,0.4) -- ++(0.05,-0.174);
\draw (-5.8,0.6) -- ++(0.05,-0.116);
\draw (-5.8,0.8) -- ++(0.05,-0.058);
\draw (-5.8,1.0) -- ++(0.05,0.0);
\draw (-5.8,1.2) -- ++(0.05,0.058);
\draw (-5.8,1.4) -- ++(0.05,0.116);
\draw (-5.8,1.6) -- ++(0.05,0.174);
\draw (-5.8,1.8) -- ++(0.05,0.232);
\draw (-5.8,2.0) -- ++(0.05,0.29);
\draw (-5.8,2.2) -- ++(0.05,0.348);
\draw (-5.8,2.4) -- ++(0.05,0.406);
\draw (-5.8,2.6) -- ++(0.05,0.464);
\draw (-5.8,2.8) -- ++(0.05,0.522);
\draw (-5.8,3.0) -- ++(0.05,0.58);
\draw (-5.8,3.2) -- ++(0.05,0.638);
\draw (-5.8,3.4) -- ++(0.05,0.696);
\draw (-5.8,3.6) -- ++(0.05,0.754);
\draw (-5.8,3.8) -- ++(0.05,0.812);
\draw (-5.6,0.0) -- ++(0.05,-0.28);
\draw (-5.6,0.2) -- ++(0.05,-0.224);
\draw (-5.6,0.4) -- ++(0.05,-0.168);
\draw (-5.6,0.6) -- ++(0.05,-0.112);
\draw (-5.6,0.8) -- ++(0.05,-0.056);
\draw (-5.6,1.0) -- ++(0.05,0.0);
\draw (-5.6,1.2) -- ++(0.05,0.056);
\draw (-5.6,1.4) -- ++(0.05,0.112);
\draw (-5.6,1.6) -- ++(0.05,0.168);
\draw (-5.6,1.8) -- ++(0.05,0.224);
\draw (-5.6,2.0) -- ++(0.05,0.28);
\draw (-5.6,2.2) -- ++(0.05,0.336);
\draw (-5.6,2.4) -- ++(0.05,0.392);
\draw (-5.6,2.6) -- ++(0.05,0.448);
\draw (-5.6,2.8) -- ++(0.05,0.504);
\draw (-5.6,3.0) -- ++(0.05,0.56);
\draw (-5.6,3.2) -- ++(0.05,0.616);
\draw (-5.6,3.4) -- ++(0.05,0.672);
\draw (-5.6,3.6) -- ++(0.05,0.728);
\draw (-5.6,3.8) -- ++(0.05,0.784);
\draw (-5.4,0.0) -- ++(0.05,-0.27);
\draw (-5.4,0.2) -- ++(0.05,-0.216);
\draw (-5.4,0.4) -- ++(0.05,-0.162);
\draw (-5.4,0.6) -- ++(0.05,-0.108);
\draw (-5.4,0.8) -- ++(0.05,-0.054);
\draw (-5.4,1.0) -- ++(0.05,0.0);
\draw (-5.4,1.2) -- ++(0.05,0.054);
\draw (-5.4,1.4) -- ++(0.05,0.108);
\draw (-5.4,1.6) -- ++(0.05,0.162);
\draw (-5.4,1.8) -- ++(0.05,0.216);
\draw (-5.4,2.0) -- ++(0.05,0.27);
\draw (-5.4,2.2) -- ++(0.05,0.324);
\draw (-5.4,2.4) -- ++(0.05,0.378);
\draw (-5.4,2.6) -- ++(0.05,0.432);
\draw (-5.4,2.8) -- ++(0.05,0.486);
\draw (-5.4,3.0) -- ++(0.05,0.54);
\draw (-5.4,3.2) -- ++(0.05,0.594);
\draw (-5.4,3.4) -- ++(0.05,0.648);
\draw (-5.4,3.6) -- ++(0.05,0.702);
\draw (-5.4,3.8) -- ++(0.05,0.756);
\draw (-5.2,0.0) -- ++(0.05,-0.26);
\draw (-5.2,0.2) -- ++(0.05,-0.208);
\draw (-5.2,0.4) -- ++(0.05,-0.156);
\draw (-5.2,0.6) -- ++(0.05,-0.104);
\draw (-5.2,0.8) -- ++(0.05,-0.052);
\draw (-5.2,1.0) -- ++(0.05,0.0);
\draw (-5.2,1.2) -- ++(0.05,0.052);
\draw (-5.2,1.4) -- ++(0.05,0.104);
\draw (-5.2,1.6) -- ++(0.05,0.156);
\draw (-5.2,1.8) -- ++(0.05,0.208);
\draw (-5.2,2.0) -- ++(0.05,0.26);
\draw (-5.2,2.2) -- ++(0.05,0.312);
\draw (-5.2,2.4) -- ++(0.05,0.364);
\draw (-5.2,2.6) -- ++(0.05,0.416);
\draw (-5.2,2.8) -- ++(0.05,0.468);
\draw (-5.2,3.0) -- ++(0.05,0.52);
\draw (-5.2,3.2) -- ++(0.05,0.572);
\draw (-5.2,3.4) -- ++(0.05,0.624);
\draw (-5.2,3.6) -- ++(0.05,0.676);
\draw (-5.2,3.8) -- ++(0.05,0.728);
\draw (-5.0,0.0) -- ++(0.05,-0.25);
\draw (-5.0,0.2) -- ++(0.05,-0.2);
\draw (-5.0,0.4) -- ++(0.05,-0.15);
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\draw (5.2,0.0) -- ++(0.05,0.26);
\draw (5.2,0.2) -- ++(0.05,0.208);
\draw (5.2,0.4) -- ++(0.05,0.156);
\draw (5.2,0.6) -- ++(0.05,0.104);
\draw (5.2,0.8) -- ++(0.05,0.052);
\draw (5.2,1.0) -- ++(0.05,0.0);
\draw (5.2,1.2) -- ++(0.05,-0.052);
\draw (5.2,1.4) -- ++(0.05,-0.104);
\draw (5.2,1.6) -- ++(0.05,-0.156);
\draw (5.2,1.8) -- ++(0.05,-0.208);
\draw (5.2,2.0) -- ++(0.05,-0.26);
\draw (5.2,2.2) -- ++(0.05,-0.312);
\draw (5.2,2.4) -- ++(0.05,-0.364);
\draw (5.2,2.6) -- ++(0.05,-0.416);
\draw (5.2,2.8) -- ++(0.05,-0.468);
\draw (5.2,3.0) -- ++(0.05,-0.52);
\draw (5.2,3.2) -- ++(0.05,-0.572);
\draw (5.2,3.4) -- ++(0.05,-0.624);
\draw (5.2,3.6) -- ++(0.05,-0.676);
\draw (5.2,3.8) -- ++(0.05,-0.728);
\draw (5.4,0.0) -- ++(0.05,0.27);
\draw (5.4,0.2) -- ++(0.05,0.216);
\draw (5.4,0.4) -- ++(0.05,0.162);
\draw (5.4,0.6) -- ++(0.05,0.108);
\draw (5.4,0.8) -- ++(0.05,0.054);
\draw (5.4,1.0) -- ++(0.05,0.0);
\draw (5.4,1.2) -- ++(0.05,-0.054);
\draw (5.4,1.4) -- ++(0.05,-0.108);
\draw (5.4,1.6) -- ++(0.05,-0.162);
\draw (5.4,1.8) -- ++(0.05,-0.216);
\draw (5.4,2.0) -- ++(0.05,-0.27);
\draw (5.4,2.2) -- ++(0.05,-0.324);
\draw (5.4,2.4) -- ++(0.05,-0.378);
\draw (5.4,2.6) -- ++(0.05,-0.432);
\draw (5.4,2.8) -- ++(0.05,-0.486);
\draw (5.4,3.0) -- ++(0.05,-0.54);
\draw (5.4,3.2) -- ++(0.05,-0.594);
\draw (5.4,3.4) -- ++(0.05,-0.648);
\draw (5.4,3.6) -- ++(0.05,-0.702);
\draw (5.4,3.8) -- ++(0.05,-0.756);
\draw (5.6,0.0) -- ++(0.05,0.28);
\draw (5.6,0.2) -- ++(0.05,0.224);
\draw (5.6,0.4) -- ++(0.05,0.168);
\draw (5.6,0.6) -- ++(0.05,0.112);
\draw (5.6,0.8) -- ++(0.05,0.056);
\draw (5.6,1.0) -- ++(0.05,0.0);
\draw (5.6,1.2) -- ++(0.05,-0.056);
\draw (5.6,1.4) -- ++(0.05,-0.112);
\draw (5.6,1.6) -- ++(0.05,-0.168);
\draw (5.6,1.8) -- ++(0.05,-0.224);
\draw (5.6,2.0) -- ++(0.05,-0.28);
\draw (5.6,2.2) -- ++(0.05,-0.336);
\draw (5.6,2.4) -- ++(0.05,-0.392);
\draw (5.6,2.6) -- ++(0.05,-0.448);
\draw (5.6,2.8) -- ++(0.05,-0.504);
\draw (5.6,3.0) -- ++(0.05,-0.56);
\draw (5.6,3.2) -- ++(0.05,-0.616);
\draw (5.6,3.4) -- ++(0.05,-0.672);
\draw (5.6,3.6) -- ++(0.05,-0.728);
\draw (5.6,3.8) -- ++(0.05,-0.784);
\draw (5.8,0.0) -- ++(0.05,0.29);
\draw (5.8,0.2) -- ++(0.05,0.232);
\draw (5.8,0.4) -- ++(0.05,0.174);
\draw (5.8,0.6) -- ++(0.05,0.116);
\draw (5.8,0.8) -- ++(0.05,0.058);
\draw (5.8,1.0) -- ++(0.05,0.0);
\draw (5.8,1.2) -- ++(0.05,-0.058);
\draw (5.8,1.4) -- ++(0.05,-0.116);
\draw (5.8,1.6) -- ++(0.05,-0.174);
\draw (5.8,1.8) -- ++(0.05,-0.232);
\draw (5.8,2.0) -- ++(0.05,-0.29);
\draw (5.8,2.2) -- ++(0.05,-0.348);
\draw (5.8,2.4) -- ++(0.05,-0.406);
\draw (5.8,2.6) -- ++(0.05,-0.464);
\draw (5.8,2.8) -- ++(0.05,-0.522);
\draw (5.8,3.0) -- ++(0.05,-0.58);
\draw (5.8,3.2) -- ++(0.05,-0.638);
\draw (5.8,3.4) -- ++(0.05,-0.696);
\draw (5.8,3.6) -- ++(0.05,-0.754);
\draw (5.8,3.8) -- ++(0.05,-0.812);
\draw[domain=-6:6, color=rouge, smooth] plot(\x,{1 - exp(-\x*\x/2)});
\draw[domain=-6:6, color=vert, smooth] plot(\x,{1 + 2*exp(-\x*\x/2)});
\draw[domain=-6:6, color=bleu, smooth] plot(\x,{1 + exp(-\x*\x/2)});
\draw (0,0) node[below left]{$0$};
\draw (1,2pt) -- ++(0,-4pt) node[below]{$1$};
\draw (2pt,1) -- ++(-4pt,0) node[left]{$1$};
\end{scriptsize} \end{tikzpicture} \end{center}

%}}}
Les trois courbes représentent des solutions du problème de Cauchy avec comme condition initiale $y(0) = 0$, $y(0) = 2$ et $y(0) = 3$. \\

Nous verrons dans la section suivante que les solutions de $(E)$ sont de la forme $y:\R \to \R$ définie par 
\[ \forall \, t \in \R: \qquad  y(t) = 1 + \lambda e^{-\frac {t^2} 2} \]
où $\lambda \in \R$.
\end{exemple}

\begin{cadre} \begin{defi}
Soit $a\in \bar{I}$. On dit que $f$ admet {\red une limite $\ell$ lorsque $x$ tend vers $a$} si $f$ admet une limite par la gauche et une limite par la droite en $a$ et si 
$\displaystyle\lim_{x\to a^-}\: f(x)=\ell=\displaystyle\lim_{x\to a^+}\: f(x)$
On note alors $\displaystyle\lim_{x\to a}\: f(x)=\ell$
\end{defi} \end{cadre}

\begin{mybox} \begin{proposition}
Soit $a\in \bar{I}$. La fonction $f$ admet une {\red limite} $\ell$ lorsque $x$ tend vers $a$ si et seulement si
$$\forall\: \epsilon>0,\: \exists\: \delta>0,\: \forall\: x\in I\setminus\{a\},\quad |x-a|\leq \delta\quad \Rightarrow \quad  \hide{|f(x)-\ell|\leq \epsilon} $$
\begin{center} \begin{tikzpicture} \begin{scriptsize}
\draw[->] (-2,0) -- (8.2,0) node[right]{$x$};
\draw[->] (0,-.1) -- (0,3.1) node[right]{$y$};
\draw[color=vert2] (3,-.5) -- (3,2);
\draw[color=vert2] (5,-.5) -- (5,1.06);
\draw[fill=rouge!10] (3,2) rectangle (5,1.06);
\draw[domain=-.5:8, thick, color=bleu] plot(\x,{((0.04792*\x - 0.54375)* \x + 1.53333)*\x + 1});
\draw (8,3) node[left]{$\mathcal C_f \; $};
\draw[color=rouge] (-1.1,2) -- (8,2);
\draw[color=rouge] (-1.1,1.06) -- (8,1.06);
\draw[fill=white, color=white] (4,1.5) circle (.6pt);
\draw[->] (-1,1.5) ++(0,1pt) -- (-1,2) node[pos=.5, left]{$+\epsilon$};
\draw[->] (-1,1.5) ++(0,-1pt) -- (-1,1.06) node[pos=.5, left]{$-\epsilon$};
\draw[<->] (-1.7,1.06) -- (-1.7,2) node[pos=.5, left]{$\forall$ voisinage de $l$};
\draw (4,2pt) -- ++(0,-4pt) node[below]{$a$};
\draw (2pt,1.5) -- (-2pt,1.5) node[left]{$l$};
\draw[->] (4,8pt) ++(1pt,0) -- (5,8pt) node[pos=.5, above]{$+\delta$};
\draw[->] (4,8pt) ++(-1pt,0) -- (3,8pt) node[pos=.5, above]{$-\delta$};
\draw[<->] (3,-.5) -- (5,-.5) node[pos=.5, below]{$\exists$ voisinage de $a$};
\end{scriptsize} \end{tikzpicture} \end{center}
\end{proposition} \end{mybox}

\begin{rmq}
La fonction $f: I \to \R$ admet pour limite $l$ lorsque $x$ tend vers $a$ si quel que soit le voisinage $\mathcal V$ de $l$ dans $\R$, l'ensemble d'arrivée de $f$, on peut trouver un voisinage $\mathcal U$ de $a$ dans $I$, l'ensemble de départ, tel que tous les éléments du voisinage $\mathcal U$ soient envoyés par $f$ dans $\mathcal V$ (c'est-à-dire: $f(\mathcal U) \subset \mathcal V$).
\end{rmq}

\begin{mybox} \begin{thm}[des valeurs intermédiaires]
Soit $f:[a;b] \to \R$, avec $a< b$,  une fonction continue $[a,b]$ et $k$ un nombre réel compris entre $f(a)$ et $f(b)$ alors l'équation
\[ f(x) = k \]
sur l'intervalle $[a;\,b]$, admet au moins une solution.
\end{thm} \end{mybox}

\begin{rmq}
Le théorème des valeurs intermédiaires nous dit entre autre que si l'on trace la courbe représentative d'une fonction {\red  continue} $f$ en partant du point de coordonnées $(a;~f(a))$ pour aller au point de coordonnées $(b,f(b))$ sans \og lever le crayon \fg{} alors quel que soit $k$ un nombre réel compris entre $f(a)$ et $f(b)$, la courbe coupe l'axe horizontale d'équation $y = k$. 
\begin{center} \begin{tikzpicture}[x=.8cm, y=.4cm] \begin{scriptsize}
\draw[->] (-4,0) -- (11,0) node[right]{$x$};
\draw[->] (0,-4) -- (0,5.5) node[right]{$y$};
\draw (0,0) node[below left]{$0$};
\draw[dashed] (-3,-2pt) node[below]{$a$} -- (-3,4) -- (2pt,4) node[right]{$f(a)$};
\draw[dashed] (10,2pt) node[above]{$b$} -- (10,-3) -- (-2pt,-3) node[left]{$f(b)$};
\draw[thick, color=black] (-3,4) .. controls +(30:5) and +(120:1) .. (4,1.5);
\draw[thick, color=black] (4,1.5) .. controls +(30:1) and +(180:.5) .. (5.5,3.2) node[above right]{$\mathcal C_f$};
\draw[thick, color=black] (5.5,3.2) .. controls +(0:2) and +(120:.5) .. (7.7,1);
\draw[thick, color=black] (7.7,1) .. controls +(-60:.5) and +(-120:2.5) .. (10,-3);
\draw[color=bleu, dash pattern=on 6pt off 2pt] (-4,1) -- (11,1) node[pos=.9,above]{$y = k$};
\draw (0,1) node[above left, color=vert2]{\small $\forall \, k$};
\draw[thick] (2pt,1) -- ++(-4pt,0);
\draw[line width=2pt, color=vert2] (0,-3) -- (0,4);
\draw[thick, dashed, color=rouge2] (7.7,-2pt) node[below]{\small $\exists \, x_0$} -- (7.7,1);
\draw[fill=rouge2] (7.7,1) circle (1.5pt);
\draw[fill=rouge2] (-3,4) circle (1.5pt);
\draw[fill=rouge2] (10,-3) circle (1.5pt);
\draw[line width=2pt, color=rouge2] (-3,0) --(10,0);
\end{scriptsize} \end{tikzpicture} \end{center}
\end{rmq}

\begin{mybox2} \begin{thm}[Accroissements Finis]
Soit $f$ une fonction continue sur $[a;b]$ et dérivable sur $]a;b[$ (avec $a < b$). Alors il existe $c\in ]a;b[$ tel que 
$$ f'(c) = \frac{ f(b) - f(a) }{b-a} $$
\end{thm} \end{mybox2}

\begin{center} \begin{tikzpicture}[x=2.2cm, y=1.2cm]
\draw[->] (-.4,0) -- (5,0) node[above]{$x$};
\draw[->] (0,-.5) -- (0,3.5) node[right]{$y$};
\draw[thick, color=vert2] (-.4,.1) .. controls +(60:.5) and +(-120:1) .. (1,1);
\draw[thick, color=vert2] (1,1) .. controls +(60:1) and +(-135:1) .. (1.8,2.5);
\draw[thick, color=vert2] (1.8, 2.5) .. controls +(45:1) and +(-170:.3) .. (3,3);
\draw[thick, color=vert2] (3,3) .. controls +(10:1) and +(-120:1) .. (4,2.6);
\def\a{1};
\def\b{3};
\def\c{1.8};
\begin{small}
\draw (\a,2pt) -- ++(0,-4pt) node[below]{$a$};
\draw (\b,2pt) -- ++(0,-4pt) node[below]{$b$};
\draw (\c,2pt) -- ++(0,-4pt) node[below]{$\exists c$};
\draw[dashed] (\a,0) -- (\a,1) node[above left]{$A$};
\draw[dashed] (\b,0) -- (\b,3) node[above left]{$B$};
\draw[dashed] (\c,0) -- (\c,2.5);
\draw[color=rouge2] (\a,1) ++(-1.2,-1.2)  -- ++(4.2,4.2) node[pos=.9, above, sloped]{$(AB)$};
\draw[<->, color=rouge2] (\c,2.5) ++(-1,-1) -- ++(2,2) node[pos=.9, above, sloped]{$T_c (f)$};
\end{small}
\draw (4,2.6) node[above]{$\mathcal C_f$};
\end{tikzpicture} \end{center}



\subsection{Suites}

Pour une suite d\'efinie par r\'ecurrence, donc de la forme $u_{n+1}=f(u_n)$ et $u_0=a$, on trace la courbe représentative de la fonction $f$ et la droite d'équation $y=x$. On place en abscisse le point $u_0=a$, puis 
\renewcommand{\labelitemi}{$\bullet$}
\begin{itemize}
\item on trace un segment verticale depuis ce point jusqu'à la courbe, pour obtenir le point de la courbe de coordonnées $(u_0,f(u_0))$;
\item on trace un segment horizontale depuis ce point jusqu'à la première bissectrice (d'équation $y = x$), ce point à pour abscisse $u_1$.
\end{itemize}
Puis, on recommence indéfiniment pour déterminer les termes suivants de la suite (voir l'illustration ci-dessous).

\begin{center} 
\begin{tikzpicture}[x=7cm,y=8cm] \begin{scriptsize}
\draw[domain=0:1, thick, color=rouge] plot (\x,{2*\x*(1- \x)});
\foreach \x in {0.8}
	\draw (\x,{2*\x*(1- \x)}) node[above right]{$\mathcal C_f$};
\draw[color=gris] (0,0) -- (.6,.6) node[pos=.9, above, sloped]{$y =x$};
\draw[->, thick] (-4pt,0) -- (1.01,0);
\draw[->, thick] (0,-4pt) -- (0,.6);
\draw (0,0) node[below left]{$0$};
\vprof{\draw (0.1,2pt) -- ++(0,-2pt) node[below]{$u_{0}$};
\draw[dashed] (0.1,0.0) -- (0.1,0.1);
\draw (0.1,0.1) -- (0.1,0.18) -- (0.18,0.18);
\draw (0.18,2pt) -- ++(0,-2pt) node[below]{$u_{1}$};
\draw[dashed] (0.18,0.0) -- (0.18,0.18);
\draw (0.18,0.18) -- (0.18,0.2952) -- (0.2952,0.2952);
\draw (0.2952,2pt) -- ++(0,-2pt) node[below]{$u_{2}$};
\draw[dashed] (0.2952,0.0) -- (0.2952,0.2952);
\draw (0.2952,0.2952) -- (0.2952,0.4161) -- (0.4161,0.4161);
\draw (0.4161,2pt) -- ++(0,-2pt) node[below]{$u_{3}$};
\draw[dashed] (0.4161,0.0) -- (0.4161,0.4161);
\draw (0.4161,0.4161) -- (0.4161,0.4859) -- (0.4859,0.4859);
\draw (0.4859,2pt) -- ++(0,-2pt) node[below]{$u_{4}$};
\draw[dashed] (0.4859,0.0) -- (0.4859,0.4859);
\draw (0.4859,0.4859) -- (0.4859,0.4996) -- (0.4996,0.4996);}
\end{scriptsize} \end{tikzpicture}  \quad
\begin{tikzpicture}[x=7cm,y=6cm] \begin{scriptsize}
\draw[domain=0:1, thick, color=rouge] plot (\x,{3/4*(\x-1)*(\x-1) + 1/4});
\foreach \x in {0.9}
	\draw (\x,{3/4*(\x-1)*(\x-1) + 1/4}) node[above right]{$\mathcal C_f$};
\draw[color=gris] (0,0) -- (1,1) node[pos=.9, above, sloped]{$y =x$};
\draw[->, thick] (-4pt,0) -- (1.05,0);
\draw[->, thick] (0,-4pt) -- (0,1.05);
\draw (0,0) node[below left]{$0$};
\vprof{\draw (0.1,0.1) -- (0.1,0.8575) -- (0.8575,0.8575);
\draw (0.8575,0.8575) -- (0.8575,0.2652) -- (0.2652,0.2652);
\draw (0.2652,0.2652) -- (0.2652,0.6549) -- (0.6549,0.6549);
\draw (0.6549,0.6549) -- (0.6549,0.3393) -- (0.3393,0.3393);
\draw (0.3393,0.3393) -- (0.3393,0.5774) -- (0.5774,0.5774);
\draw (0.5774,0.5774) -- (0.5774,0.384) -- (0.384,0.384);
\draw (0.384,0.384) -- (0.384,0.5346) -- (0.5346,0.5346);
\draw (0.5346,0.5346) -- (0.5346,0.4124) -- (0.4124,0.4124);
\draw (0.4124,0.4124) -- (0.4124,0.5089) -- (0.5089,0.5089);
\draw (0.5089,0.5089) -- (0.5089,0.4309) -- (0.4309,0.4309);
\draw (0.4309,0.4309) -- (0.4309,0.4929) -- (0.4929,0.4929);
\draw (0.4929,0.4929) -- (0.4929,0.4428) -- (0.4428,0.4428);
\draw (0.4428,0.4428) -- (0.4428,0.4828) -- (0.4828,0.4828);
\draw (0.4828,0.4828) -- (0.4828,0.4506) -- (0.4506,0.4506);
\draw (0.4506,0.4506) -- (0.4506,0.4764) -- (0.4764,0.4764);
\draw (0.4764,0.4764) -- (0.4764,0.4556) -- (0.4556,0.4556);
\draw (0.4556,0.4556) -- (0.4556,0.4723) -- (0.4723,0.4723);
\draw (0.4723,0.4723) -- (0.4723,0.4589) -- (0.4589,0.4589);
\draw (0.4589,0.4589) -- (0.4589,0.4696) -- (0.4696,0.4696);
\draw (0.4696,0.4696) -- (0.4696,0.461) -- (0.461,0.461);
\draw (0.1,2pt) -- ++(0,-2pt) node[below]{$u_{0}$};
\draw[dashed] (0.1,0.0) -- (0.1,0.1);
\draw (0.8575,2pt) -- ++(0,-2pt) node[below]{$u_{1}$};
\draw[dashed] (0.8575,0.0) -- (0.8575,0.8575);
\draw (0.2652,2pt) -- ++(0,-2pt) node[below]{$u_{2}$};
\draw[dashed] (0.2652,0.0) -- (0.2652,0.2652);
\draw (0.6549,2pt) -- ++(0,-2pt) node[below]{$u_{3}$};
\draw[dashed] (0.6549,0.0) -- (0.6549,0.6549);
\draw (0.3393,2pt) -- ++(0,-2pt) node[below]{$u_{4}$};
\draw[dashed] (0.3393,0.0) -- (0.3393,0.3393);
\draw (0.5774,2pt) -- ++(0,-2pt) node[below]{$u_{5}$};
\draw[dashed] (0.5774,0.0) -- (0.5774,0.5774);
\draw[dashed] (0.384,0.0) -- (0.384,0.384);}
\end{scriptsize} \end{tikzpicture} \end{center}

\begin{cadre} \begin{defi}
Une suite $(u_n)_{n \geq p}$ a pour {\red limite} le nombre {\red $\ell$ quand $n$ tend vers l'infini}, si quelle que soit la distance $\epsilon$ qu'on se fixe, il existe un rang $N$ à partir duquel tous les termes $u_n$ de la suite sont à une distance inférieure ou égale à $\epsilon$ du nombre $\ell$:
\begin{center} \begin{tikzpicture}[x=.8cm, y=.7cm] \begin{scriptsize}
\draw[->] (-.5,1) -- (14.5,1);
\draw[->] (0,.7) -- (0,7);
\draw[color=white,fill=vert!30, opacity=.3] (6,5.5) -- (14.5,5.5) -- (14.5,6.5) -- (6,6.5) -- (6,5.5);
\draw[thick] (1,1.25)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (2,2.25)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (3,2.75)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (4,3.25)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (5,4.25)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (6,5.59)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (7,5.75)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (8,6.25)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (9,5.8)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (10,6.3)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (11,6.2)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (12,6.15)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (13,6.1)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[thick] (14,5.95)-- ++(-2pt,-2pt) -- ++(4pt,4pt) ++(-0pt,-4pt) -- ++(-4pt,4pt);
\draw[color = vert2, dashed] (-2pt,6) node[left]{$\ell$} -- (14.5,6);
\draw[color = bleu] (-2pt,6.5) -- (14.5,6.5);
\draw[color = bleu] (-2pt,5.5) -- (14.5,5.5);
\draw[<->] (-30pt,5.5) -- (-30pt, 6) node[pos=.5,left]{$\epsilon$};
\draw[<->] (-30pt,6) -- (-30pt, 6.5) node[pos=.5,left]{$\epsilon$};
\draw[dashed] (6,6.5) -- (6,1) node[below]{$N$};
\draw (10,1) ++(0,2pt) -- ++(0,-5pt) node[below]{$n$};
\draw (8,1) ++(0,-3pt) node[below]{$\leq$};
\draw (1pt,6.3) -- (-3pt,6.3) node[left]{$u_n$};
\end{scriptsize} \end{tikzpicture} \end{center}

On dit que $(u_n)$ converge vers $l$ et on note $\limite u_n = l$.
\end{defi} \end{cadre}


\subsection{Probabilités}

Soit $A$ et $B$ deux parties (ou sous-populations) d'une population $E$.
\begin{center} \begin{tikzpicture}[scale=.7] \begin{scriptsize}
\clip (-2.5, -2.6) rectangle (3.1,2.6);
\draw[thick, color=rouge] (0,1) .. controls +(60:3) and +(-30:2) .. (0.2,-1) node[pos=.5, right]{$A$};
\draw[thick, color=bleu] (0,1) .. controls +(-70:1) and +(70:1) .. (0.2,-1);
\draw[thick, color=bleu] (0,1) .. controls +(110:4) and +(-110:4) .. (0.2,-1) node[pos=.5, left]{$B$};
\draw[thick, color=rouge] (0,1) .. controls +(-120:1) and +(150:1) .. (0.2,-1);
\draw[thick] (-2,-1) .. controls +(-60:3) and +(-100:4) .. (3,1) node[pos=.7, right]{$E$};
\draw[thick] (-2,-1) .. controls +(120:4) and +(80:3) .. (3,1);
\end{scriptsize} \end{tikzpicture} \end{center}
Par exemple, considérons la population $E$ des élèves d'une classe de lycée et désignons par $A$ ceux qui possèdent un smartphone et par $B$ ceux qui ont une calculatrice scientifique.

\begin{cadre} \begin{defi} \qquad 
\begin{enumerate}
\item L'{\red intersection}, notée $A \cap B$, est constituée des éléments qui sont dans $A$ {\red et} dans $B$:
\begin{center} \begin{tikzpicture}[scale=.9] \begin{scriptsize}
\clip (-2.5, -2.6) rectangle (4.5,2.6);
\draw[color=violet!80] (0,1) -- (0.2,-1);
\draw[thick, color=rouge] (0,1) .. controls +(60:3) and +(-30:2) .. (0.2,-1) node[pos=.5, right]{$A$};
\draw[thick, color=bleu, fill=violet!80] (0,1) .. controls +(-70:1) and +(70:1) .. (0.2,-1);
\draw[thick, color=bleu] (0,1) .. controls +(110:4) and +(-110:4) .. (0.2,-1) node[pos=.5, left]{$B$};
\draw[thick, color=rouge, fill=violet!80] (0,1) .. controls +(-120:1) and +(150:1) .. (0.2,-1);
\draw (0,0) -- (3,0)  node[right, color=violet!80]{\small $A \cap B$};
\draw[thick] (-2,-1) .. controls +(-60:3) and +(-100:4) .. (3,1) node[pos=.8, right]{$E$};
\draw[thick] (-2,-1) .. controls +(120:4) and +(80:3) .. (3,1);
\end{scriptsize} \end{tikzpicture} \end{center}
\item L'{\red union},  notée $A \cup B$, est constituée des éléments qui appartiennent à {\it au moins l'une} des deux parties $A$ {\red ou} $B$:
\begin{center} \begin{tikzpicture}[scale=.9] \begin{scriptsize}
\clip (-2.5, -2.6) rectangle (4.5,2.6);
\draw[thick, color=rouge, fill=violet!30] (0,1) .. controls +(60:3) and +(-30:2) .. (0.2,-1) node[pos=.5, right]{$A$};
\draw[color=violet!30] (0,1) -- (0.2,-1);
\draw[thick, color=bleu, fill=violet!30] (0,1) .. controls +(-70:1) and +(70:1) .. (0.2,-1);
\draw[thick, color=bleu, fill=violet!30] (0,1) .. controls +(110:4) and +(-110:4) .. (0.2,-1) node[pos=.5, left]{$B$};
\draw[thick, color=rouge, fill=violet!30] (0,1) .. controls +(-120:1) and +(150:1) .. (0.2,-1);
\draw (1,1) -- (3.2,1.5) node[right, color=violet!30]{\small $A \cup B$};
\draw[thick] (-2,-1) .. controls +(-60:3) and +(-100:4) .. (3,1) node[pos=.8, right]{$E$};
\draw[thick] (-2,-1) .. controls +(120:4) and +(80:3) .. (3,1);
\end{scriptsize} \end{tikzpicture} \end{center}
\end{enumerate}
\end{defi} \end{cadre}

Considérons maintenant $n$ événements $A_1$, $A_2$, \ldots, $A_n$ de probabilité non nulle, formant une partition de $\Omega$. C'est-à-dire que les événements $A_i$ sont deux à deux incompatibles et leur union est l'univers entier. Soit $B$ un événement, on a $B =  (B \cap A_1) \cup \ldots \cup (B \cap A_n)$ et l'union est disjointe. D'où
\[ \p(B) = \p(B \cap A_1) + \ldots \p(B \cap A_n)  \]
%{{{ Une partition: 
\begin{center} \begin{tikzpicture} \begin{scriptsize}
\coordinate (O) at (3.5,3.5);
\draw[fill=bleu2!10, color=bleu2!10] (O) circle (1.8);
\coordinate (A) at (4,6);
\coordinate (E) at (2.5,5.5);
\coordinate (F) at (1,4);
\coordinate (B) at (1.5,2);
\coordinate (C) at (4,1);
\coordinate (D) at (6,4.2);
\coordinate (I) at (2.75,4);
\draw[color=rouge2] (A) .. controls +(180:1) and +(10:1) .. (E);
\draw[color=rouge2] (E) .. controls +(190:1) and +(90:1) .. (F);
\draw[color=rouge2] (F) .. controls +(-90:1) and +(150:1) .. (B);
\draw[color=rouge2] (B) .. controls +(-30:1) and +(180:1) .. (C);
\draw[color=rouge2] (C) .. controls +(0:1) and +(-80:1) .. (D);
\draw[color=rouge2] (D) .. controls +(100:1) and +(0:1) .. (A);
\draw (A) -- (B);
\draw (I) .. controls +(120:.5) and +(-90:.5) .. (E);
\draw (I) .. controls +(180:.5) and +(-10:1) .. (F);
\draw (B) .. controls +(30:2) and +(150:1) .. (C);
\coordinate (G) at (4.5,2.5);
\coordinate (H) at (5,3);
\draw (C) .. controls +(90:.5) and +(-100:1) .. (G) .. controls +(80:.5) and +(-90:.5) .. (H) .. controls +(90:1) and +(200:1) .. (D);
\draw (I) .. controls +(-50:1) and +(170:1) .. (G);
\draw (I) .. controls +(50:1) and +(150:1) .. (H);
\end{scriptsize} 
\draw[<-, shift={(3.5,3.5)}, color=bleu2] (30:1.8) --++(30:1.2) node[right]{$B$};
\draw[color=rouge2] (B) ++(150:1) node{$\Omega$};
\draw (I) ++(-25:1) node{$\ldots$};
\draw (I) ++(80:1) node{$A_1$};
\draw (I) ++(150:1) node{$A_2$};
\draw (I) ++(210:1.2) node{$A_3$};
\draw (I) ++(-70:1.4) node{$A_5$};
\draw (I) ++(20:1.7) node{$A_{n-1}$};
\draw (B) ++(0:1) node{$A_4$};
\draw (G) ++(0:.6) node{$A_n$};
\end{tikzpicture} \end{center}
%}}}
En d'autres termes, une partition $A_1$, $A_2$, \ldots, $A_n$ est une manière de classer les issues possibles de l'expérience en $n$ catégories d'événements notées $A_1$ à $A_n$. La probabilité d'un événement $B$ s'obtient en additionnant les probabilités des événements des $B \cap A_i$, c'est-à-dire, en distinguant les différents cas.

\begin{cadre} \begin{defi}
Un {\red système complet} (ou exhaustif) d'événements est une famille $(A_i)_{i \in I}$ d'événements deux à deux incompatibles telle que $\displaystyle\bigcup_{i \in I} A_i=\Omega$.
\end{defi} \end{cadre}  

De la remarque précédente, on déduit

\begin{mybox2} \begin{thm}[Formule des probabilités totales]
Soit $n$ un entier et $(A_i)_{i \in \llbracket 1;n\rrbracket}$ un système complet d'événements de probabilités toutes non nulles. Pour tout événement $B$, on a 
$$\p(B)=\p(A_1)\p_{A_1}(B) + \ldots + \p(A_n) \p_{A_n}(B)$$
\end{thm} \end{mybox2}

On peut aussi illustrer la formule des probabilités totales à l'aide des arbres pondérés: Sur un {\red arbre pondéré de probabilités} (réalisé ci-dessous pour $n=4$ dans la figure \ref{18figure1}), une {\red branche} est représentée par un segment (portant une probabilité), un {\red noeud} est la jonction de deux ou plusieurs branches, et un {\red chemin} est une succession de branches allant du noeud initial de l'arbre à l'une de ses extrémités. 

Chaque chemin correspond à l'évènement intersection des événements figurant sur ce chemin (par exemple $A_2 \cap B$ pour le $3^e$ chemin de la figure \ref{18figure1}). 

\begin{center} \begin{figure}[h!] \begin{tikzpicture}[x=2cm, y=1cm]
\node (O) at (0,0) {};
\foreach \x in {1,4}{
	\node (A\x) at ($(1,5-2*\x)$) {$A_\x$};
	\draw (O) -- (A\x) node[pos=.5, above, sloped]{\tiny $\p(A_\x)$}; 

	\node (B\x) at ($(2,5.5-2*\x)$) {$B$};
	\draw (A\x) -- (B\x) node[pos=.5, above, sloped]{\tiny $\p_{A_\x}(B)$};
	
	\node (nB\x) at ($(2,4.5-2*\x)$) {$\bar B$};
	\draw (A\x) -- (nB\x) node[pos=.5, below, sloped]{\tiny $\p_{A_\x}(\bar B)$}; 
}

\draw (1, 4.3) node[right] {Arbre};
\draw (3.5, 4.3) node[right] {Événement};
\draw (5, 4.3) node[right] {Probabilité};


\foreach \x in{2}{
	\node (A\x) at ($(1,5-2*\x)$) {$A_\x$};
	\draw[color=rouge] (O) -- (A\x) node[pos=.5, above, sloped]{\tiny $\p(A_\x)$}; 
	\node (B\x) at ($(2,5.5-2*\x)$) {$B$};
	\draw[color=rouge] (A\x) -- (B\x) node[pos=.5, above, sloped]{\tiny $\p_{A_\x}(B)$};
	\node (nB\x) at ($(2,4.5-2*\x)$) {$\bar B$};
	\draw (A\x) -- (nB\x) node[pos=.5, below, sloped]{\tiny $\p_{A_\x}(\bar B)$}; 

	\draw[color=rouge] ($(3.5,5.5-2*\x)$) node[right]{$A_\x \cap B$};
	\draw[color=rouge] ($(5,5.5-2*\x)$) node[right]{$\p(A_\x \cap B) = \p(A_\x) \times \p(B \,|\, {A_\x} ) $};
}


\foreach \x in{3}{
	\node (A\x) at ($(1,5-2*\x)$) {$A_\x$};
	\draw[color=bleu] (O) -- (A\x) node[pos=.5, above, sloped]{\tiny $\p(A_\x)$}; 
	\node (B\x) at ($(2,5.5-2*\x)$) {$B$};
	\draw (A\x) -- (B\x) node[pos=.5, above, sloped]{\tiny $\p_{A_\x}(B)$};
	\node (nB\x) at ($(2,4.5-2*\x)$) {$\bar B$};
	\draw[color=bleu] (A\x) -- (nB\x) node[pos=.5, below, sloped]{\tiny $\p_{A_\x}(\bar B)$}; 

	\draw[color=bleu] ($(3.5,4.5-2*\x)$) node[right]{$A_\x \cap \bar B$};
	\draw[color=bleu] ($(5,4.5-2*\x)$) node[right]{$\p(A_\x \cap \bar B) = \p(A_\x) \times \p( \bar B  \,|\, A_\x ) $};
}
\end{tikzpicture} 
\caption{Système complet de 4 événements} 
\label{18figure1}
\end{figure} \end{center}

À partir de là, pour les calculs, on utilise les règles suivantes:
\begin{enumerate}
\item La {\it somme des probabilités} portées sur les branches issues d'un même noeud est égale à 1. 
\item La {\it probabilité de l'événement correspondant à un chemin} est le produit des probabilité portées sur ses branches.
\item La {\it probabilité d'un événement} est la somme des probabilités des événements correspondant aux chemins qui y aboutissent.
\end{enumerate}
Le premier point est une conséquence de la proposition \ref{18probacondi}, le second point de la formule des probabilités composées et le troisième point de la formule des probabilités totales.

Avec $n=4$, on retrouve:
\[ \p(B) = \p(A_1) \times \p(B \,|\, A_1) + \p(A_2) \times \p(B  \,|\, A_2 ) + \p(A_3) \times \p(B \,|\, A_3 ) + \p(A_4) \times \p(B \,|\, A_4  )  \]

La figure \ref{18figure2} représente un second exemple lorsque $n = 2$.
\begin{center} \begin{figure}[h!] \begin{tikzpicture}[x=2cm, y=1cm] \begin{small}
\node (O) at (0,0) {};
\foreach \x in{1}{
	\node (A\x) at ($(1,3-2*\x)$) {$A$};
	\draw[color=bleu] (O) -- (A\x) node[pos=.5, above, sloped]{\tiny $\p(A)$}; 
	\node (B\x) at ($(2,3.5-2*\x)$) {$B$};
	\draw[color=bleu] (A\x) -- (B\x) node[pos=.5, above, sloped]{\tiny $p_{A}(B)$};
	\node (nB\x) at ($(2,2.5-2*\x)$) {$\bar B$};
	\draw (A\x) -- (nB\x) node[pos=.5, below, sloped]{\tiny $p_{A}(\bar B)$}; 
	\draw[color=bleu] ($(3.5,3.5-2*\x)$) node[right]{$A\cap B$};
	\draw[color=bleu] ($(5,3.5-2*\x)$) node[right]{$\p(A \cap B) = \p(A) \times \p(B \,|\, A ) $};
}

\foreach \x in{2}{
	\node (A\x) at ($(1,3-2*\x)$) {$\bar A$};
	\draw[color=bleu] (O) -- (A\x) node[pos=.5, above, sloped]{\tiny $\p(\bar A)$}; 
	\node (B\x) at ($(2,3.5-2*\x)$) {$B$};
	\draw[color=bleu] (A\x) -- (B\x) node[pos=.5, above, sloped]{\tiny $p_{\bar A}(B)$};
	\node (nB\x) at ($(2,2.5-2*\x)$) {$\bar B$};
	\draw (A\x) -- (nB\x) node[pos=.5, below, sloped]{\tiny $p_{\bar A}(\bar B)$}; 

	\draw[color=bleu] ($(3.5,3.5-2*\x)$) node[right]{$\bar A \cap B$};
	\draw[color=bleu] ($(5,3.5-2*\x)$) node[right]{$\p(\bar A \cap B) = \p(\bar A) \times \p(B \,|\, {\bar A}) $};
}
\draw (2.5, -2.3) node[right]{$B = (A \cap B) \cup (\bar A \cap B)$};
\draw (5, -2.3) node[right]{$\p(B) = \p(A \cap B) + \p(\bar A \cap B)$};

\end{small} \end{tikzpicture} 
\caption{Formule des probabilités totales (version 1)}
\label{18figure2}
\end{figure} \end{center}

\subsection{Géométrie}

\begin{mybox} \begin{propriété}[Valeurs remarquables] \qquad
\begin{center} \begin{tabular}{|*6{c|}} \hline
$\alpha $ & $0$ & $\frac \pi 6$ & $\frac \pi 4 $ & $\frac \pi 3$ & $\frac \pi 2$ \\ \hline
$\cos(\alpha)$ & $1$ & $\frac {\sqrt 3} 2$ & $\frac {\sqrt 2} 2$ & $\frac 1 2$ & $0$ \\ \hline
$\sin(\alpha)$ & $0$ & $\frac 1 2$ & $\frac {\sqrt 2} 2$ & $\frac {\sqrt 3} 2$ & $1$ \\ \hline
\end{tabular} \end{center}
\begin{center} \begin{tikzpicture}[x=4.5cm, y=4.5cm] 
\coordinate (I) at (0:1);
\draw[thick] (0,0) circle (1);
\draw[->, thick] (20:1.3) arc (20:40:1.3);
\draw[->] (-1.1,0) -- (1.2,0) node[right]{$\cos(\textcolor{bleu}{\alpha})$};
\draw[->] (0,-1.1) -- (0,1.2) node[above]{$\sin(\textcolor{bleu}{\alpha})$};
\draw (0,0) node[below left]{$O$};
%\draw (0,1) node[below left]{$J$};
%\draw[fill=black] (I) circle (.5pt) node[above left]{$I$};
\coordinate (A1) at (30:1);
\coordinate (A2) at (45:1);
\coordinate (A3) at (60:1);
\coordinate (A4) at (90:1);
\coordinate (A5) at (120:1);
\coordinate (A6) at (135:1);
\coordinate (A7) at (150:1);
\coordinate (A8) at (180:1);
\coordinate (B1) at (-30:1);
\coordinate (B2) at (-45:1);
\coordinate (B3) at (-60:1);
\coordinate (B4) at (-90:1);
\coordinate (B5) at (-120:1);
\coordinate (B6) at (-135:1);
\coordinate (B7) at (-150:1);
\draw (0:1) ++(180:2pt) -- ++(0:4pt) ++(45:7pt) node{\textcolor{bleu}{ $0$}}; 
\draw (A1) ++(210:2pt) -- ++(30:4pt) ++(30:7pt) node{\textcolor{bleu}{ $\frac{\pi}6$}}; 
\draw (A3) ++(240:2pt) -- ++(60:4pt) ++(60:7pt) node{\textcolor{bleu}{ $\frac{\pi}3$}}; 
\draw (A2) ++(225:2pt) -- ++(45:4pt) ++(45:7pt) node{\textcolor{bleu}{ $\frac{\pi}4$}}; 
\draw (A4) ++(270:2pt) -- ++(90:4pt) ++(150:7pt) node{\textcolor{bleu}{ $\frac{\pi}2$}}; 
\draw (A5) ++(300:2pt) -- ++(120:4pt) ++(120:7pt) node{\textcolor{bleu}{ $\frac{2\pi}3$}}; 
\draw (A6) ++(315:2pt) -- ++(135:4pt) ++(135:7pt) node{\textcolor{bleu}{ $\frac{3\pi}4$}}; 
\draw (A7) ++(330:2pt) -- ++(150:4pt) ++(150:7pt) node{\textcolor{bleu}{ $\frac{5\pi}6$}}; 
\draw (A8) ++(0:2pt) -- ++(180:4pt) ++(150:7pt) node{\textcolor{bleu}{ $\pi$}}; 
\draw (B1) ++(-210:2pt) -- ++(-30:4pt) ++(-30:9pt) node{\textcolor{bleu}{ $\frac{-\pi}6$}}; 
\draw (B3) ++(-240:2pt) -- ++(-60:4pt) ++(-60:9pt) node{\textcolor{bleu}{ $\frac{-\pi}3$}}; 
\draw (B2) ++(-225:2pt) -- ++(-45:4pt) ++(-45:9pt) node{\textcolor{bleu}{ $\frac{-\pi}4$}}; 
\draw (B4) ++(-270:2pt) -- ++(-90:4pt) ++(-150:9pt) node{\textcolor{bleu}{ $\frac{-\pi}2$}}; 
\draw (B5) ++(-300:2pt) -- ++(-120:4pt) ++(-60:9pt) node{\textcolor{bleu}{ $\frac{-2\pi}3$}}; 
\draw (B6) ++(-315:2pt) -- ++(-135:4pt) ++(-135:9pt) node{\textcolor{bleu}{ $\frac{-3\pi}4$}}; 
\draw (B7) ++(-330:2pt) -- ++(-150:4pt) ++(-150:9pt) node{\textcolor{bleu}{ $\frac{-5\pi}6$}}; 
\draw[loosely dashed] (A2) -- (B6);
\draw[loosely dashed] (B2) -- (A6);
\foreach \x in {1,2,...,7}
	\draw[dashed] (A\x) -- (B\x);
\draw[dashed] (A1) -- (A7);
\draw[dashed] (A2) -- (A6);
\draw[dashed] (A3) -- (A5);
\draw[dashed] (B1) -- (B7);
\draw[dashed] (B2) -- (B6);
\draw[dashed] (B3) -- (B5);
\draw (.5,0) ++(5pt,-9pt) node{$\frac 1 2$};
\draw (-.5,0) ++(7pt,-9pt) node{$\frac {-1} 2$};
\draw (0,.5) ++(-5pt,-9pt) node{$\frac 1 2$};
\draw (0,-.5) ++(-7pt,-9pt) node{$\frac {—1} 2$};
\draw (.70711,0) ++(7pt,-9pt) node{$\frac {\sqrt 2} 2$};
\draw (-.70711,0) ++(8pt,-9pt) node{$\frac {-\sqrt 2} 2$};
\draw (0,.70711) ++(-7pt,-9pt) node{$\frac {\sqrt 2} 2$};
\draw (0,-.70711) ++(-10pt,-9pt) node{$\frac {—\sqrt 2} 2$};
\draw (.86603,0) ++(7pt,-9pt) node{$\frac {\sqrt 3} 2$};
\draw (-.86603,0) ++(8pt,-9pt) node{$\frac {-\sqrt 3} 2$};
\draw (0,.86603) ++(-7pt,-9pt) node{$\frac {\sqrt 3} 2$};
\draw (0,-.86603) ++(-10pt,-9pt) node{$\frac {—\sqrt 3} 2$};
\end{tikzpicture} \end{center}
\end{propriété} \end{mybox}

Soit $\alpha \in [0;~\frac \pi 2]$. Dans la figure ci-dessous, on a représenté l'angle $\alpha$ sur le cercle trigonométrique ainsi que son cosinus, son sinus et les angles $-\alpha$, $\alpha + \frac \pi 2$, $\alpha + \pi$. Il n'est pas bien difficile de vérifier que les triangles 1, 2, 3 et 4 en bleu sont rectangles et ont les mêmes dimensions (en particulier l'hypoténuse vaut 1). Ainsi, en revenant à la définition du cosinus et du sinus, l'abscisse et l'ordonnée du point sur le cercle trigonométrique et en comparant les triangles entre eux, on note les relations suivantes: \\
\renewcommand{\labelitemi}{$\bullet$}
\begin{itemize}
\item des triangles 1 et 2: $\cos(-\alpha) = \cos(\alpha)$ et $\sin(-\alpha) = - \sin(\alpha)$.
\item des triangles 1 et 3: $\cos(\alpha + \frac \pi 2) = -\sin(\alpha)$ et $\sin(\alpha + \frac \pi 2) = \cos(\alpha)$.
\item des triangles 1 et 4: $\cos(\alpha + \pi) = - \cos(\alpha)$ et $\sin(\alpha + \pi) = - \sin(\alpha)$.
\end{itemize}
\begin{center} \begin{tikzpicture}[x=3.5cm, y=3.5cm] \begin{scriptsize}
\draw[->] (-1.1,0) -- (1.3,0) node[right]{$x$};
\draw[->] (0,-1.1) -- (0,1.3) node[right]{$y$};
\draw (0,0) node[below left]{$O$};
\draw (-1,0) node[below left]{$-1$};
\draw (1,0) node[below right]{$1$};
\draw (0,-1) node[above left]{$-1$};
\draw (0,1) node[above left]{$1$};
\draw[thick] (-1,2pt) -- ++(0,-4pt);
\draw[thick] (1,2pt) -- ++(0,-4pt);
\draw[thick] (-2pt,-1) -- ++(4pt,0);
\draw[thick] (-2pt,1) -- ++(4pt,0);
\draw (0,0) circle (1);
\def\x{0.87631} % angle = 28.8° = 4 pi / 25
\def\y{0.48175}
\draw[dashed] (\x,-2pt) node[below]{$\cos(\alpha)$} -- (\x,\y) node[above right]{$\alpha$} -- (-2pt,\y) node[left]{$\sin(\alpha)$};
\draw[color=bleu] (0,0) -- (\x,0) -- (\x,\y) -- (0,0);
\draw[color=bleu] (0,0) -- (\x,0) -- (\x,-\y) -- (0,0);
\draw[color=bleu] (0,0) -- (-\x,0)  -- (-\x,-\y) -- (0,0);
\draw[color=bleu] (0,0) -- (0,\x) -- (-\y,\x) -- (0,0);
\draw[dashed] (-\y,\x) -- (-\y,-2pt);
\draw[dashed] (-\x,-\y) -- (2pt,-\y);
\draw (-\y,\x) node[above left]{$\alpha + \frac \pi 2$};
\draw (-\x,-\y) node[below left]{$\alpha + \pi$};
\draw (\x,-\y) node[below right]{$-\alpha$};
\node[draw, circle] (A) at (0.6*\x,0.3*\y) {1};
\node[draw, circle] (B) at (0.6*\x,-0.3*\y) {2};
\node[draw, circle] (C) at (-0.3*\y,0.7*\x) {3};
\node[draw, circle] (D) at (-0.6*\x,-0.3*\y) {4};
\end{scriptsize} \end{tikzpicture} \end{center}
Ces relations restent vraies quel que soit $\alpha$.

\begin{mybox} \begin{propriété} \label{formulescosetsin} Pour tout réel $\alpha$, on a
\begin{enumerate}
\item $\cos( - \alpha) = \cos( \alpha )$ et $\sin(-\alpha) = - \sin(\alpha)$; \hfill (le cosinus est pair et le sinus impair)
\item $\cos( \pi + \alpha) =  - \cos( \alpha )$ et $\sin(\pi + \alpha) = - \sin(\alpha)$; 
\item $\cos( \pi - \alpha) =  - \cos( \alpha )$ et $\sin(\pi - \alpha) = \sin(\alpha)$;
\item $\cos( \frac \pi 2 + \alpha) =  - \sin( \alpha )$ et $\sin( \frac \pi 2  + \alpha) = \cos(\alpha)$;
\item $\cos( \frac \pi 2 - \alpha) =  \sin( \alpha )$ et $\sin( \frac \pi 2 - \alpha) = \cos(\alpha)$.
\end{enumerate}
\end{propriété} \end{mybox}

\begin{proof}
Les relations 1, 2 et 4. ont été vu avant. \\
Soit $\alpha$ un nombre réel, \\
{\it 3.} Si on applique les relations 2. et 1. avec le nombre réel $-\alpha$, on a
\[ \cos(\pi - \alpha) = \cos(\pi + (-\alpha)) \overset{2.} = - \cos(-\alpha) \overset{1.} = - \cos(\alpha) \]
De même, on montre la seconde identité. \\
{\it 5.} Nous allons appliquer les identités 4. et 1.,
\[ \cos(\frac \pi 2 - \alpha) = \cos(\frac \pi 2 + (-\alpha)) \overset{4.} =  - \sin(-\alpha) \overset{1.} =  \sin(\alpha) \]
et de même, on déduit la seconde identité.
\end{proof}

\begin{cadre} \begin{defi}
La fonction cosinus, notée $\cos$, est la fonction définie sur $\R$ par $\cos: x \mapsto \cos(x)$.
La fonction sinus, notée $\sin$, est la fonction définie sur $\R$ par $\sin: x \mapsto \sin(x)$.
\end{defi} \end{cadre}

\begin{mybox} \begin{propriété}[tableau de variations] \qquad
\begin{center}
\begin{tikzpicture}[y=.8cm]
\tkzTabInit[lgt=1.7, espcl=1.8]{$x$ /0.75,$\cos(x)$ /1.2}{$0$, $\pi$, $2\pi$}
%\tkzTabLine{,-,z,+,}
\tkzTabVar{+/$1$,-/$-1$, +/$1$}
\end{tikzpicture}
\; \begin{tikzpicture}[y=.8cm]
\tkzTabInit[lgt=1.6, espcl=1.8]{$x$ /0.75,$\sin(x)$ /1.2}{$0$, $\frac{\pi}{2}$, $\pi$, $\frac{3\pi}{2}$, $2\pi$}
%\tkzTabLine{,+,z,,-,,z,+,}
\tkzTabVar{-/$0$,+/$1$,R/,-/$-1$, +/$0$}
\tkzTabVal{1}{5}{0.5}{}{0}
\end{tikzpicture}
\end{center}

\begin{center}
{\shorthandoff{:}
\begin{tikzpicture}[x=1cm, y=1cm]
\draw[->] (-7,0) -- (7,0) node[right]{$x$};
\draw[->] (0,-1.5) -- (0,1.5) node[right]{$y$};
\draw (0,0) node[below right]{{\tiny $O$}};
\def\pii{3.14159}
\draw[color=white] (1.5*\pii,2pt) -- ++(45:1) node[pos=.6, sloped, above, color=black]{$y = \cos(x)$};
\foreach \x/\n in {3.14159/\pi,6.2832/2\pi}
	\draw (\x,2pt) -- ++(0,-4pt) node[below]{$\n$};
\foreach \x in {-2*\pii,-1.5*\pii,-\pii,-.5*\pii,0.5*\pii,1.5*\pii}
	\draw (\x,2pt) -- ++(0,-4pt);
\foreach \y in {-1,1}
	\draw (2pt,\y) -- ++(-4pt,0) node[left]{$\y$};
\draw[domain=-7:7,samples=100,smooth,thick, color=rouge2] plot(\x,{cos(\x r)});
\end{tikzpicture}
}
{\shorthandoff{:}
\begin{tikzpicture}[x=1cm, y=1cm]
\draw[->] (-7,0) -- (7,0) node[right]{$x$};
\draw[->] (0,-1.5) -- (0,1.5) node[right]{$y$};
\draw (0,0) node[below right]{{\tiny $O$}};
\def\pii{3.14159}
\draw[color=white] (.7*\pii,1) ++(0,2pt) -- ++(-45:1) node[pos=.8, sloped, above, color=black]{$y = \sin(x)$};
\foreach \x/\n in {3.14159/\pi,6.2832/2\pi}
	\draw (\x,2pt) -- ++(0,-4pt) node[below]{$\n$};
\foreach \x in {-2*\pii,-1.5*\pii,-\pii,-.5*\pii,0.5*\pii,1.5*\pii}
	\draw (\x,2pt) -- ++(0,-4pt);
\foreach \y in {-1,1}
	\draw (2pt,\y) -- ++(-4pt,0) node[left]{$\y$};
\draw[domain=-7:7,samples=100,smooth,thick, color=rouge2] plot(\x,{sin(\x r)});
\draw[dashed] (-1.5,-1.5) -- (1.5,1.5);
\end{tikzpicture}

}
\end{center}
\end{propriété} \end{mybox}

\begin{cadre} \begin{defi}
Soit $\vect{u}$, $\vect{v}$ et $\vect{w}$ trois vecteurs. Le {\red produit mixte} de $\vect{u}$, $\vect{v}$ et $\vect{w}$ est le nombre $$[\vect{u},\vect{v},\vect{w}]=(\vect{u}\wedge\vect{v})\cdot \vect{w}$$
\end{defi} \end{cadre}

Soit $\vect u,\; \vect v$ et $\vect w$ trois vecteurs comme représentés ci-dessous
\begin{center} \begin{tikzpicture}[x=.7cm, y=.6cm] \begin{scriptsize}
\draw[fill=gris!20, color=gris!20] (0,0) -- ++(9,0) -- ++(3,1) -- ++(-9,0) -- cycle;
\draw[->, thick, color=rouge2] (0,0) -- ++(9,0) node[pos=.5, color=black, below]{$\vect u$};
\draw[->, dashed, thick, color=rouge2] (0,0) -- ++(3,1) node[pos=.7, color=black, above left]{$\vect v$};
\draw[->, thick, color=rouge2] (0,0) -- ++(1,4) node[pos=.7, color=black, right]{$\vect w$};
\draw (9,0) -- ++(1,4);
\draw (1,4) -- ++(9,0) -- ++(3,1) -- ++(-9,0) -- ++(-3,-1);
\draw (9,0) -- ++(3,1) -- ++(1,4);
\draw[dashed] (3,1) -- ++(9,0);
\draw[dashed] (3,1) -- ++(1,4);
\draw[color=bleu2, thick, ->] (0,0) -- (0,6) node[color=black, pos=.8, left]{$\vect u \wedge \vect v$};
\draw[dashed] (0,0) -- ++(-2,0);
\draw[dashed] (1,4) -- ++(-3,0);
\draw[<->] (-1.5,0) -- (-1.5,4) node[pos=.5, left]{$h$};
\draw (0,0) ++(75.96:1) arc (75.96:90:1);
\draw (82:1.2) node[above]{$\theta$};
\end{scriptsize} \end{tikzpicture} \end{center}
L'aire du parallélogramme engendré par $\vect u$ et $\vect v$ est $\|\vect u\| \|\vect v\||\sin(\vect u,\vect v)| = \|\vect u \wedge \vect v\|$ par définition du produit vectoriel. La hauteur du parallélépipède est 
\[ h = \|\vect w\| \cos(\theta) = \|\vect w\| |\cos(\vect u \wedge \vect v,\vect w)|  \]
De plus, le volume $V$ du parallélépipède engendré par les trois vecteurs est égal à l'aire de la base fois la hauteur, d'où
\[ V = \|\vect u \wedge \vect v\| \times h = 
 \|\vect u \wedge \vect v\| \|\vect w\| |\cos(\vect u \wedge \vect v,\vect w)| = | (\vect u \wedge \vect v) \cdot \vect w | = |[\vect u,\vect v, \vect w]|
 \]

\begin{thm}
Soit $\vect u,\; \vect v$ et $\vect w$ trois vecteurs. Le volume du parallélépipède formé par les trois vecteurs est égale à la valeur absolue du produit mixte des trois vecteurs.
\end{thm}


\end{document}
