<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\langchapter{Trigonométrie}{Trigonometry}
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\langsection{Cercle unitaire}{Unit circle}
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Le cercle unitaire est un cercle de centre $(0,0)$ et de rayon 1.

$\forall x \in \R, \cos^2 x + \sin^2 x = 1$

\subsection{cos}
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Formule d'Euler

$\forall \theta \in \R, cos(\theta) = \frac{e^{i\theta} + e^{-i \theta}}{2}$

$\cos 0 = 1$

$\cos \frac{\pi}{2} = 0$

$\cos \pi = -1$

$\cos(-\frac{\pi}{2}) = 0$

$\cos(\frac{\pi}{2} + t) = -\sin(t)$

$\cos(\pi + t) = -\cos(t)$

$\cos(\pi - t) = -\cos(t)$

$\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$

$\cos\frac{\pi}{3} = \frac{1}{2}$

$\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$

$\forall (a,b) \in \R$

$\cos(a + b) = \cos a \cos b + \sin a \sin b$

$\cos(a - b) = \cos a \cos b - \sin a \sin b$

$\cos a + \cos b = 2 \cos(\frac{a + b}{2}) \cos(\frac{a - b}{2} )$

$\cos a - \cos b = -2 \sin(\frac{a + b}{2}) \sin(\frac{a - b}{2} )$

$\cos a \cos b = \frac{\cos(a + b) + \cos(a - b)}{2}$

$\forall t \in \R, \cos(\frac{\pi}{2} - t) = \sin(t)$

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$

$\frac{d}{dx} \cos x = -\sin x$

$\forall x \in [-1, 1], \cos(\arcsin(x)) = \sqrt{1-x^{2}}$

$\forall x \in [-1, 1], \cos(\arccos(x)) = x$

$\forall x\in\R, \cos^2 x = \frac{1 + \cos(2x)}{2}$

\subsection{sin}
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Formule d'Euler

$\forall \theta \in \R, sin(\theta) = \frac{e^{i\theta} - e^{-i \theta}}{2i}$

$\sin 0 = 0$

$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$

$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$

$\sin \frac{\pi}{6} = \frac{1}{2}$

$\sin \frac{\pi}{2} = 1$

$\sin(\frac{\pi}{2} + t) = -\cos(t)$

$\sin(\pi - t) = \sin(t)$

$\sin(\pi + t) = -\sin(t)$

$\sin(\frac{\pi}{2} - t) = \cos(t)$

$\forall (a,b) \in \R$

$\sin(a + b) = \sin a \cos b + \sin b \cos a$

$\sin(a - b) = \sin a \cos b - \sin b \cos a$

$\sin a - \sin b = 2 \cos (\frac{a+b}{2}) \sin (\frac{a-b}{2})$

$\sin a + \sin b = 2 \sin (\frac{a+b}{2}) \cos (\frac{a-b}{2})$

$\sin a\sin b = \frac{\cos(a - b) - \cos(a + b)}{2}$

$\sin a \cos b = \frac{\sin(a - b) - \sin(a + b)}{2}$

$\lim\limits_{t \to 0} \frac{\sin t}{t} = 1$

$\frac{d}{dx} \sin x = \cos x$

$\forall x \in [-1, 1], \sin(\arcsin(x)) = x$

$\forall x \in [-1, 1], \sin(\arccos(x)) = \sqrt{1-x^{2}}$

$\forall x\in\R, \sin^2 x = \frac{1 - \cos(2x)}{2}$

\subsection{tan}
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$\tan 0 = 0$

$\tan \frac{\pi}{3} = \sqrt{3}$

$\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$

$\tan \frac{\pi}{4} = 1$

$\tan(\frac{\pi}{2} - x) = \frac{1}{\tan x}$

$\tan(\frac{\pi}{2} + x) = -\frac{1}{\tan x}$

$\tan(a + b) = \frac{\tan(a) + \tan(b)}{1- \tan(a)\tan(b)}$

$\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)}$

\subsection{Combinaisons}
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$\forall (a,b) \in \R$

$\sin a \cos b = \frac{\sin(a + b) + \sin(a - b)}{2}$

\langsection{Fonctions hyperboliques}{Hyperbolic functions}

\subsection{cosh}

$cosh\ x = \frac{e^x + e^{-x}}{2} = \frac{e^{2x} + 1}{2e^x} = \frac{1 + e^{-2x}}{2e^{-x}}$

\subsection{sinh}

$sinh\ x = \frac{e^x - e^{-x}}{2} = \frac{e^{2x} - 1}{2e^x} = \frac{1 - e^{-2x}}{2e^{-x}}$

\subsection{tanh}

$tanh\ x = \frac{sinh\ x}{cosh\ x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x} - 1}{e^{2x} + 1}$
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