Trigonometrie : formules

Un peu de géométrie

Périodicité et symétrie :
$\begin{array}[]{rcl}\cos(x)&=&\cos(x+2k\pi)\\ \sin(x)&=&\sin(x+2k\pi)\\ \tan(x)&=&\tan(x+k\pi)\\ \cos(-x)&=&\cos(x)\\ \sin(-x)&=&-\sin(x)\\ \tan(-x)&=&-\tan(x)\\ \cos(x+\frac{\pi}{2})&=&-\sin(x)\\ \sin(x+\frac{\pi}{2})&=&\cos(x)\\ \cos(\frac{\pi}{2}-x)&=&\sin(x)\\ \sin(\frac{\pi}{2}-x)&=&\cos(x)\\ \cos(x+\pi)&=&-\cos(x)\\ \sin(x+\pi)&=&-\sin(x)\\ \cos(\pi-x)&=&-\cos(x)\\ \sin(\pi-x)&=&\sin(x)\\ \end{array}$

Additions et soustractions des angles :
$\begin{array}[]{rcl}\cos(a+b)&=&\cos(a)\cos(b)-\sin(a)\sin(b)\\ \cos(a-b)&=&\cos(a)\cos(b)+\sin(a)\sin(b)\\ \sin(a+b)&=&\sin(a)\cos(b)+\cos(a)\sin(b)\\ \sin(a-b)&=&\sin(a)\cos(b)-\cos(a)\sin(b)\\ \tan(a+b)&=&\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\\ \tan(a-b)&=&\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}\\ \cos(2a)&=&2\cos^{2}(a)-1=1-2\sin^{2}(a)\\ \sin(2a)&=&2\cos(a)\sin(a)\\ \tan(2a)&=&\frac{2\tan(a)}{1-\tan^{2}(a)}\\ \end{array}$

Additiond et produits des fonctions :
$\begin{array}[]{rcl}\cos(a)\cos(b)&=&\frac{1}{2}\left(\cos(a+b)+\cos(a-b)% \right)\\ \sin(a)\sin(b)&=&\frac{1}{2}\left(\cos(a-b)-\cos(a+b)\right)\\ \sin(a)\cos(b)&=&\frac{1}{2}\left(\sin(a+b)+\sin(a-b)\right)\\ \cos^{2}(a)&=&\frac{1+\cos(2a)}{2}\\ \sin^{2}(a)&=&\frac{1-\cos(2a)}{2}\\ \end{array}$
$\begin{array}[]{rcl}\cos(a)+\cos(b)&=&2\cos\left(\frac{a+b}{2}\right)\cos\left% (\frac{a-b}{2}\right)\\ \cos(a)-\cos(b)&=&-2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}% \right)\\ \sin(a)+\sin(b)&=&2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right% )\\ \sin(a)-\sin(b)&=&2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right% )\\ \end{array}$

Fonctions réciproques :
$\begin{array}[]{rcl}\arcsin(-x)&=&-\arcsin(x)\\ \arccos(x)&=&\frac{\pi}{2}-\arcsin(x)\\ \arctan(x)&=&\arcsin\left(\frac{x}{\sqrt{x^{2}+1}}\right)\\ \end{array}$

Relations entre fonctions et fonctions réciproques :
$\begin{array}[]{rcl}\cos(\arcsin(x))&=&\sqrt{1-x^{2}}\\ \cos(\arctan(x))&=&\frac{1}{\sqrt{1+x^{2}}}\\ \sin(\arccos(x))&=&\sqrt{1-x^{2}}\\ \sin(\arctan(x))&=&\frac{x}{\sqrt{1+x^{2}}}\\ \tan(\arccos(x))&=&\frac{\sqrt{1-x^{2}}}{x}\\ \tan(\arcsin(x))&=&\frac{x}{\sqrt{1-x^{2}}}\\ \end{array}$

Avec cette liste :

\begin{array}[]{lcl}\cos(-x)&=&\cos(x)\\ \sin(-x)&=&-\sin(x)\\ \cos^{2}(x)+\sin^{2}(x)&=&1\\ \tan(x)&=&\frac{\sin(x)}{\cos(x)}\\ \cos(a+b)&=&\cos(a)\cos(b)-\sin(a)\sin(b)\\ \sin(a+b)&=&\sin(a)\cos(b)+\cos(a)\sin(b)\\ \arctan(x)&=&\arcsin\left(\frac{x}{\sqrt{x^{2}+1}}\right)\\ \end{array}

$\displaystyle\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$
$\displaystyle\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)$
$\displaystyle\cos(2a)=2\cos^{2}(a)-1=1-2\sin^{2}(a)$
$\displaystyle\sin(2a)=2\cos(a)\sin(a)$
$\displaystyle\cos(a)\cos(b)=\frac{1}{2}\left(\cos(a+b)+\cos(a-b)\right)$
$\displaystyle\sin(a)\sin(b)=\frac{1}{2}\left(\cos(a-b)-\cos(a+b)\right)$
$\displaystyle\sin(a)\cos(b)=\frac{1}{2}\left(\sin(a+b)+\sin(a-b)\right)$
$\displaystyle\cos(a)+\cos(b)=2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-% b}{2}\right)$
$\displaystyle\cos(a)-\cos(b)=-2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a% -b}{2}\right)$

$\displaystyle\sin(a)+\sin(b)=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-% b}{2}\right)$
$\displaystyle\sin(a)-\sin(b)=2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-% b}{2}\right)$
$\displaystyle\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$
$\displaystyle\cos(\arcsin(x))=\sqrt{1-x^{2}}$
$\displaystyle\cos(\arctan(x))=\frac{1}{\sqrt{1+x^{2}}}$
$\displaystyle\sin(\arctan(x))=\frac{x}{\sqrt{1+x^{2}}}$
$\displaystyle\sin(\arccos(x))=\sqrt{1-x^{2}}$
$\displaystyle\tan(\arccos(x))=\frac{\sqrt{1-x^{2}}}{x}$
$\displaystyle\tan(\arcsin(x))=\frac{x}{\sqrt{1-x^{2}}}$

Les Principales formules

Les formules