Additionstheoreme

$\displaystyle \sin ( x \pm y )$	$\displaystyle =$	$\displaystyle \sin x \; \cos y \pm \sin y \; \cos x$
$\displaystyle \cos ( x \pm y )$	$\displaystyle =$	$\displaystyle \cos x \; \cos y \mp \sin x \; \sin y$
$\displaystyle \tan ( x + y )$	$\displaystyle =$	$\displaystyle \frac{ \tan x + \tan y }{ 1 - \tan x \; \tan y } = \frac{ \sin (x + y) }{ \cos (x + y) }$
$\displaystyle \tan ( x - y )$	$\displaystyle =$	$\displaystyle \frac{ \tan x - \tan y }{ 1 + \tan x \; \tan y } = \frac{ \sin (x - y) }{ \cos (x - y) }$
$\displaystyle \cot \left( x+y\right)$	$\displaystyle =$	$\displaystyle \frac{\cot x\cot y-1}{\cot x+\cot y} = \frac{ \cos (x + y) }{ \sin (x + y) }$
$\displaystyle \sin ( x \pm y ) \cdot \sin ( x - y )$	$\displaystyle =$	$\displaystyle \cos ^2 y \mp \cos^2 x$
$\displaystyle \cos ( x + y ) \cdot \cos ( x - y )$	$\displaystyle =$	$\displaystyle \cos ^2 y - \sin^2 x$

$\displaystyle \arcsin x \pm \arcsin y$	$\displaystyle =$	$\displaystyle \arcsin\left(x\sqrt{1-y^2}\pm y\sqrt{1-x^2}\right)$
$\displaystyle \arccos x \pm \arccos y$	$\displaystyle =$	$\displaystyle \arccos\left(xy\mp \sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right)$
$\displaystyle \arctan x \pm \arctan y$	$\displaystyle =$	$\displaystyle \arctan\left(\frac{x\pm y}{1\mp xy}\right)$
$\displaystyle \arcsin x \pm \arccos y$	$\displaystyle =$	$\displaystyle \arcsin \left(xy\pm \sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right) = \arccos \left(y\sqrt{1-x^2}\mp x \sqrt{1-y^2}\right)$
$\displaystyle \arctan x \pm arc\cot y$	$\displaystyle =$	$\displaystyle \arctan\left(\frac{xy\pm 1}{x\mp y}\right) = \arctan\left(\frac{x\mp y}{xy\pm 1}\right)$