XSMath

Mathematics and Statistics Resources

$\sin x=\frac{1}{\csc x}$		$\csc x=\frac{1}{\sin x}$
$\cos x=\frac{1}{\sec x}$		$\sec x=\frac{1}{\cos x}$
$\tan x=\frac{\sin x}{\cos x}=\frac{1}{\cot x}$		$\cot x=\frac{\cos x}{\sin x}=\frac{1}{\tan x}$

$\sin^2x+\cos^2x=1$
$\therefore$ $\sin^2x=1-\cos^2x$		and $\cos^2x=1-\sin^2x$
$\tan^2x+1=\sec^2x$		$1+\cot^2x=\csc^2x$

$\sin 2x=2\sin x\cos x$		$\cos 2x=\cos^2x-\sin^2x$ $=2\cos^2x-1$ $=1-2\sin^2x$

$\tan 2x=\frac{2\tan x}{1-\tan^2x}$

$\sin x=\cos (\frac{\pi}{2}-x)$		$\cos x=\sin (\frac{\pi}{2}-x)$
$\tan x=\cot (\frac{\pi}{2}-x)$
$\csc x=\sec (\frac{\pi}{2}-x)$		$\sec x=\csc (\frac{\pi}{2}-x)$
$\cot x=\tan (\frac{\pi}{2}-x)$

$\sin (-x)=-\sin x$
$\cos (-x)=\cos x$
$\tan (-x)=-\tan x$

$\sin (x+2\pi)=\sin x$		$\sin (\pi-x)=\sin x$
$\cos (x+2\pi)=\cos x$		$\cos (\pi-x)=-\cos x$
$\tan (x+\pi)=\tan x$		$\tan (\pi-x)=-\tan x$

$\sin (x+y)$ $=\sin x\cos y+\cos x\sin y$		$\sin (x-y)$ $=\sin x\cos y-\cos x\sin y$
$\cos (x+y)$ $=\cos x\cos y-\sin x\sin y$		$\cos (x-y)$ $=\cos x\cos y+\sin x\sin y$
$\tan (x+y)$ $=\frac{\tan x+\tan y}{1-\tan x\tan y}$		$\tan (x-y)$ $=\frac{\tan x-\tan y}{1+\tan x\tan y}$

$\sin\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{2}}$
$\cos\frac{x}{2}=\pm\sqrt{\frac{1+\cos x}{2}}$
$\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{\sin x}$