三角函数#

基本概念#

正弦：$sin \alpha = \frac{a}{c}$
余弦：$cos \alpha = \frac{b}{c}$
正切：$tan \alpha = \frac{a}{b}$

正割：$sec \alpha = \frac{c}{b}$
余割：$csc \alpha = \frac{c}{a}$
余切: $cot \alpha = \frac{b}{a}$

基本关系#

$sin \alpha \cdot csc \alpha = 1$
$cos \alpha \cdot sec \alpha = 1$
$tan \alpha \cdot cot \alpha = 1$

$\csc \alpha = \frac{1}{\sin \alpha}$

$\sec \alpha = \frac{1}{\cos \alpha}$

平方公式#

$\sin^2 \alpha + \cos^2 \alpha = 1$
$1+\cot^2 \alpha = \csc^2 \alpha$
$1+\tan^2 \alpha = \sec^2 \alpha$

求导公式#

$(\sin x)' = \cos x$
$(\cos x)' = -\sin x$
$(\tan x)' = \sec^2x$
$(\cot x)' = -\csc^2x$
$(\sec x)' = \tan x \cdot \sec x$
$(\csc x)' = -\cot x \cdot \csc x$
$(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}$
$(\arccos x)' = - \frac{1}{\sqrt{1-x^2}}$
$(\arctan x)' = \frac{1}{1 + x^2}$
$(\arcctg x)' = - \frac{1}{1+ x^2}$

加法减法公式#

$\sin(\alpha + \beta)=\sin\alpha\cos\beta + \sin\beta\cos\alpha$
$\sin(\alpha - \beta)=\sin\alpha\cos\beta - \sin\beta\cos\alpha$
$\cos(\alpha + \beta)=\cos\alpha\cos\beta - \sin\alpha\cos\alpha$
$\cos(\alpha - \beta)=\cos\alpha\cos\beta + \sin\alpha\sin\beta$

$\tan(\alpha + \beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$\tan(\alpha - \beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$

$\cot(\alpha + \beta)=\frac{\cot\alpha\cot\beta - 1}{\cot\alpha+\cot\beta}$

$\cot(\alpha - \beta)=\frac{\cot\alpha\cot\beta + 1} {\cot\alpha-\cot\beta}$

二倍角公式#

$\sin2\alpha=2\sin\alpha\cos\alpha$
$\cos2\alpha=\cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha -1 = 1 - 2\sin^2\alpha$

$\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}$

$\cot2\alpha=\frac{\cot^2\alpha - 1}{2\cot\alpha}$

$\sec2\alpha=\frac{\sec^2\alpha}{1-\tan^2\alpha}=\frac{\cot\alpha+\tan\alpha}{\cot\alpha-tan\alpha}$

$\csc2\alpha=\frac{1}{2}\sec\alpha\csc\alpha=\frac{1}{2}(\tan\alpha+\cot\alpha)$

和差化积#

$\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$

$\cos\alpha+\cos\beta=$

诱导公式#

$\sin(2k\pi + \alpha) = \sin(\alpha)(k \in Z)$
$\cos(2k\pi + \alpha) = \cos(\alpha)(k \in Z)$
$\tan(2k\pi + \alpha) = \tan(\alpha)(k \in Z)$