常用数学公式(不定期更新)

\[ \newcommand{\arccot}{\mathrm{arccot}\,} \newcommand{\arcsec}{\mathrm{arcsec}\,} \newcommand{\arccsc}{\mathrm{arccsc}\,} \newcommand{\d}{\mathrm{d}\,} \]

三角函数公式

\[ \begin{aligned} \sin(A+B)&=\sin A\cos B+\cos A\sin B\\ \sin(A-B)&=\sin A\cos B-\cos A\sin B\\ \cos(A+B)&=\cos A\cos B-\sin A\sin B\\ \cos(A-B)&=\cos A\cos B+\sin A\sin B\\ \sin 2A&=2\sin A\cos A\\ \cos 2A&=\cos^2A-\sin^2A=1-2\sin^2A=2\cos^2A-1\\ \sin\frac{A}{2}&=\sqrt{\frac{1-\cos A}{2}}\\ \cos\frac{A}{2}&=\sqrt{\frac{1+\cos A}{2}}\\ \tan\frac{A}{2}&=\frac{1-\cos A}{\sin A}=\frac{\sin A}{1+\cos A}\\ \sin A+\sin B&=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\\ \sin A-\sin B&=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}\\ \cos A+\cos B&=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}\\ \cos A-\cos B&=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\\ \tan A+\tan B&=\frac{\sin (A+B)}{\cos A\cos B}\\ \sin A\sin B&=\frac{1}{2}[\cos(A+B)-\cos(A-B)]\\ \cos A\cos B&=\frac{1}{2}[\cos(A+B)+\cos(A-B)]\\ \sin A\cos B&=\frac{1}{2}[\sin(A+B)+\sin(A-B)]\\ \end{aligned} \]

\[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\\ \cos A=\frac{b^2+c^2-a^2}{2bc} \]

\[ \sin^2A=\frac{1-\cos 2A}{2}\\ \cos^2A=\frac{1+\cos 2A}{2} \]

导数公式

\[ \begin{aligned} (u\pm v)'&=u'\pm v'\\ (uv)'&=u'v+uv'\\ (cu)'&=cu'\\ (\frac{u}{v})'&=\frac{u'v-uv'}{v^2}\\ c'&=0 \end{aligned} \]

\[ \begin{aligned} (x^n)'&=nx^{n-1}\\ (a^x)'&=a^x\ln x\\ (\log_ax)'&=\frac{1}{x\ln a}\\ (\sin x)'&=\cos x\\ (\cos x)'&=-\sin x\\ (\tan x)'&=\sec^2x\\ (\cot x)'&=-\csc^2x\\ (\sec x)'&=\sec x\tan x\\ (\arcsin x)'&=\frac{1}{\sqrt{1-x^2}} \end{aligned} \]

积分公式

\[ \begin{aligned} \int k \d x&=kx+c\\ \int x^n \d x&=\frac{1}{n+1}x^{n+1}+c\\ \int \frac{1}{x}\d x&=\ln |x|+c\\ \int a^x \d x&=\frac{a^x}{\ln a}+c\\ \int \sin x\d x&=-\cos x+c\\ \int \cos x\d x&=\sin x+c\\ \int \sec^2x\d x&=\tan x+c\\ \int \csc^2x\d x&=-\cot x+c\\ \int \sec x\tan x\d x&=\sec x+c\\ \int \cot x\csc x\d x&=\csc x+c\\ \int \frac{1}{\sqrt{1-x^2}}\d x&=\arcsin x+c\\ \int \frac{1}{1+x^2}\d x&=\arctan x+c \end{aligned} \]

泰勒展开公式

\[ f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\cdots \]

\[ \frac{1}{1-ax}=\sum\limits_{i=0}^\infty a^ix^i \]

posted @ 2019-09-05 20:31 chy_2003 阅读(...) 评论(...) 编辑 收藏