\[\newcommand{\arccot}{\mathrm{arccot}\,}
\newcommand{\arcsec}{\mathrm{arcsec}\,}
\newcommand{\arccsc}{\mathrm{arccsc}\,}
\newcommand{\d}{\mathrm{d}\,}
\]
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
\[f(x)=f(a)+\frac{f\'(a)}{1!}(x-a)+\frac{f\'\'(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\cdots
\]
\[\frac{1}{1-ax}=\sum\limits_{i=0}^\infty a^ix^i
\]