三角函数公式
\[ \tan A=\frac{\sin A}{\cos A},\cot A=\frac{\cos A}{\sin A}\\ \sec A=\frac{1}{\cos A},\csc A=\frac{1}{\sin A}\\ 1=\sin^2 A+\cos^2 A\\ \]
\[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]
\[ \sin(A-B)=\sin A\cos B-\cos A\sin B \]
\[ \cos(A+B)=\cos A\cos B-\sin A\sin B \]
\[ \cos(A-B)=\cos A\cos B+\sin A\sin B \]
\[ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \]
\[ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} \]
\[ \cot (A+B)=\frac{\cot A\cot B-1}{\cot B+\cot A} \]
\[ \cot (A-B)=\frac{\cot A\cot B+1}{\cot B-\cot A} \]
\[ \tan 2A=\frac{2\tan A}{1-\tan^2A} \]
\[ \sin 2A=2\sin A\cos A \]
\[ \cos 2A=\cos^2A-\sin^2A \]
\[ \sin \frac{A}{2}=\sqrt{\frac{1-\cos A}{2}} \]
\[ \cos \frac{A}{2}=\sqrt{\frac{1+\cos A}{2}} \]
\[ \tan \frac{A}{2}=\sqrt{\frac{1-\cos A}{1+\cos A}} \]
\[ \cot \frac{A}{2}=\sqrt{\frac{1+\cos A}{1-\cos A}} \]
\[ \sin A+\sin B=2\sin \frac{A+B}{2}\cos\frac{A-B}{2} \]
\[ \sin A-\sin B=2\sin \frac{A-B}{2}\cos\frac{A+B}{2} \]
\[ \cos A+\cos B=2\cos \frac{A+B}{2}\cos\frac{A-B}{2} \]
\[ \cos A-\cos B=-2\sin \frac{A+B}{2}\sin\frac{A-B}{2} \]
\[ \tan A+\tan B=\frac{\sin (A+B)}{\cos A \cos B} \]
等价无穷小
\[ 当 \ x\to 0 \ 时 \]
\[ \sin x \sim x \]
\[ \tan x \sim x \]
\[ \ln(1+x)\sim x \]
\[ e^x-1\sim x \]
\[ \arcsin x\sim x \]
\[ \arctan x\sim x \]
\[ \log_a(1+x)\sim\frac{x}{\ln a} \]
\[ a^x-1\sim x\ln a \]
\[ 1-\cos x\sim\frac{1}{2}x^2 \]
\[ \ln(x+\sqrt{1+x^2})\sim x \]
\[ x-\sin x\sim \frac{1}{6}x^3 \]
\[ \tan x-x\sim\frac{1}{3}x^3 \]
\[ (1+x)^a-1\sim ax \]
\[ \arcsin x-x\sim\frac{1}{6}x^3 \]
\[ x-\arctan x\sim\frac{1}{3}x^3 \]
\[ \tan x-\sim x\sim\frac{1}{2}x^3 \]
导数表
\[ (f(x)\pm g(x))'=f'(x)\pm g'(x) \]
\[ (Cf(x))'=Cf'(x) \]
\[ (f(x)g(x))'=f'(x)g(x)+f(x)g'(x) \]
\[ (\frac{f(x)}{g(x)})'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)} \]
\[ (C)'=0 \]
\[ (x^a)'=ax^{a-1} \]
\[ (\sin x)'=\cos x \]
\[ (\cos x)'=-\sin x \]
\[ (\tan x)'=\sec^2 x \]
\[ (\cot x)'=-\csc^2x \]
\[ (\sec x)'=\sec x\tan x \]
\[ (\csc x)'=-\csc x\cot x \]
\[ (a^x)'=a^x\ln a \]
\[ (e^x)'=e^x \]
\[ (\log_ax)'=\frac{1}{x\ln a} \]
\[ (\ln x)'=\frac{1}{x} \]
\[ (\arcsin x)'=\frac{1}{\sqrt{1-x^2}} \]
\[ (\arccos x)'=-\frac{1}{\sqrt{1-x^2}} \]
\[ (\arctan x)'=\frac{1}{1+x^2} \]
\[ (arccot\ x)'=-\frac{1}{1+x^2} \]