数学要背的内容
数学要背的内容
常用泰勒公式
\(\sin x = x - {\Large\frac{x^3}{3!}} + o(x^3)\)
\(\cos x = 1 - {\Large \frac{x^2}{2!} + \frac{x^4}{4!}} + o(x^4)\)
\(\tan x = x + {\Large\frac{x^3}{3!}} + o(x^3)\)
\(\arcsin x = x + {\Large\frac{x^3}{3!}} + o(x^3)\)
\(\arctan x = x - {\Large\frac{x^3}{3!}} + o(x^3)\)
\(\ln (1 + x) = x - {\Large \frac{x^2}{2!} + \frac{x^3}{3!}} + o(x^3)\)
\(e^x = 1 + x + {\Large \frac{x^2}{2!} + \frac{x^3}{3!}} + o(x^3)\)
\((1 + x)^a = 1 + ax + {\Large \frac{a(a - 1)}{2!}}x^2 + o(x^2)\)
常用等价替换
当 x->0 时，有：
\(\sin x \sim x，\tan x \sim x，\arcsin x \sim x，\arctan x \sim x\)
\(1-\cos x \sim {\Large \frac{x^2}{2}} ， a^x-1 \sim x\ln a ，e^x-1 \sim x\)
\(\ln(1+x) \sim x ，(1+x)^a-1 \sim ax\)
当 \(x\rightarrow\infty\) 时，有：
重要极限:
\(\underset{x\rightarrow \infty}{\lim}(1 + \frac{1}{x})^x = e\)
\(\underset{x\rightarrow \infty}{\lim}{\Large \frac{\sin x}{x}} = 1\)
三角函数公式
在直角三角形中：
\(\sin \alpha\) ：对边除以斜边, 音标[saɪn]
\(\cos \alpha\) ：邻边除以斜边, 音标[ˈkəʊsaɪn]
\(\tan \alpha = \Large \frac{\sin \alpha}{\cos \alpha}\) ： 对边除以邻边, 音标[ˈtændʒənt]
\(\cot \alpha = \Large \frac{1}{\tan \alpha} = \frac{\cos \alpha}{\sin \alpha}\) : 邻边除以对边, 音标['kəʊ'tændʒənt]
\(\sec \alpha = \Large \frac{1}{\cos \alpha}\) : 斜边除以邻边, 音标['si:kənt]
\(\csc \alpha = \Large \frac{1}{\sin \alpha}\) : 斜边除以对边, 音标['kəʊ'si:kənt]
三角函数的变换
半径为 1，圆心在 (0，0) 的圆可以推导下面公式：
\(\sin^2 \alpha + \cos^2 \alpha = 1\)
\(\sec^2 \alpha - 1 = \tan^2 \alpha\)
\(\sin({\Large\frac{\pi}{2}}+\alpha)=\cos \alpha\)
\(\sin({\Large\frac{\pi}{2}}-\alpha)=\cos \alpha\)
\(\cos({\Large\frac{\pi}{2}}+\alpha)=-\sin \alpha\)
\(\cos({\Large\frac{\pi}{2}}-\alpha)=\sin \alpha\)
\(\tan({\Large\frac{\pi}{2}}+\alpha)=-\cot \alpha\)
\(\tan({\Large\frac{\pi}{2}}-\alpha)=\cot \alpha\)
\(\cot({\Large\frac{\pi}{2}}+\alpha)=-\tan \alpha\)
\(\cot({\Large\frac{\pi}{2}}-\alpha)=\tan \alpha\)
和差角公式
\(\sin(\alpha \pm \beta)=\sin \alpha\cos \beta \pm \cos \alpha \sin \beta\)
\(\cos(\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin \alpha\sin \beta\)
\(\tan(\alpha \pm \beta)= \Large \frac{\tan \alpha \pm tan \beta}{1 \mp \tan \alpha \cdot \tan \beta}\)
\(\cot(\alpha \pm \beta)= \Large \frac{\cot \alpha \cdot \cot \beta \mp 1}{\cot \beta \pm \cot \alpha}\)
和差化积公式
\(\sin \alpha + \sin \beta = 2\sin{\Large \frac{\alpha + \beta}{2}}\cdot \cos {\Large \frac{\alpha - \beta}{2}}\)
\(\sin \alpha - \sin \beta = 2\cos{\Large \frac{\alpha + \beta}{2}}\cdot \sin {\Large \frac{\alpha - \beta}{2}}\)
\(\cos \alpha + \cos \beta = 2\cos{\Large \frac{\alpha + \beta}{2}}\cdot \cos {\Large \frac{\alpha - \beta}{2}}\)
\(\cos \alpha - \cos \beta = -2\sin{\Large \frac{\alpha + \beta}{2}}\cdot \sin {\Large \frac{\alpha - \beta}{2}}\)
$\sin \alpha + \cos \alpha = \sqrt{2} \cos (\alpha - {\Large\frac{\pi}{4})} = -\sqrt{2}\sin(\alpha + {\Large\frac{\pi}{4})} $
积化和差公式
\(\sin \alpha \cos \beta = {\Large\frac{1}{2}}[\sin(\alpha + \beta)+\sin(\alpha - \beta)]\)
\(\cos \alpha \sin \beta = {\Large\frac{1}{2}}[\sin(\alpha + \beta)-\sin(\alpha - \beta)]\)
\(\cos \alpha \cos \beta = {\Large\frac{1}{2}}[\cos(\alpha + \beta)+\cos(\alpha - \beta)]\)
\(\sin \alpha \sin \beta = {\Large\frac{1}{2}}[\cos(\alpha + \beta)-\cos(\alpha - \beta)]\)
倍角公式
\(\sin 2 \alpha = 2\sin \alpha \cos \alpha=\Large \frac{2\tan \alpha}{1 + \tan^2 \alpha}\)
\(\cos 2 \alpha = \cos^2 \alpha - \sin^2 \alpha \\ \qquad = 2\cos^2 \alpha - 1 \\\qquad = 1 - 2\sin^2 \alpha \\\qquad = \Large \frac{1-\tan^2 \alpha}{1 + \tan^2 \alpha}\)
\(\tan 2 \alpha = {\Large \frac{2\tan \alpha}{1-\tan^2 \alpha}}\cdot \cot 2 \alpha = \Large \frac{\cot^2 \alpha - 1}{2\cot \alpha}\)
\(\sin 3 \alpha = 3\sin \alpha - 4\sin^3 \alpha\)
\(\cos 3 \alpha = 4\cos^3 \alpha - 3\cos \alpha\)
基本求导公式
\((x^a)' = ax^{a - 1}, a为常数\)
\((a^x)' = a^x\ln a,(a>0, a\neq 1)\)
\((e^x)' = e^x\)
\((\log _ax)' = {\Large \frac{1}{x\ln a}},(a>0, a\neq 1）\)
\((\ln|x|)' = \Large \frac{1}{x}\)
\((\sin x)' = \cos x\)
\((\cos x)' = -\sin x\)
\((\arcsin x)'=\Large \frac{1}{\sqrt{1 - x^2}}\)
\((\arccos x)' = -\Large \frac{1}{\sqrt{1 - x^2}}\)
\((\tan x)' = \sec ^2x\)
\((\cot x)' = -\csc ^2x\)
\((\arctan x)' = \Large \frac{1}{1 + x^2}\)
\((arccot x)' = -\Large \frac{1}{1 + x^2}\)
\((\sec x)'= \sec x \tan x\)
\((\csc x)' = -\csc x \cot x\)
\((\ln(x + \sqrt{x^2 + 1}))' = \Large \frac{1}{\sqrt{x^2 + 1}}\)
\((\ln(x + \sqrt{x^2 - 1}))' = \Large \frac{1}{\sqrt{x^2 - 1}}\)
