考研高数常用基础公式自用版

2019-06-20 本文已影响0人木秋阳

三角函数

$\csc x =\frac{1}{\sin x}$ $\sec x = \frac{1}{\cos x}$ $1+\tan^2 x = \sec x$ $1+\cot^2x = \csc^2x$

$\sin 2x =2\sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2x=2\cos^2x -1 = 1-2\sin^2x$

$\tan 2x =\frac{2\tan x}{1-\tan^2x}$ $\cot 2x = \frac{1-\cot^2x}{2\cot x}$

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

$\sin^2 \alpha = \frac{1-\cos2\alpha}{2}$ $\cos^2\alpha = \frac{1+\cos2\alpha}{2}$

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

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

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

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

$\sin\alpha \cos\beta = \frac{1}{2}[\sin(\alpha +\beta) + \sin(\alpha - \beta)]$

$\cos\alpha\cos\beta = \frac{1}{2}[\cos(\alpha + \beta) +\cos(\alpha - \beta)]$

$\cos\alpha \sin\beta = \frac{1}{2}[\sin(\alpha +\beta) - \sin(\alpha +\beta)]$

$\sin\alpha \cos\beta = \frac{1}{2}[\cos(\alpha +\beta) + \cos(\alpha -\beta)]$

常用的无穷小替换

当 $x\rightarrow 0$ 时

$\sin x$ $\arcsin x$ $\tan x$ $\arctan x$ $\ln(1+x)$ $e^x -1$ 等价于 $x$

$1- \cos x \leftrightarrow \frac{1}{2}x^2$

$(1+x)^\frac{1}{n} -1$ ~~ $\frac{1}{n}x$

几个常用的极限

$\lim_{x\to∞} \sqrt[n]{\alpha} = 1$ 特别的 $\lim_{n\to+∞}\sqrt[n]{n} = 1$

上一篇下一篇

猜你喜欢

热点阅读