考研高数背诵知识点
诱导公式sin(π2±α)=cosαsin(π±α)=∓sinαcos(π2±α)=∓sinαsin(π±α)=−cosα\sin(\frac{\pi}{2}\pm\alpha) = \cos\alpha\\ \sin(\pi\pm\alpha) = \mp\sin\alpha\\ \cos(\frac{\pi}{2}\pm\alpha) = \mp\sin\alpha\\ \sin(\pi\pm\alpha) = -\cos\alpha\\ sin(2π±α)=cosαsin(π±α)=∓sinαcos(2π±α)=∓sinαsin(π±α)=−cosα
三倍角公式
sin3α=−4sin3α+3sinαcos3α=4cos3α−3cosα\sin 3\alpha = -4\sin^3\alpha+3\sin\alpha\\ \cos 3\alpha=4\cos^3\alpha-3\cos\alpha sin3α=−4sin3α+3sinαcos3α=4cos3α−3cosα
半角公式
sin2α2=12(1−cosα)cos2α2=12(1+cosα)tanα2=1−cosαsinα=sinα1+cosα=±1−cosα1+cosαcotα2=sinα1−cosα=1+cosαsinα=±1+cosα1−cosα\sin^2\frac{\alpha}{2} = \frac{1}{2}(1-\cos\alpha)\\ \cos^2\frac{\alpha}{2} = \frac{1}{2}(1+\cos\alpha)\\ \tan\frac{\alpha}{2} = \frac{1-\cos\alpha}{\sin\alpha} = \frac{\sin\alpha}{1+\cos\alpha} = \pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}\\ \cot\frac{\alpha}{2} = \frac{\sin\alpha}{1-\cos\alpha} = \frac{1+\cos\alpha}{\sin\alpha} = \pm\sqrt{\frac{1+\cos\alpha}{1-\cos\alpha}}\\ sin22α=21(1−cosα)cos22α=21(1+cosα)tan2α=sinα1−cosα=1+cosαsinα=±1+cosα1−cosαcot2α=1−cosαsinα=sinα1+cosα=±1−cosα1+cosα
cot\cotcot和差公式
cot(α+β)=cotαcotβ−1cotβ+cotα\cot(\alpha+\beta) = \frac{\cot\alpha\cot\beta-1}{\cot\beta+\cot\alpha} cot(α+β)=cotβ+cotαcotαcotβ−1
cot(α−β)=cotαcotβ+1cotβ−cotα\cot(\alpha-\beta) = \frac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha} cot(α−β)=cotβ−cotαcotαcotβ+1
积化和差公式[^考前背,基本不考]
sinαcosβ=sin(α+β)+sin(α−β)2cosαsinβ=sin(α+β)−sin(α−β)2cosαcosβ=cos(α+β)+cos(α−β)2sinαsinβ=−cos(α+β)−cos(α−β)2\sin\alpha\cos\beta=\frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}\\ \cos\alpha\sin\beta=\frac{\sin(\alpha+\beta)-\sin(\alpha-\beta)}{2}\\ \cos\alpha\cos\beta=\frac{\cos(\alpha+\beta)+cos(\alpha-\beta)}{2}\\ \sin\alpha\sin\beta=-\frac{\cos(\alpha+\beta)-\cos(\alpha-\beta)}{2} sinαcosβ=2sin(α+β)+sin(α−β)cosαsinβ=2sin(α+β)−sin(α−β)cosαcosβ=2cos(α+β)+cos(α−β)sinαsinβ=−2cos(α+β)−cos(α−β)
和差化积公式[^考前背,基本不考]
sinα+sinβ=2sinα+β2cosα−β2sinα−sinβ=2cosα+β2sinα−β2cosα+cosβ=2cosα+β2cosα−β2cosα−cosβ=−2sinα+β2sinα−β2\sin\alpha+\sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\ \sin\alpha-\sin\beta = 2\cos\frac{\alpha+\beta}{2}\sin\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} sinα+sinβ=2sin2α+βcos2α−βsinα−sinβ=2cos2α+βsin2α−βcosα+cosβ=2cos2α+βcos2α−βcosα−cosβ=−2sin2α+βsin2α−β
万能公式
若u=tanx2(−π<x<π)u=\tan\frac{x}{2}(-\pi<x<\pi)u=tan2x(−π<x<π),则sinx=2u1+u2\sin x = \frac{2u}{1+u^2}sinx=1+u22u,cosx=1−u21+u2\cos x = \frac{1-u^2}{1+u^2}cosx=1+u21−u2
常见的等差数列前n项和
∑k=1nk2=12+22+32+⋯+n2=n(n+1)(2n+1)6\sum_{k=1}^nk^2 = 1^2+2^2+3^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6} k=1∑nk2=12+22+32+⋯+n2=6n(n+1)(2n+1)
根的公式
x1,2=−b±b2−4ac2ax_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a} x1,2=2a−b±b2−4ac
根与系数的关系(韦达定理)
x1+x2=−bax1x2=cax_1+x_2=-\frac{b}{a}\\ x_1x_2 = \frac{c}{a} x1+x2=−abx1x2=ac
因式分解公式
(a+b)3=a3+3a2b+3ab2+b3(a−b)3=a3−3a2b+3ab2−b3a3−b3=(a−b)(a2+ab+b2)a3+b3=(a+b)(a2−ab+b2)an−bn=(a−b)∑i=0n−1an−1−ibian+bn=(a−b)∑i=0n−1(−1)ian−1−ibi(n为正奇数)(a+b)^3=a^3+3a^2b+3ab^2+b^3\\ (a-b)^3=a^3-3a^2b+3ab^2-b^3\\ a^3-b^3 = (a-b)(a^2+ab+b^2)\\ a^3+b^3 = (a+b)(a^2-ab+b^2)\\ a^n-b^n = (a-b)\sum_{i=0}^{n-1}a^{n-1-i}\ b^i\\ a^n+b^n = (a-b)\sum_{i=0}^{n-1}(-1)^{i}a^{n-1-i}\ b^i(n 为正奇数) (a+b)3=a3+3a2b+3ab2+b3(a−b)3=a3−3a2b+3ab2−b3a3−b3=(a−b)(a2+ab+b2)a3+b3=(a+b)(a2−ab+b2)an−bn=(a−b)i=0∑n−1an−1−ibian+bn=(a−b)i=0∑n−1(−1)ian−1−ibi(n为正奇数)
双阶乘
(2n)!!=2∗4∗6∗8∗10∗⋯∗(2n)=2n∗n!(2n−1)!!=1∗3∗5∗7∗⋯∗(2n−1)(2n)!! = 2*4*6*8*10*\dots*(2n)=2^n*n!\\ (2n-1)!! = 1 * 3* 5*7*\dots*(2n-1) (2n)!!=2∗4∗6∗8∗10∗⋯∗(2n)=2n∗n!(2n−1)!!=1∗3∗5∗7∗⋯∗(2n−1)
常用不等式
∣a±b∣≤∣a∣+∣b∣∣∣a∣−∣b∣∣≤∣a−b∣∣a1±a2±⋯±an∣≤∣a1∣+∣a2∣+⋯+∣an∣|a\pm b|\leq|a|+|b|\\ ||a|-|b||\leq|a-b|\\ |a_1\pm a_2\pm \dots\pm a_n|\leq|a_1|+|a_2|+\dots+|a_n|\\ ∣a±b∣≤∣a∣+∣b∣∣∣a∣−∣b∣∣≤∣a−b∣∣a1±a2±⋯±an∣≤∣a1∣+∣a2∣+⋯+∣an∣
ab≤a+b2≤a2+b22(a,b>0)abc3≤a+b+c3≤a2+b2+c23(a,b,c>0)\sqrt{ab}\leq \frac{a+b}{2}\leq\sqrt{\frac{a^2+b^2}{2}}\ \ \ \ \ (a,b>0)\\ \sqrt[3]{abc}\leq\frac{a+b+c}{3}\leq\sqrt{\frac{a^2+b^2+c^2}{3}}\ \ \ \ (a,b,c>0)\\ ab≤2a+b≤2a2+b2(a,b>0)3abc≤3a+b+c≤3a2+b2+c2(a,b,c>0)
若0<a<x<b,0<c<y<d,则cb<yx<da若0<a<x<b,0<c<y<d,则\frac{c}{b}<\frac{y}{x}<\frac{d}{a} 若0<a<x<b,0<c<y<d,则bc<xy<ad
1x+1<ln(1+1x)<1x(x>0)\frac{1}{x+1}<\ln(1+\frac{1}{x})<\frac{1}{x}\ \ \ \ \ (x>0) x+11<ln(1+x1)<x1(x>0)