考研高数背诵知识点

sin⁡(π2±α)=cos⁡αsin⁡(π±α)=∓sin⁡αcos⁡(π2±α)=∓sin⁡αsin⁡(π±α)=−cos⁡α\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

sin⁡3α=−4sin⁡3α+3sin⁡αcos⁡3α=4cos⁡3α−3cos⁡α\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

sin⁡2α2=12(1−cos⁡α)cos⁡2α2=12(1+cos⁡α)tan⁡α2=1−cos⁡αsin⁡α=sin⁡α1+cos⁡α=±1−cos⁡α1+cos⁡αcot⁡α2=sin⁡α1−cos⁡α=1+cos⁡αsin⁡α=±1+cos⁡α1−cos⁡α\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−cosα​​cot2α​=1−cosαsinα​=sinα1+cosα​=±1−cosα1+cosα​​

cot⁡\cotcot和差公式

cot⁡(α+β)=cot⁡αcot⁡β−1cot⁡β+cot⁡α\cot(\alpha+\beta) = \frac{\cot\alpha\cot\beta-1}{\cot\beta+\cot\alpha} cot(α+β)=cotβ+cotαcotαcotβ−1​

cot⁡(α−β)=cot⁡αcot⁡β+1cot⁡β−cot⁡α\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]

sin⁡αcos⁡β=sin⁡(α+β)+sin⁡(α−β)2cos⁡αsin⁡β=sin⁡(α+β)−sin⁡(α−β)2cos⁡αcos⁡β=cos⁡(α+β)+cos(α−β)2sin⁡αsin⁡β=−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]

sin⁡α+sin⁡β=2sin⁡α+β2cos⁡α−β2sin⁡α−sin⁡β=2cos⁡α+β2sin⁡α−β2cos⁡α+cos⁡β=2cos⁡α+β2cos⁡α−β2cos⁡α−cos⁡β=−2sin⁡α+β2sin⁡α−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+β​cos2α−β​sinα−sinβ=2cos2α+β​sin2α−β​cosα+cosβ=2cos2α+β​cos2α−β​cosα−cosβ=−2sin2α+β​sin2α−β​

万能公式

​ 若u=tan⁡x2(−π<x<π)u=\tan\frac{x}{2}(-\pi<x<\pi)u=tan2x​(−π<x<π),则sin⁡x=2u1+u2\sin x = \frac{2u}{1+u^2}sinx=1+u22u​,cos⁡x=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∑n​k2=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​=−ab​x1​x2​=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−1​an−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)