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勾股定理
\[a^2+b^2=c^2 \]一元二次方程组
\[ax^2 + bx + c = 0 \]\[x = {-b \pm \sqrt{b^2-4ac} \over 2a} \]圆的标准方程
\[(x-a)^2+(y-b)^2=r^2 \]椭圆公式
\[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \]球的体积公式
\[V_球=\frac{4\pi r^3}{3}, (r为半径) \]球的表面积公式
\[S_球=4\pi r^2, (r为半径) \]指数
\[a^{m+n}=a^m \cdot a^n \tag{1} \]\[a^{m-n}=\frac{a^m}{a^n} \tag{2} \]\[(ab)^{n}=a^n \cdot b^n \tag{3} \]\[a^{\frac{m}{n}}=\sqrt[n]{a^m} \tag{4} \]对数
\[a^N = b \Rightarrow \log_{a}{b} = N \tag{1} \]\[\log_a(MN) = \log_aM + \log_aN \tag{2} \]\[\log_a(\frac{M}{N}) = \log_aM - \log_aN \tag{3} \]\[\log_a N = \frac{\log_bN}{\log_ba} (b>0,b\neq1) \tag{4} \]\[\log_a M^n = n \cdot \log_a M \tag{5} \]\[\log_{a^n} M = \frac{1}{n} \cdot \log_a M \tag{6} \]三角函数常用值
\[\sin 30° = \frac{1}{2},\sin45°=\frac{\sqrt{2}}{2},\sin60°=\frac{\sqrt{3}}{2} \tag{1} \]\[\cos 30° = \frac{\sqrt{3}}{2},\cos 45°=\frac{\sqrt{2}}{2},\cos 60°=\frac{1}{2} \tag{2} \]\[\tan 30° = \sqrt{3},\cos 45°=1,\cos 60°=\frac{\sqrt{3}}{3} \tag{3} \]三角函数
\[\tan x = \frac{\sin x}{\cos x},\cot x = \frac{\cos x}{\sin x} \]反三角函数
\[y=\sin x,x=\arcsin y \]\[y=\cos x,x=\arccos y \]和差角公式
\[\sin(\alpha+\beta) = \sin \alpha \cos \beta + \cos\alpha\sin\beta \]\[\sin(\alpha-\beta) = \sin \alpha \cos \beta - \cos\alpha\sin\beta \]\[\cos(\alpha+\beta) = \cos \alpha \cos \beta - \sin\alpha\sin\beta \]\[\cos(\alpha-\beta) = \cos \alpha \cos \beta + \sin\alpha\sin\beta \]和差化积公式
\[\sin \alpha + \sin \beta =2 \sin \frac{\alpha + \beta}{2}\cos \frac{\alpha - \beta}{2} \]倍角公式
\[\sin2\alpha=2\sin\alpha\cos\beta \]\[\cos2\alpha=2\cos \alpha^2-1=1-2\sin^2\alpha =cos^2\alpha-\sin^2\alpha \]\[\tan 2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha} \]半角公式
微分公式（导数公式）
\[(c)^\prime = 0,(c为常数) \tag{1} \]\[(a^x)^\prime = a^x \ln a \tag{2} \]\[(e^x)^\prime = e^x \tag{3} \]\[(x^n)^\prime = nx^{n-1} \tag{4} \]\[(\log_ax)^\prime = \frac{1}{x\ln a} \tag{5} \]\[(\ln x)^\prime = \frac{1}{x} \tag{6} \]\[(\sin x)^\prime = \cos x \tag{7} \]\[(\cos x)^\prime = -\sin x \tag{8} \]\[(\tan x)^\prime = \frac{1}{\cos^2x} = \sec^2x \tag{9} \]\[(\arcsin x)^\prime = \frac{1}{\sqrt{1-x^2}} \tag{10} \]\[(\cot x)^\prime = -\frac{1}{\sin^2x} \tag{11} \]\[(uv)^\prime=u^\prime \cdot v + u \cdot v^\prime \tag{12} \]高阶导数公式
\[(uv)^{(n)}=\sum_{k=0}^{n} C_n^k u^{(n-k)} v^k \]常用积分公式
\[\int k\mathrm{d}x = kx+C \tag{1} \]\[\int x^\mu \mathrm{d}x = \frac{x^{\mu+1}}{\mu+1}+C \tag{2} \]\[\int \frac{1}{x}\mathrm{d}x= \ln \left| x \right| +C \tag{3} \]\[\int \frac{1}{1+x^{2}}\mathrm{d}x= \arctan x +C \tag{4} \]\[\int \frac{1}{\sqrt{1-x^{2}}}\mathrm{d}x= \arcsin x +C \tag{5} \]\[\int \cos x \mathrm{d}x= \sin x +C \tag{6} \]\[\int \sin x \mathrm{d}x= -\cos x +C \tag{7} \]\[\int \frac{1}{\cos^2x} \mathrm{d}x= \tan x +C \tag{8} \]\[\int \frac{1}{\sin^2x} \mathrm{d}x= -\cot x +C \tag{9} \]\[\int \sec x \tan x \mathrm{d}x= \sec x +C \tag{10} \]\[\int \csc x \cot x \mathrm{d}x= -\csc x +C \tag{11} \]\[\int e^x \mathrm{d}x= e^x +C \tag{12} \]\[\int a^x \mathrm{d}x= \frac{a^x}{\ln a} +C \tag{13} \]\[\int sh x \mathrm{d}x= ch x +C \tag{14} \]\[\int ch x \mathrm{d}x= sh x +C \tag{15} \]不定积分
\[\int f(x)\mathrm{d}x = F(x) + c \]牛顿-莱布尼兹公式
如果函数(f(x))在区间([a,b])上连续，并且存在原函数(F(x))，则
\[\int_a^b f(x)\mathrm{d}x = F(b)-F(a)=F(x)|_a^b \]三角函数积分
\[\sin x = { 2u \over 1 + u^ 2} \]\[\cos x = { 1-u^2 \over 1 + u^ 2} \]全微分公式
\[dz = {{\partial z} \over {\partial x}}dx + { {\partial z} \over {\partial y}}dy \]\[du = {{\partial u} \over {\partial x}}dx + { {\partial u} \over {\partial y}}dy + { {\partial u} \over {\partial z}}dz \]两个重要极限
\[\lim_{x \to 0} \frac{\sin x}{x}=1 \tag{1} \]\[\lim_{x \to \infty} (1+\frac{1}{x} )^x = e \tag{2} \]常用极限
\[\lim_{x \to \infty} \sqrt[x]{a} = 1 \]\[\lim_{x \to \infty} \sqrt[x]{x} = 1 \]分部积分法
\[\int u \frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x=uv-\int \frac{\mathrm{d}u}{\mathrm{d}x}v\,\mathrm{d}x \]\[\int u \mathrm{d}v = u \cdot v - \int v \mathrm{d}u \]旋转体的体积计算
\[\int \pi f^2(x) \mathrm{d}x \]二重积分
\[\iint_{a}^{b} f(x,y) \mathrm{d}x \mathrm{d}y \]\[\int_{0}^{1} \mathrm{d}y \int_{0}^{1} f(x,y) \mathrm{d}x \]\[\iint\limits_D dx\,dy \]常数项级数
\[\sum_{n=1}^{\infty}{u_n} \]幂级数
\[a_0+a_1x+a_2x^2+\cdots+a_nx^n+\cdots= \sum_{n=1}^{\infty}{a_nx^n} \]收敛半径
\[R=\frac{1}{\rho}=\frac{u_{n+1}}{u_n} \]傅里叶级数
行列式计算
\[\begin{vmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 &9 \end{vmatrix}=1×5×9+2×6×7+3×4×8-(3×5×7+2×4×9+6×8×1)=n \]矩阵
\[\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 &9 \end{bmatrix} \]二项式系数
\[\dbinom{n}{r}=\binom{n}{n-r}=\mathrm{C}_n^r=\mathrm{C}_n^{n-r} \]齐次线性方程组
\[\begin{cases} 3x + 5y + z \\ 7x - 2y + 4z \\ -6x + 3y + 2z \end{cases} \]\[\left\{\begin{aligned} 3x + 5y + z \\ 7x - 2y + 4z \\ -6x + 3y + 2z \end{aligned}\right. \]\[f(x)=\begin{cases} 1, & x>0\\ 0, & x=0\\ -1, & x<0 \end{cases} \]等差数列公式
\[a_{n}=a_{1}+ \left( n-1 \left) d\right. \right. \]等差数列的前N项和
\[S_{n}=\frac{n \left( a_{1}+a_{n}\right)}{2}=na_{1}+\frac{n \left( n-1 \right)}{{2}}d \]等比数列公式
\[a_{n}=a_{1}q^{n-1} \]等比数列的前N项和
\[S_{n}=\frac{a_1(1-q^n)}{1-q}=\frac{a_1+a_nq}{1-q} \]等差中项
\[2b=a+c,(a,b,c为等差数列) \]等比中项
\[b^2=a \cdot c,(a,b,c为等比数列) \]裂项相消法
\[若 \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1},则有T_n=1-\frac{1}{n+1}=\frac{n}{n+1} \tag{1} \]二项式定理
\[(a+b)^n=C_{n}^{0}a^n+C_{n}^{1}a^{n-1}b^1+C_{n}^{2}a^{n-2}b^2+\cdots+C_{n}^{k}a^{n-k}b^k+\cdots+C_{n}^{n}b^n \]\[(a+b)^n=\sum_{r=0}^{n}{C_{n}^{r}a^{n-r}b^r} \]二项展开式的通项
\[T_{r+1}=C_{n}^{r}a^{n-r}b^r \]平面方程
\[Ax+By+Cz+D=0 \]空间两点的距离公式
\[d=|M_1M_2|=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \]平面的点法式方程
\[A(x-x_0)+B(y-y_0)+C(z-z_0)=0 \]- 法向量(\vec n={A,B,C})
直线与平面的夹角
\[\sin \theta=\frac{Al+Bm+Cn}{\sqrt{A^2+B^2+C^2} \cdot \sqrt{l^2+m^2+n^2}} \]两平面的夹角
\[\cos \theta=\frac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{A_1^2+B_1^2+C_1^2} \cdot \sqrt{A_2^2+B_2^2+C_2^2}} \]点到平面的距离公式
\[d=\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}} \]空间直线方程
\[\frac{x-x_0}{m}=\frac{y-y_0}{n}=\frac{z-z_0}{p}=t \]点到直线的距离
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点：\(M(x_0,y_0,z_0)\)
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直线：\({L_1：\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}}\)
极坐标公式
\[\left\{\begin{matrix} x=a + r\text{cos}\theta \\ y=b + r\text{sin}\theta \end{matrix}\right. \tag{圆的参数方程} \]双曲线
- a>0,b>0