球 表 面 积 公 式 : S = 4 π R 2 球 体 积 公 式 : V = 4 3 π R 3 圆 锥 体 积 公 式 : V = 1 3 s h ( s 为 圆 锥 底 面 积 , h 为 圆 锥 的 高 ) 椭 圆 面 积 公 式 : S = π a b 扇 形 面 积 公 式 : S = 1 2 l r = 1 2 r 2 θ ( 其 中 l 为 弧 长 , r 为 半 径 , θ 为 夹 角 ( 用 π 表 示 ) ) \begin{aligned} & \\ & 球表面积公式:S= 4\pi R^2 \\ \\ & 球体积公式:V = \frac{4}{3}\pi R^3 \\ \\ & 圆锥体积公式:V=\frac{1}{3} sh ~~~~~(s为圆锥底面积,h为圆锥的高) \\\\ & 椭圆面积公式: S=\pi ab \\ \\ & 扇形面积公式: S= \frac{1}{2}l r = \frac{1}{2}r^2\theta ~~~~~(其中l为弧长,r为半径,\theta为夹角(用\pi表示)) \end{aligned} 球表面积公式:S=4πR2球体积公式:V=34πR3圆锥体积公式:V=31sh (s为圆锥底面积,h为圆锥的高)椭圆面积公式:S=πab扇形面积公式:S=21lr=21r2θ (其中l为弧长,r为半径,θ为夹角(用π表示))
一 元 二 次 方 程 : a x 2 + b x + c = 0 ( a ≠ 0 ) 根 的 公 式 x 1 , 2 = − b ± b 2 − 4 a c 2 a 韦 达 定 理 : x 1 + x 2 = − b a x 1 x 2 = c a 判 别 式 : Δ = b 2 − 4 a c ⟹ { Δ > 0 , 两 个 不 等 实 根 Δ = 0 , 两 个 相 等 实 根 Δ < 0 , 两 个 共 轭 的 复 根 ( 无 实 根 ) 抛 物 线 y = a x 2 + b x + c 的 顶 点 : ( − b 2 a , c − b 2 4 a ) \begin{aligned} & \\ & 一元二次方程:ax^2 + bx + c =0 ~~~~~(a \ne 0) \\ \\ & 根的公式 ~~~~ x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ & 韦达定理: x_1 + x_2 = -\frac{b}{a} ~~~~~~~ x_1 x_2 = \frac{c}{a} \\ \\ & 判别式: \Delta=b^2 - 4ac \implies \begin{cases} \Delta >0,两个不等实根 \\ \Delta =0,两个相等实根 \\ \Delta <0,两个共轭的复根(无实根) \\ \end{cases} \\\\ & 抛物线~ y=ax^2 + bx + c 的顶点:(-\frac{b}{2a}, c-\frac{b^2}{4a}) \end{aligned} 一元二次方程:ax2+bx+c=0 (a=0)根的公式 x1,2=2a−b±b2−4ac韦达定理:x1+x2=−ab x1x2=ac判别式:Δ=b2−4ac⟹⎩⎪⎨⎪⎧Δ>0,两个不等实根Δ=0,两个相等实根Δ<0,两个共轭的复根(无实根)抛物线 y=ax2+bx+c的顶点:(−2ab,c−4ab2)
直 角 坐 标 化 极 坐 标 { x = ρ cos θ y = ρ sin θ ⟹ x 2 + y 2 = ρ 2 极 坐 标 化 直 角 坐 标 : ρ 2 = x 2 + y 2 ⟹ tan θ = y x \begin{aligned} &直角坐标化极坐标 \begin{cases} x = \rho \cos \theta \\ y = \rho \sin \theta \end{cases} \implies x^2+y^2=\rho ^2 \\\\ &极坐标化直角坐标 :\rho ^2 = x^2+y^2 \implies \tan \theta = \frac{y}{x} \\\\ \end{aligned} 直角坐标化极坐标{ x=ρcosθy=ρsinθ⟹x2+y2=ρ2极坐标化直角坐标:ρ2=x2+y2⟹tanθ=xy
切 线 方 程 : y − y 0 x − x 0 = f ′ ( x 0 ) , 即 y − y 0 = f ′ ( x 0 ) ( x − x 0 ) 法 线 方 程 : y − y 0 x − x 0 = − 1 f ′ ( x 0 ) , 即 y − y 0 = − 1 f ′ ( x 0 ) ( x − x 0 ) \begin{aligned} & 切线方程: \frac{y - y_0}{x - x_0} = f'(x_0) ~~~~,即 y-y_0 = f'(x_0)(x-x_0) \\ \\ & 法线方程: \frac{y - y_0}{x - x_0} = -\frac{1}{f'(x_0)}~~~~~,即y-y_0 = -\frac{1}{f'(x_0)}(x-x_0) \end{aligned} 切线方程:x−x0y−y0=f′(x0) ,即y−y0=f′(x0)(x−x0)法线方程:x−x0y−y0=−f′(x0)1 ,即y−y0=−f′(x0)1(x−x0)
( a + b ) 2 = a 2 + 2 a b + b 2 ( a − b ) 2 = a 2 − 2 a b + b 2 ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 a c + 2 b c ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 ( a + b ) ( a − b ) = a 2 − b 2 a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) a n − b n = ( a − b ) ( a n − 1 + a n − 2 b + ⋯ + a b n − 2 + b n − 1 ) ( a + b ) n = ∑ k = 0 n C n k a k b n − k = a n + n a n − 1 b + n ( n − 1 ) 2 ! a n − 1 b 2 + ⋯ + n ( n − 1 ) ⋯ ( n − k + 1 ) k ! a n − k b k + ⋯ + n a b n − 1 + b n \begin{aligned} & \\ & (a+b)^2 = a^2 + 2ab+b^2 \\ \\ & (a-b)^2 = a^2 - 2ab + b^2 \\ \\ & (a+b+c)^2 =a^2+b^2+c^2 + 2ab+2ac+2bc \\\\ & (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \\ \\ & (a-b)^3 = a^3 - 3a^2b+3ab^2-b^3 \\ \\ & (a+b)(a-b) = a^2 - b^2 \\ \\ & 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)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1}) \\ \\ & (a+b)^n = \sum_{k=0}^n C_n^ka^kb^{n-k} = a^n + na^{n-1}b + \frac{n(n-1)}{2!}a^{n-1}b^2 + \cdots + \frac{n(n-1)\cdots(n-k+1)}{k!}a^{n-k}b^k + \cdots + nab^{n-1} + b^n \end{aligned} (a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2(a+b+c)2=a2+b2+c2+2ab+2ac+2bc(a+b)3=a3+3a2b+3ab2+b3(a−b)3=a3−3a2b+3ab2−b3(a+b)(a−b)=a2−b2a3+b3=(a+b)(a2−ab+b2)a3−b3=(a−b)(a2+ab+b2)an−bn=(a−b)(an−1+an−2b+⋯+abn−2+bn−1)(a+b)n=k=0∑nCnkakbn−k=an+nan−1b+2!n(n−1)an−1b2+⋯+k!n(n−1)⋯(n−k+1)an−kbk+⋯+nabn−1+bn
n ! = 1 × 2 × 3 × . . . × n ( 规 定 0 ! = 1 ) ( 2 n ) ! ! = 2 × 4 × 6 × . . . × ( 2 n ) = 2 n ⋅ n ! ( 2 n − 1 ) ! ! = 1 × 3 × 5... × ( 2 n − 1 ) \begin{aligned} & n! = 1\times2\times3\times ... \times n ~~~~~(规定0!=1) \\ \\ & (2n)!! = 2\times4\times6\times ... \times (2n) = 2^n \cdot n! \\ \\ & (2n-1)!! = 1\times3\times5...\times(2n-1) \end{aligned} n!=1×2×3×...×n (规定0!=1)(2n)!!=2×4×6×...×(2n)=2n⋅n!(2n−1)!!=1×3×5...×(2n−1)
定 义 在 [ − a , a ] 上 的 任 一 函 数 , 可 以 表 示 为 一 个 奇 函 数 与 一 个 偶 函 数 之 和 : f ( x ) = 1 2 [ f ( x ) − f ( − x ) ] + 1 2 [ f ( x ) + f ( − x ) ] \begin{aligned} & 定义在[-a,a]上的任一函数,可以表示为一个奇函数与一个偶函数之和: \\ \\ & f(x) = \frac{1}{2}[f(x)-f(-x)] + \frac{1}{2}[f(x) + f(-x)] \end{aligned} 定义在[−a,a]上的任一函数,可以表示为一个奇函数与一个偶函数之和:f(x)=21[f(x)−f(−x)]+21[f(x)+f(−x)]
A n m = n ( n − 1 ) ( n − 2 ) ⋯ ( n − m + 1 ) = n ! ( n − m ) ! C n m = A n m m ! = n ( n − 1 ) ⋯ ( n − m + 1 ) m ! = n ! m ! ( n − m ) ! \begin{aligned} A_n^m & = n(n-1)(n-2)\cdots(n-m +1) \\\\ & = \frac{n!}{(n-m)!} \\\\ \\ C_n^m & = \frac{A_n^m}{m!} = \frac{n(n-1)\cdots(n-m + 1)}{m!} \\\\ & = \frac{n!}{m!(n-m)!} \end{aligned} AnmCnm=n(n−1)(n−2)⋯(n−m+1)=(n−m)!n!=m!Anm=m!n(n−1)⋯(n−m+1)=m!(n−m)!n!
a n = a 1 + ( n − 1 ) d S n = n a 1 + n ( n − 1 ) 2 d n ∈ N ∗ S n = n ( a 1 + a n ) 2 \begin{aligned} & a_n = a_1 + (n-1)d \\ \\ & S_n = na_1 + \frac{n(n-1)}{2}d ~~~~~~~~ n \in N^* \\ \\ & S_n = \frac{n(a_1+a_n)}{2} \\ \\ \end{aligned} an=a1+(n−1)dSn=na1+2n(n−1)d n∈N∗Sn=2n(a1+an)
a n = a 1 ⋅ q n − 1 S n = a 1 ( 1 − q n ) 1 − q ( q ≠ 1 ) \begin{aligned} & a_n = a_1 \cdot q^{n-1} \\ \\ & S_n = \frac{a_1(1-q^n)}{1-q} ~~~~~~(q\neq 1) \end{aligned} an=a1⋅qn−1Sn=1−qa1(1−qn) (q=1)
∑ k = 1 n k = 1 + 2 + 3 + ⋯ + n = n ( n + 1 ) 2 ∑ k = 1 n ( 2 k − 1 ) = 1 + 3 + 5 + ⋯ + ( 2 n − 1 ) = n 2 ∑ k = 1 n k 2 = 1 2 + 2 2 + 3 2 + ⋯ + n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 ∑ k = 1 n k 3 = 1 3 + 2 3 + 3 3 + ⋯ + n 3 = [ n ( n + 1 ) 2 ] 2 = ( ∑ k = 1 n k ) 2 ∑ k = 1 n k ( k + 1 ) = 1 × 2 + 2 × 3 + 3 × 4 + ⋯ + n ( n + 1 ) = n ( n + 1 ) ( n + 2 ) 3 ∑ k = 1 n 1 k ( k + 1 ) = 1 1 × 2 + 1 2 × 3 + 1 3 × 4 + ⋯ + 1 n ( n + 1 ) = n n + 1 \begin{aligned} & \\ & \sum_{k=1}^n k = 1 + 2+3+\cdots + n=\frac{n(n+1)}{2} \\ \\ & \sum_{k=1}^n (2k-1) = 1+ 3 + 5 + \cdots + (2n-1) = n^2 \\ \\ & \sum_{k=1}^n k^2 = 1^2+2^2+3^2+\cdots +n^2 = \frac{n(n+1)(2n+1)}{6} \\ \\ & \sum_{k=1}^n k^3 = 1^3 + 2^3 +3^3 +\cdots + n^3 = [\frac{n(n+1)}{2}]^2 = (\sum_{k=1}^n k)^2 \\ \\ & \sum_{k=1}^n k(k+1) = 1 \times 2 + 2 \times 3 + 3 \times 4 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3} \\\\ & \sum_{k=1}^n \frac{1}{k(k+1)} = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1} \end{aligned} k=1∑nk=1+2+3+⋯+n=2n(n+1)k=1∑n(2k−1)=1+3+5+⋯+(2n−1)=n2k=1∑nk2=12+22+32+⋯+n2=6n(n+1)(2n+1)k=1∑nk3=13+23+33+⋯+n3=[2n(n+1)]2=(k=1∑nk)2k=1∑nk(k+1)=1×2+2×3+3×4+⋯+n(n+1)=3n(n+1)(n+2)k=1∑nk(k+1)1=1×21+2×31+3×41+⋯+n(n+1)1=n+1n
2 ∣ a b ∣ ≤ a 2 + b 2 ∣ a ± b ∣ ≤ ∣ a ∣ + ∣ b ∣ ∣ ∣ a ∣ − ∣ b ∣ ∣ ≤ ∣ a − b ∣ ∣ a 1 ± a 2 ± ⋅ ⋅ ⋅ ⋅ ± a n ∣ ≤ ∣ a 1 ∣ + ∣ a 2 ∣ + ⋅ ⋅ ⋅ + ∣ a n ∣ ∣ ∫ a b f ( x ) d x ∣ ≤ ∫ a b ∣ f ( x ) ∣ d x ( a < b ) a b ≤ a + b 2 ≤ a 2 + b 2 2 ( a , b > 0 ) a b c 3 ≤ a + b + c 3 ≤ a 2 + b 2 + c 2 3 ( a , b , c > 0 ) a 1 a 2 ⋅ ⋅ ⋅ a n n ≤ a 1 + a 2 + . . . + a n n ≤ a 1 2 + a 2 2 + . . . + a n 2 n ( a 1 , a 2 , . . . a n > 0 , 等 号 当 且 仅 当 a 1 = a 2 = . . . = a n 时 成 立 ) x y ≤ x p p + x q q ( x , y , p , q > 0 , 1 p + 1 q = 1 ) ( a c + b d ) 2 ≤ ( a 2 + b 2 ) ( c 2 + d 2 ) ( a 1 b 1 + a 2 b 2 + a 3 b 3 ) 2 ≤ ( a 1 2 + a 2 2 + a 3 2 ) ( b 1 2 + b 2 2 + b 3 2 ) [ ∫ a b f ( x ) ⋅ g ( x ) d x ] 2 ≤ ∫ a b f 2 ( x ) d x ⋅ ∫ a b g 2 ( x ) d x sin x < x < tan x ( 0 < x < π 2 ) arctan x ≤ x ≤ arcsin x ( 0 ≤ x ≤ 1 ) x + 1 ≤ e x ln x ≤ x − 1 1 1 + x < ln ( 1 + 1 x ) < 1 x ( x > 0 ) \begin{aligned} & \\ & 2 |ab| \le a ^ 2 + b^2 \\ \\ & |a \pm b| \le |a| + |b| \\ \\ & | |a| - |b| | \le |a-b| \\ \\ & |a_1 \pm a_2 \pm \cdot\cdot\cdot\cdot \pm a_n| \le |a_1| + |a_2| + \cdot\cdot\cdot + |a_n| \\ \\ & |\int_a^b f(x) dx| \le \int_a^b |f(x)| dx ~~~~~(a0) \\ \\ & \sqrt[3]{abc} \le \frac{a+b+c}{3} \le \sqrt{\frac{a^2+b^2+c^2}{3}} ~~~~~(a,b,c>0) \\ \\ & \sqrt[n]{a_1a_2\cdot\cdot\cdot a_n} \le \frac{a_1+a_2+...+a_n}{n} \le \sqrt{\frac{ {a_1}^2+{a_2}^2 + ... + {a_n}^2}{n}} ~~~~(a_1,a_2,...a_n > 0,等号当且仅当 a_1 = a_2 = ... = a_n时成立) \\ \\ & xy \le \frac{x^p}{p} + \frac{x^q}{q} ~~~~~(x,y,p,q>0, \frac{1}{p}+\frac{1}{q}=1) \\ \\ & (ac+bd)^2 \le (a^2+b^2)(c^2+d^2) \\ \\ & (a_1 b_1 + a_2 b_2 + a_3 b_3)^2 \le ({a_1}^2 + {a_2}^2 + {a_3}^2)({b_1}^2 + {b_2}^2 + {b_3}^2) \\ \\ & [\int_a^b f(x)\cdot g(x) dx]^2 \le \int_a^bf^2(x)dx \cdot\int_a^bg^2(x)dx \\ \\ & \sin x < x < \tan x ~~~~~(0
sin ( − α ) = − sin α cos ( − α ) = cos α sin ( π 2 − α ) = cos α cos ( π 2 − α ) = sin α sin ( π 2 + α ) = cos α cos ( π 2 + α ) = − sin α sin ( π − α ) = sin α cos ( π − α ) = − cos α sin ( π + α ) = − sin α cos ( π + α ) = − cos α 奇 变 偶 不 变 , 符 号 看 象 限 奇 指 k ⋅ π 2 中 k 看 象 限 是 指 : 将 α 看 成 锐 角 , 然 后 看 sin ( 变 之 前 的 , 或 c o s ) ( k ⋅ π 2 ± α ) 的 符 号 \begin{aligned} & \sin (-\alpha) = -\sin \alpha \\ \\ & \cos (-\alpha) = \cos \alpha \\ \\ & \sin (\frac{\pi}{2} - \alpha) = \cos \alpha \\ \\ & \cos (\frac{\pi}{2} - \alpha) = \sin \alpha \\ \\ & \sin (\frac{\pi}{2} + \alpha) = \cos \alpha \\ \\ & \cos (\frac{\pi}{2} + \alpha) = - \sin \alpha \\ \\ & \sin (\pi - \alpha) = \sin \alpha \\ \\ & \cos (\pi - \alpha) = - \cos \alpha \\ \\ & \sin (\pi + \alpha) = - \sin \alpha \\ \\ & \cos (\pi + \alpha) = - \cos \alpha \\ \\ \\ & 奇变偶不变,符号看象限 \\ & 奇指 ~~ k\cdot\frac{\pi}{2} ~~ 中 k \\ & 看象限是指:将 \alpha 看成锐角,然后看 \sin_{_{(变之前的,或cos)}}(k\cdot\frac{\pi}{2} \pm \alpha) 的符号 \end{aligned} sin(−α)=−sinαcos(−α)=cosαsin(2π−α)=cosαcos(2π−α)=sinαsin(2π+α)=cosαcos(2π+α)=−sinαsin(π−α)=sinαcos(π−α)=−cosαsin(π+α)=−sinαcos(π+α)=−cosα奇变偶不变,符号看象限奇指 k⋅2π 中k看象限是指:将α看成锐角,然后看sin(变之前的,或cos)(k⋅2π±α)的符号
1 + tan 2 α = sec 2 α 1 + cot 2 α = csc 2 α sin 2 α + cos 2 α = 1 \begin{aligned} & \\ & 1 + \tan ^2 \alpha = \sec^2 \alpha \\ \\ & 1 + \cot^2 \alpha = \csc ^2 \alpha \\ \\ & \sin^2 \alpha + \cos ^2 \alpha = 1 \end{aligned} 1+tan2α=sec2α1+cot2α=csc2αsin2α+cos2α=1
sin ( α + β ) = sin α cos β + cos α sin β cos ( α + β ) = cos α cos β − sin α sin β sin ( α − β ) = sin α cos β − cos α sin β cos ( α − β ) = cos α cos β + sin α cos β tan ( α + β ) = tan α + tan β 1 − tan α tan β tan ( α − β ) = tan α − tan β 1 + tan α tan β \begin{aligned} & \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \\ & \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \\ & \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \\ & \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \cos \beta \\ \\ & \tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{ 1- \tan \alpha \tan \beta} \\ \\ & \tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1+ \tan \alpha \tan \beta} \end{aligned} sin(α+β)=sinαcosβ+cosαsinβcos(α+β)=cosαcosβ−sinαsinβsin(α−β)=sinαcosβ−cosαsinβcos(α−β)=cosαcosβ+sinαcosβtan(α+β)=1−tanαtanβtanα+tanβtan(α−β)=1+tanαtanβtanα−tanβ
cos α cos β = 1 2 [ cos ( α + β ) + c o s ( α − β ) ] cos α sin β = 1 2 [ sin ( α + β ) − sin ( α − β ) ] sin α cos β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] sin α sin β = − 1 2 [ cos ( α + β ) − cos ( α − β ) ] \begin{aligned} & \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} [\sin (\alpha + \beta) + \sin (\alpha - \beta)] \\ \\ & \sin \alpha \sin \beta = - \frac{1}{2} [\cos (\alpha + \beta) - \cos (\alpha - \beta)] \end{aligned} cosαcosβ=21[cos(α+β)+cos(α−β)]cosαsinβ=21[sin(α+β)−sin(α−β)]sinαcosβ=21[sin(α+β)+sin(α−β)]sinαsinβ=−21[cos(α+β)−cos(α−β)]
记忆口诀:
积化和差得和差,余弦在后要想加。异名函数取正弦,正弦相乘取负号。
sin α + sin β = 2 sin α + β 2 cos α − β 2 sin α − sin β = 2 cos α + β 2 sin α − β 2 cos α + cos β = 2 cos α + β 2 cos α − β 2 cos α − cos β = − 2 sin α + β 2 sin α − β 2 \begin{aligned} & \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} \end{aligned} sinα+sinβ=2sin2α+βcos2α−βsinα−sinβ=2cos2α+βsin2α−βcosα+cosβ=2cos2α+βcos2α−βcosα−cosβ=−2sin2α+βsin2α−β
记忆口诀:
正加正,正在前,正减正,余在前;余加余,余并肩;余减余,负正弦
sin 2 α = 2 sin α cos α cos 2 α = cos 2 α − sin 2 α = 1 − 2 sin 2 α = 2 cos 2 α − 1 sin 3 α = − 4 sin 3 α + 3 sin α cos 3 α = 4 cos 3 α − 3 cos α sin 2 α = 1 − cos 2 α 2 cos 2 α = 1 + cos 2 α 2 tan 2 α = 2 tan α 1 − tan 2 α cot 2 α = cot 2 α − 1 2 cot α \begin{aligned} & \sin 2\alpha = 2\sin \alpha \cos \alpha \\ \\ & \cos 2\alpha = \cos ^2 \alpha - \sin ^2 \alpha = 1- 2\sin^2 \alpha = 2 \cos^2\alpha -1 \\ \\ & \sin 3\alpha = -4 \sin^3 \alpha + 3\sin \alpha \\ \\ & \cos 3 \alpha = 4\cos^3\alpha -3 \cos \alpha \\ \\ & \sin^2 \alpha = \frac{1-\cos 2\alpha}{2} \\ \\ & \cos^2 \alpha = \frac{1+\cos 2\alpha}{2} \\ \\ & \tan 2\alpha = \frac{2\tan \alpha}{1-\tan ^2\alpha} \\ \\ & \cot 2\alpha = \frac{\cot ^2 \alpha -1}{2\cot \alpha} \end{aligned} sin2α=2sinαcosαcos2α=cos2α−sin2α=1−2sin2α=2cos2α−1sin3α=−4sin3α+3sinαcos3α=4cos3α−3cosαsin2α=21−cos2αcos2α=21+cos2αtan2α=1−tan2α2tanαcot2α=2cotαcot2α−1
sin 2 α 2 = 1 − cos α 2 cos 2 α 2 = 1 + cos α 2 sin α 2 = ± 1 − cos α 2 cos α 2 = ± 1 + cos α 2 tan α 2 = 1 − cos α sin α = sin α 1 + cos α = ± 1 − cos α 1 + cos α cot α 2 = sin α 1 − cos α = 1 + cos α sin α = ± 1 + cos α 1 − cos α \begin{aligned} & \sin^2 \frac{\alpha}{2} = \frac{1- \cos \alpha}{2} \\ \\ & \cos^2 \frac{\alpha}{2} = \frac{1+\cos \alpha}{2} \\ \\ & \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1- \cos \alpha}{2}} \\ \\ & \cos \frac{\alpha}{2} = \pm \sqrt{\frac{1+ \cos \alpha}{2}} \\ \\ & \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}} \end{aligned} sin22α=21−cosαcos22α=21+cosαsin2α=±21−cosαcos2α=±21+cosαtan2α=sinα1−cosα=1+cosαsinα=±1+cosα1−cosαcot2α=1−cosαsinα=sinα1+cosα=±1−cosα1+cosα
sin α = 2 tan α 2 1 + tan 2 α 2 cos α = 1 − tan 2 α 2 1 + tan 2 α 2 \begin{aligned} & \sin \alpha = \frac{2 \tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} \\ \\ & \cos \alpha = \frac{1- \tan^2 \frac{\alpha}{2}}{1+ \tan^2 \frac{\alpha}{2}} \end{aligned} sinα=1+tan22α2tan2αcosα=1+tan22α1−tan22α
1 + sin α = ( sin α 2 + cos α 2 ) 2 1 − sin α = ( sin α 2 − cos α 2 ) 2 \begin{aligned} & 1 + \sin \alpha = (\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}) ^2 \\ \\ & 1 - \sin \alpha = (\sin \frac{\alpha}{2} - \cos \frac{\alpha}{2}) ^2 \end{aligned} 1+sinα=(sin2α+cos2α)21−sinα=(sin2α−cos2α)2
arcsin x + arccos x = π 2 arctan x + a r c c o t x = π 2 sin ( arccos x ) = 1 − x 2 cos ( arcsin x ) = 1 − x 2 sin ( arcsin x ) = x arcsin ( sin x ) = x cos ( arccos x ) = x arccos ( cos x ) = x arccos ( − x ) = π − arccos x \begin{aligned} & \arcsin x + \arccos x = \frac{\pi}{2} \\ \\ & \arctan x + arccot ~x= \frac{\pi}{2} \\ \\ & \sin(\arccos x) = \sqrt{1-x^2} \\ \\ & \cos(\arcsin x) = \sqrt{1- x^2} \\ \\ & \sin(\arcsin x) = x \\ \\ & \arcsin (\sin x) = x \\ \\ & \cos (\arccos x) = x \\ \\ & \arccos (\cos x) =x \\ \\ & \arccos (-x) = \pi - \arccos x \end{aligned} arcsinx+arccosx=2πarctanx+arccot x=2πsin(arccosx)=1−x2cos(arcsinx)=1−x2sin(arcsinx)=xarcsin(sinx)=xcos(arccosx)=xarccos(cosx)=xarccos(−x)=π−arccosx
a n + 1 = f ( a n ) 结 论 一 : f ′ ( x ) > 0 , { a 2 > a 1 ⟹ { a n } ↗ 单 调 递 增 a 2 < a 1 ⟹ { a n } ↘ 单 调 递 减 结 论 二 ( 压 缩 映 像 原 理 ) : ∃ k ∈ ( 0 , 1 ) , 使 得 ∣ f ′ ( x ) ∣ ≤ k ⟹ a n 收 敛 \begin{aligned} & a_{n+1} = f(a_n) \\\\ 结论一: & f'(x) > 0 , \begin{cases} a_2 > a_1 \implies \{ a_n \} \nearrow单调递增 \\ a_2 < a_1 \implies \{ a_n \} \searrow单调递减 \end{cases} \\\\ 结论二(压缩映像原理):& \exist k \in (0,1),使得 |f'(x)| \le k \implies {a_n} 收敛 \end{aligned} 结论一:结论二(压缩映像原理):an+1=f(an)f′(x)>0,{ a2>a1⟹{ an}↗单调递增a2<a1⟹{ an}↘单调递减∃k∈(0,1),使得∣f′(x)∣≤k⟹an收敛
lim x → 0 + x α ln x = 0 其 中 α > 0 lim x → 0 + x α ( ln x ) k = 0 其 中 α > 0 , k > 0 lim x → + ∞ x α e − δ x = 0 其 中 α > 0 , δ > 0 lim x → 0 sin x x = 1 ⟹ lim ϕ ( x ) → 0 sin ϕ ( x ) ϕ ( x ) = 1 其 中 ϕ ( x ) ≠ 0 lim x → 0 ( 1 + x ) 1 x = e ⟹ lim ϕ ( x ) → 0 ( 1 + ϕ ( x ) ) 1 ϕ ( x ) = e 其 中 ϕ ( x ) ≠ 0 lim n → ∞ n n = 1 lim n → ∞ a n = 1 ( 常 数 a > 0 ) \begin{aligned} & \lim_{x \to 0^+} x^\alpha \ln x = 0 ~~~~~ 其中 \alpha >0 \\\\ & \lim_{x \to 0^+} x^\alpha (\ln x)^k = 0 ~~~~~ 其中 \alpha >0 ,k>0 \\\\ & \lim_{x \to +\infty} x^\alpha e^{-\delta x} = 0 ~~~~~ 其中 \alpha >0 ,\delta >0 \\\\ & \lim_{x\to 0} \frac{\sin x}{x} = 1 ~~ \implies \lim_{\phi (x) \to 0} \frac{\sin \phi (x)}{\phi (x)} =1 ~~~~~其中\phi (x) \neq 0 \\ \\ & \lim_{x \to 0} (1+x)^{\frac{1}{x}} = e ~~ \implies \lim_{\phi (x) \to 0} (1+\phi (x))^{\frac{1}{\phi (x)}} = e ~~~~~ 其中\phi (x) \neq 0 \\\\ & \lim_{n \to \infty} \sqrt[n]{n} = 1 \\\\ & \lim_{n \to \infty} \sqrt[n]{a} = 1 ~~~(常数a>0)\\\\ \end{aligned} x→0+limxαlnx=0 其中α>0x→0+limxα(lnx)k=0 其中α>0,k>0x→+∞limxαe−δx=0 其中α>0,δ>0x→0limxsinx=1 ⟹ϕ(x)→0limϕ(x)sinϕ(x)=1 其中ϕ(x)=0x→0lim(1+x)x1=e ⟹ϕ(x)→0lim(1+ϕ(x))ϕ(x)1=e 其中ϕ(x)=0n→∞limnn=1n→∞limna=1 (常数a>0)
x → 0 时 , sin x ∼ tan x ∼ arcsin x ∼ arctan x ∼ ( e x − 1 ) ∼ ln ( 1 + x ) ∼ x , 1 − cos x ∼ 1 2 x 2 , ( 1 + x ) a − 1 ∼ a x , a x − 1 ∼ x ln a ( a > 0 , a ≠ 1 ) \begin{aligned} & x \to 0 时,\\\\ & \sin x \sim \tan x \sim \arcsin x \sim \arctan x \sim (e^x - 1) \sim \ln(1+x) \sim x ~~,~~ 1- \cos x \sim \frac{1}{2} x^2 ~~, \\ \\ & (1+x)^a - 1 \sim ax ~~,~~ a^x - 1 \sim x\ln a ~~~(a>0,a\neq1) \end{aligned} x→0时,sinx∼tanx∼arcsinx∼arctanx∼(ex−1)∼ln(1+x)∼x , 1−cosx∼21x2 ,(1+x)a−1∼ax , ax−1∼xlna (a>0,a=1)
lim u v = e lim ( u − 1 ) v ( 其 中 lim u = 1 , lim v = ∞ , 即 1 ∞ 型 ) \lim u^v = e^{\lim (u-1)v} ~~~~~~(其中 \lim u=1,\lim v=\infty,即 1^\infty 型) limuv=elim(u−1)v (其中limu=1,limv=∞,即1∞型)
f ′ ( x 0 ) = lim Δ x → 0 f ( x 0 + Δ x ) − f ( x 0 ) Δ x f ′ ( x 0 ) = lim x → x 0 f ( x ) − f ( x 0 ) x − x 0 \begin{aligned} & f'(x_0) = \lim\limits_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} \\ \\ & f'(x_0) = \lim\limits_{x \to x_0} \frac{f(x) - f(x_0)}{x-x_0} \end{aligned} f′(x0)=Δx→0limΔxf(x0+Δx)−f(x0)f′(x0)=x→x0limx−x0f(x)−f(x0)
Δ y = f ( x 0 + Δ x ) − f ( x 0 ) Δ y = A Δ x + o ( Δ x ) A Δ x = f ′ ( x 0 ) Δ x \begin{aligned} & \Delta y = f(x_0 + \Delta x) - f(x_0) \\ \\ & \Delta y = A\Delta x + o(\Delta x) \\ \\ & A\Delta x = f'(x_0) \Delta x \end{aligned} Δy=f(x0+Δx)−f(x0)Δy=AΔx+o(Δx)AΔx=f′(x0)Δx
[ u ( x ) ± v ( x ) ] ′ = u ′ ( x ) ± v ′ ( x ) [ u ( x ) v ( x ) ] ′ = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) [ u ( x ) v ( x ) w ( x ) ] ′ = u ′ ( x ) v ( x ) w ( x ) + u ( x ) v ′ ( x ) w ( x ) + u ( x ) v ( x ) w ′ ( x ) [ u ( x ) v ( x ) ] ′ = u ′ ( x ) v ( x ) − u ( x ) v ′ ( x ) [ v ( x ) ] 2 \begin{aligned} & [u(x) \pm v(x)]' = u'(x) \pm v'(x) \\ \\ & [u(x)v(x)]' = u'(x)v(x) + u(x)v'(x) \\ \\ & [u(x)v(x)w(x)]' = u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x) \\ \\ & \begin{bmatrix}\frac{u(x)}{v(x)} \end{bmatrix}' = \frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \\ \\ \end{aligned} [u(x)±v(x)]′=u′(x)±v′(x)[u(x)v(x)]′=u′(x)v(x)+u(x)v′(x)[u(x)v(x)w(x)]′=u′(x)v(x)w(x)+u(x)v′(x)w(x)+u(x)v(x)w′(x)[v(x)u(x)]′=[v(x)]2u′(x)v(x)−u(x)v′(x)
{ f [ g ( x ) ] } ′ = f ′ [ g ( x ) ] g ′ ( x ) \{ f[g(x)] \}' = f'[g(x)]g'(x) { f[g(x)]}′=f′[g(x)]g′(x)
y = f ( x ) , x = φ ( y ) ⟹ φ ′ ( y ) = 1 f ′ ( x ) y x ′ = d y d x = 1 d x d y = 1 x y ′ y x x ′ ′ = d 2 y d x 2 = d ( d y d x ) d x = d ( 1 x y ′ ) d x = d ( 1 x y ′ ) d y ⋅ d y d x = d ( 1 x y ′ ) d y ⋅ 1 x y ′ = − x y y ′ ′ ( x y ′ ) 3 \begin{aligned} & y = f(x), x = \varphi(y) \implies \varphi ' (y) = \frac{1}{f'(x)} \\ \\ & y'_x = \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{x'_y} \\ \\ & y^{''}_{xx} = \frac{d^2 y}{dx^2} = \frac{d(\frac{dy}{dx})}{dx} = \frac{d(\frac{1}{x'_y})}{dx} = \frac{d(\frac{1}{x'_y})}{dy} \cdot \frac{dy}{dx} = \frac{d(\frac{1}{x'_y})}{dy} \cdot \frac{1}{x'_y} = \frac{-x^{''}_{yy}}{(x'_y)^3} \end{aligned} y=f(x),x=φ(y)⟹φ′(y)=f′(x)1yx′=dxdy=dydx1=xy′1yxx′′=dx2d2y=dxd(dxdy)=dxd(xy′1)=dyd(xy′1)⋅dxdy=dyd(xy′1)⋅xy′1=(xy′)3−xyy′′
{ x = φ ( t ) y = ψ ( t ) d y d x = d y / d t d x / d t = ψ ′ ( t ) φ ′ ( t ) d 2 y d x 2 = d ( d y d x ) d x = d ( d y d x ) / d t d x / d t = ψ ′ ′ ( t ) φ ′ ( t ) − ψ ′ ( t ) φ ′ ′ ( t ) [ φ ′ ( t ) ] 3 \begin{aligned} & \begin{cases} x = \varphi (t) \\ y = \psi (t) \end{cases} \\\\ & \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\psi ' (t)}{\varphi ' (t)} \\ \\ & \frac{d^2 y}{dx^2} = \frac{d(\frac{dy}{dx})}{dx} = \frac {d(\frac{dy}{dx})/dt}{dx/dt} = \frac{\psi '' (t) \varphi '(t) - \psi '(t) \varphi '' (t) }{[\varphi ' (t)]^3} \end{aligned} { x=φ(t)y=ψ(t)dxdy=dx/dtdy/dt=φ′(t)ψ′(t)dx2d2y=dxd(dxdy)=dx/dtd(dxdy)/dt=[φ′(t)]3ψ′′(t)φ′(t)−ψ′(t)φ′′(t)
设 F ( x ) = ∫ φ 1 ( x ) φ 2 ( x ) f ( t ) d t , 则 F ′ ( x ) = d d x [ ∫ φ 1 ( x ) φ 2 ( x ) f ( t ) d t ] = f [ φ 2 ( x ) ] φ 2 ′ ( x ) − f [ φ 1 ( x ) ] φ 1 ′ ( x ) \begin{aligned} & 设 F(x) = \int ^{\varphi_2(x)}_{\varphi_1(x)} f(t) dt, 则 \\ \\ & F'(x) = \frac{d}{dx}\begin{bmatrix}\int ^{\varphi_2(x)}_{\varphi_1(x)} f(t) dt \end{bmatrix} = f[\varphi _2(x)]\varphi '_2(x) - f[\varphi_1(x)]\varphi '_1(x) \end{aligned} 设F(x)=∫φ1(x)φ2(x)f(t)dt,则F′(x)=dxd[∫φ1(x)φ2(x)f(t)dt]=f[φ2(x)]φ2′(x)−f[φ1(x)]φ1′(x)
( x a ) ′ = a x a − 1 ( a 为 常 数 ) ( a x ) ′ = a x ln a ( e x ) ′ = e x ( l o g a x ) ′ = 1 x ln a ( a > 0 , a ≠ 1 ) ( ln x ) ′ = 1 x ( sin x ) ′ = cos x ( cos x ) ′ = − sin x ( arcsin x ) ′ = 1 1 − x 2 ( arccos x ) ′ = − 1 1 − x 2 ( tan x ) ′ = sec 2 x ( cot x ) ′ = − csc 2 x ( arctan x ) ′ = 1 1 + x 2 ( a r c c o t x ) ′ = − 1 1 + x 2 ( sec x ) ′ = sec x ⋅ tan x ( csc x ) ′ = − csc x ⋅ cot x [ ln ( x + x 2 + 1 ) ] ′ = 1 x 2 + 1 [ ln ( x + x 2 − 1 ) ] ′ = 1 x 2 − 1 \begin{aligned} & (x^a)' = a x^{a-1} ~~~~~(a为常数) \\ \\ & (a^x)' = a^x \ln a \\ \\ & (e^x)' = e^x \\ \\ & (log_a x)' = \frac{1}{x \ln a} ~~~~~(a>0, a \ne 1) \\ \\ & (\ln x)' = \frac{1}{x} \\ \\ & (\sin x)' = \cos x \\ \\ & (\cos x)' = -\sin x \\ \\ & (\arcsin x)' = \frac{1}{\sqrt{1-x^2}} \\ \\ & (\arccos x)' = - \frac{1}{\sqrt{1-x^2}} \\ \\ & (\tan x)' = \sec ^2 x \\ \\ & (\cot x)' = - \csc^2 x \\ \\ & (\arctan x)' = \frac{1}{1+x^2} \\ \\ & (arccot ~ x)' = - \frac{1}{1+x^2} \\ \\ & (\sec x)' = \sec x \cdot \tan x \\ \\ & (\csc x)' = - \csc x \cdot \cot x \\ \\ & [\ln (x+\sqrt{x^2+1})]' = \frac{1}{\sqrt{x^2+1}} \\ \\ & [\ln (x+\sqrt{x^2-1})]' = \frac{1}{\sqrt{x^2-1}} \\ \\ \end{aligned} (xa)′=axa−1 (a为常数)(ax)′=axlna(ex)′=ex(logax)′=xlna1 (a>0,a=1)(lnx)′=x1(sinx)′=cosx(cosx)′=−sinx(arcsinx)′=1−x21(arccosx)′=−1−x21(tanx)′=sec2x(cotx)′=−csc2x(arctanx)′=1+x21(arccot x)′=−1+x21(secx)′=secx⋅tanx(cscx)′=−cscx⋅cotx[ln(x+x2+1)]′=x2+11[ln(x+x2−1)]′=x2−11
[ u ± v ] ( n ) = u ( n ) ± v ( n ) [u \pm v ]^{(n)} = u^{(n)} \pm v^{(n)} [u±v](n)=u(n)±v(n)
( u v ) ( n ) = u ( n ) v + C n 1 u ( n − 1 ) v ′ + C n 2 u ( n − 2 ) v ′ ′ + . . . + C n k u ( n − k ) v ( k ) + . . . + C n n − 1 u ′ v ( n − 1 ) + u v ( n ) = ∑ k = 0 n C n k u ( n − k ) v ( k ) \begin{aligned} (uv)^{(n)} & = u^{(n)}v + C_n^1 u^{(n-1)}v' + C_n^2 u^{(n-2)}v'' + ... + C_n^k u^{(n-k)}v^{(k)} + ... + C_n^{n-1} u'v^{(n-1)} + uv^{(n)} \\ & = \displaystyle\sum_{k=0}^n C_n^k u^{(n-k)}v^{(k)} \end{aligned} (uv)(n)=u(n)v+Cn1u(n−1)v′+Cn2u(n−2)v′′+...+Cnku(n−k)v(k)+...+Cnn−1u′v(n−1)+uv(n)=k=0∑nCnku(n−k)v(k)
( a x ) ( n ) = a x ( ln a ) n ( e x ) ( n ) = e x ( sin k x ) ( n ) = k n sin ( k x + n ⋅ π 2 ) ( cos k x ) ( n ) = k n cos ( k x + n ⋅ π 2 ) ( ln x ) ( n ) = ( − 1 ) n − 1 ( n − 1 ) ! x n ( x > 0 ) [ ln ( 1 + x ) ] ( n ) = ( − 1 ) n − 1 ( n − 1 ) ! ( x + 1 ) n ( x > − 1 ) [ ( x + x 0 ) m ] ( n ) = m ( m − 1 ) ( m − 2 ) ⋅ ⋅ ⋅ ⋅ ( m − n + 1 ) ( x + x 0 ) m − n ( 1 x + a ) ( n ) = ( − 1 ) n ⋅ n ! ( x + a ) n + 1 \begin{aligned} & (a^x)^{(n)} = a^x (\ln a)^n \\ \\ & (e^x)^{(n)} = e^x \\ \\ & (\sin kx)^{(n)} = k^n \sin (kx + n\cdot \frac{\pi}{2}) \\ \\ & (\cos kx)^{(n)} = k^n \cos (kx + n\cdot \frac{\pi}{2}) \\ \\ & (\ln x) ^ {(n)} = (-1)^{n-1} \frac{(n-1)!}{x^n} ~~~~~(x>0) \\ \\ & [\ln(1+x)]^{(n)} = (-1)^{n-1} \frac{(n-1)!}{(x+1)^n} ~~~~~(x>-1) \\ \\ & [(x+x_0)^m]^{(n)} = m(m-1)(m-2)\cdotp\cdotp\cdotp\cdot (m-n+1)(x+x_0)^{m-n} \\ \\ & (\frac{1}{x+a})^{(n)} = \frac{(-1)^n \cdot n!}{(x+a)^{n+1}} \\ \\ \end{aligned} (ax)(n)=ax(lna)n(ex)(n)=ex(sinkx)(n)=knsin(kx+n⋅2π)(coskx)(n)=kncos(kx+n⋅2π)(lnx)(n)=(−1)n−1xn(n−1)! (x>0)[ln(1+x)](n)=(−1)n−1(x+1)n(n−1)! (x>−1)[(x+x0)m](n)=m(m−1)(m−2)⋅⋅⋅⋅(m−