极限

  • 当$X \to 0 时，时，(1+X)^\alpha -1 \sim \alpha X$
  • 当X→1X \to 1时, Xα∼α(X−1)X^\alpha \sim \alpha(X-1)

等价无穷小

  • sin⁡x∼x\sin x \sim x

  • 1−cos⁡x∼12x21-\cos x \sim \frac{1}{2}x^2

  • 1−cos⁡ax∼a2x21-\cos^a x \sim \frac{a}{2}x^2

  • tan⁡x∼x\tan x \sim x

  • arctan⁡x∼x\arctan x \sim x

  • ex−1∼xe^x -1 \sim x

  • ln⁡(1+x)∼x\ln (1+x) \sim x

  • (1+x)α−1∼αx(1+x)^\alpha -1 \sim \alpha x

  • lim⁡A→1ln⁡A∼A−1\lim_{A \to 1} \ln A \sim A-1

三角函数

两角和差公式

  • sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B

  • sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B) = \sin A\cos B - \cos A\sin B

  • cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A\cos B - \sin A\sin B

  • cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A-B) = \cos A\cos B + \sin A\sin B

通过两角和差可推出积化和差和和差化积

积化和差公式

  • sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]\sin A\cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]

  • cos⁡Asin⁡B=12[sin⁡(A+B)−sin⁡(A−B)]\cos A\sin B = \frac{1}{2}[\sin(A+B) - \sin(A-B)]

  • cos⁡Acos⁡B=12[cos⁡(A+B)+cos⁡(A−B)]\cos A\cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]

  • sin⁡Asin⁡B=−12[cos⁡(A+B)−cos⁡(A−B)]\sin A\sin B = -\frac{1}{2}[\cos(A+B) - \cos(A-B)]

和差化积公式

  • sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}

  • sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}

  • cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}

  • cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}

泰勒展开

  • sin⁡x=x−13!x3+15!x5−17!+o(x7)\sin x = x - \frac{1}{3!}x^3+ \frac{1}{5!}x^5 - \frac{1}{7!} + o(x^7)

  • cos⁡x=1−12!x2+14!x4−16!x6+o(x6)\cos x = 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+o(x^6)

  • tan⁡x=x+13x3+215x5+o(x5)\tan x = x+\frac{1}{3}x^3+\frac{2}{15}x^5+o(x^5)

  • ex=1+x+12!x2+1n!xn+o(xn)e^x = 1+x+\frac{1}{2!}x^2+\frac{1}{n!}x^n+o(x^n)