数学整理1

EveSunMaple Lv3

三角函数的基本公式

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}

cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}

sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}

csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}

sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta

cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta

tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta

cot⁡(−θ)=−cot⁡θ\cot(-\theta) = -\cot\theta

sec⁡(−θ)=sec⁡θ\sec(-\theta) = \sec\theta

csc⁡(−θ)=−csc⁡θ\csc(-\theta) = -\csc\theta

sin⁡(θ+2kπ)=sin⁡θ\sin(\theta + 2k\pi) = \sin\theta

cos⁡(θ+2kπ)=cos⁡θ\cos(\theta + 2k\pi) = \cos\theta

tan⁡(θ+kπ)=tan⁡θ\tan(\theta + k\pi) = \tan\theta

cot⁡(θ+kπ)=cot⁡θ\cot(\theta + k\pi) = \cot\theta

sec⁡(θ+2kπ)=sec⁡θ\sec(\theta + 2k\pi) = \sec\theta

csc⁡(θ+2kπ)=csc⁡θ\csc(\theta + 2k\pi) = \csc\theta

其中kk是任意整数。

sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta

cos⁡2θ=cos⁡2θ−sin⁡2θ\cos2\theta=\cos^2\theta-\sin^2\theta

tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan2\theta=\frac{2\tan \theta}{1 - \tan^2 \theta}

sin⁡(3A)=3sin⁡(A)−4sin⁡3(A)\sin(3A) = 3\sin(A) - 4\sin^3(A)

cos⁡(3A)=4cos⁡3(A)−3cos⁡(A)\cos(3A) = 4\cos^3(A) - 3\cos(A)

sin⁡(α±β)=sin⁡αcos⁡β±cos⁡αsin⁡β\sin(\alpha \pm \beta) = \sin\alpha \cos\beta \pm \cos\alpha \sin\beta

cos⁡(α±β)=cos⁡αcos⁡β∓sin⁡αsin⁡β\cos(\alpha \pm \beta) = \cos\alpha \cos\beta \mp \sin\alpha \sin\beta

tan⁡(α±β)=tan⁡α±tan⁡β1−tan⁡αtan⁡β\tan(\alpha \pm \beta) = \frac{\tan\alpha\pm\tan\beta}{1-\tan\alpha\tan\beta}

sin⁡2θ2=1−cos⁡θ2\sin^2{\frac{\theta}{2}} = {\frac{1 - \cos{\theta}}{2}}

cos⁡2θ2=1+cos⁡θ2\cos^2{\frac{\theta}{2}} = {\frac{1 + \cos{\theta}}{2}}

tan⁡2θ2=1−cos⁡θ1+cos⁡θ\tan^2{\frac{\theta}{2}} = {\frac{1 - \cos{\theta}}{1 + \cos{\theta}}}

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}}

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)]

  • 标题: 数学整理1
  • 作者: EveSunMaple
  • 创建于 : 2023-10-14 00:10:00
  • 更新于 : 2024-02-23 12:02:20
  • 链接: https://old.saroprock.com/post/824ac3d0.html
  • 版权声明: 本文章采用 CC BY-NC-SA 4.0 进行许可。
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数学整理1