数学整理1

EveSunMaple Lv3

三角函数的基本公式

sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1

tanθ=sinθcosθ\tan\theta = \frac{\sin\theta}{\cos\theta}

cotθ=cosθsinθ\cot\theta = \frac{\cos\theta}{\sin\theta}

secθ=1cosθ\sec\theta = \frac{1}{\cos\theta}

cscθ=1sinθ\csc\theta = \frac{1}{\sin\theta}

sin(θ)=sinθ\sin(-\theta) = -\sin\theta

cos(θ)=cosθ\cos(-\theta) = \cos\theta

tan(θ)=tanθ\tan(-\theta) = -\tan\theta

cot(θ)=cotθ\cot(-\theta) = -\cot\theta

sec(θ)=secθ\sec(-\theta) = \sec\theta

csc(θ)=cscθ\csc(-\theta) = -\csc\theta

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

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

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

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

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

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

其中kk是任意整数。

sin2θ=2sinθcosθ\sin2\theta=2\sin\theta\cos\theta

cos2θ=cos2θsin2θ\cos2\theta=\cos^2\theta-\sin^2\theta

tan2θ=2tanθ1tan2θ\tan2\theta=\frac{2\tan \theta}{1 - \tan^2 \theta}

sin(3A)=3sin(A)4sin3(A)\sin(3A) = 3\sin(A) - 4\sin^3(A)

cos(3A)=4cos3(A)3cos(A)\cos(3A) = 4\cos^3(A) - 3\cos(A)

sin(α±β)=sinαcosβ±cosαsinβ\sin(\alpha \pm \beta) = \sin\alpha \cos\beta \pm \cos\alpha \sin\beta

cos(α±β)=cosαcosβsinαsinβ\cos(\alpha \pm \beta) = \cos\alpha \cos\beta \mp \sin\alpha \sin\beta

tan(α±β)=tanα±tanβ1tanαtanβ\tan(\alpha \pm \beta) = \frac{\tan\alpha\pm\tan\beta}{1-\tan\alpha\tan\beta}

sin2θ2=1cosθ2\sin^2{\frac{\theta}{2}} = {\frac{1 - \cos{\theta}}{2}}

cos2θ2=1+cosθ2\cos^2{\frac{\theta}{2}} = {\frac{1 + \cos{\theta}}{2}}

tan2θ2=1cosθ1+cosθ\tan^2{\frac{\theta}{2}} = {\frac{1 - \cos{\theta}}{1 + \cos{\theta}}}

sinA+sinB=2sinA+B2cosAB2\sin A + \sin B = {2}\sin{\frac{A+B}{2}}\cos{\frac{A-B}{2}}

sinAsinB=2cosA+B2sinAB2\sin A - \sin B = {2}\cos{\frac{A+B}{2}}\sin{\frac{A-B}{2}}

cosA+cosB=2cosA+B2cosAB2\cos A + \cos B = {2}\cos{\frac{A+B}{2}}\cos{\frac{A-B}{2}}

cosAcosB=2sinA+B2sinAB2\cos A - \cos B = {-2}\sin{\frac{A+B}{2}}\sin{\frac{A-B}{2}}

sinAcosB=12[sin(A+B)+sin(AB)]\sin A \cos B = {\frac{1}{2}}[\sin(A+B)+ \sin(A-B)]

cosAsinB=12[sin(A+B)sin(AB)]\cos A \sin B = {\frac{1}{2}}[\sin(A+B)- \sin(A-B)]

cosAcosB=12[cos(A+B)+cos(AB)]\cos A \cos B = {\frac{1}{2}}[\cos(A+B)+ \cos(A-B)]

sinAsinB=12[cos(A+B)cos(AB)]\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。
 评论
此页目录
数学整理1