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[函数] sin曲线与类sin曲线

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青青子衿 Posted 2014-6-22 17:04 |Read mode
explore.renren.com/item/5123a91ac8f2c2b6bcaf4e20
谁能给出类sin曲线的解析式?

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 Author| 青青子衿 Posted 2014-6-25 16:45
回复 1# 青青子衿
original.gif
其余来个显而易见,第三个:
\[y = \left\{ \begin{gathered}
  \tan x &   \left( { - \frac{\pi }{6},\frac{\pi }{6}} \right] \\
  \frac{{2\tan x}}{{\sqrt 3 \tan x + 1}} & \left( {\frac{\pi }{6},\frac{\pi }{2}} \right]  \\
  \frac{{2\tan x}}{{\sqrt 3 \tan x - 1}} & \left( {\frac{\pi }{2},\frac{{5\pi }}{6}} \right]  \\
   - \tan x &   \left( {\frac{{5\pi }}{6},\frac{{7\pi }}{6}} \right]  \\
   - \frac{{2\tan x}}{{\sqrt 3 \tan x + 1}} & \left( {\frac{{7\pi }}{6},\frac{{3\pi }}{2}} \right]  \\
  \frac{{2\tan x}}{{1 - \sqrt 3 \tan x}} & \left( {\frac{{3\pi }}{2},\frac{{5\pi }}{3}} \right]\\
\end{gathered}  \right.\]

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hbghlyj Posted 2022-8-12 23:37
链接挂了

假设角速度为1, 右边的$x$轴正方向是左, 那么
左边的$(x,y)$映射到右边的$$\left(\operatorname{atan2}(y,x),y\right)$$逆变换: 右边的$(θ,y)$映射到左边的$$\left(y\cot θ,y\right)$$
其中atan2是Four-quadrant inverse tangent

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