Find the inverse of following function

  • #1

Main Question or Discussion Point

How do you find the inverse of following function?

p(x)=(cos2pix,sin2pix) where pi means 3.14etc NOT p times i.
 

Answers and Replies

  • #2
HallsofIvy
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This is a function from R to R2? I.e. [itex]p(t)= (cos(2\pi x), sin(2\pi x)[/itex]. Then it maps the half open interval [0, 1) in R onto the unit circle in R2. At t=1 you start around the circle again so the function is not one-to-one and you do not have an inverse outside that interval. The range of that function is, of course, {(x,y)|x2+ y2= 1} and that must be the domain of the inverse function. On that domain, [itex]p^{-1}(x,y)= arccos(x)/(2\pi)[/itex] if y>= 0, [itex]1- arcos(x)/(2\pi)[/itex] if y< 0.
 

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