# Find the inverse of following function

1. Apr 30, 2007

### sit.think.solve

How do you find the inverse of following function?

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

2. Apr 30, 2007

### HallsofIvy

Staff Emeritus
This is a function from R to R2? I.e. $p(t)= (cos(2\pi x), sin(2\pi x)$. 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, $p^{-1}(x,y)= arccos(x)/(2\pi)$ if y>= 0, $1- arcos(x)/(2\pi)$ if y< 0.