Find the inverse of following function

[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.

 

[HallsofIvy]

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.

 

[tacman]

To find the inverse of the function p(x)=(cos2πx,sin2πx), we can use the following steps:

1. Write the function in terms of x and y: p(x)=(x,y)

2. Switch the positions of x and y: p(y)=(y,x)

3. Solve for y: y=sin2πx and x=cos2πx

4. Rewrite the equations in terms of y: x=sin2πy and y=cos2πy

5. Solve for y in terms of x: y=sin^-1(x/2π) and y=cos^-1(x/2π)

6. Combine the equations to get the inverse function: p^-1(x)=(sin^-1(x/2π), cos^-1(x/2π))

The inverse function of p(x) is p^-1(x)=(sin^-1(x/2π), cos^-1(x/2π)). This can also be written as p^-1(x)=(arcsin(x/2π), arccos(x/2π)).

In general, to find the inverse of a function, we switch the positions of x and y and solve for y in terms of x. Then, we combine the equations to get the inverse function.