# Homework Help: Proving bijection between a region

1. Jun 13, 2010

### Gekko

1. The problem statement, all variables and given/known data

Prove bijection between the regions
0<x<1, 0<y<1, 0<u, 0<v, u+v<pi/2

2. Relevant equations

x=sinu/cosv y = sinv/cosu

3. The attempt at a solution

We need to show that an inverse function exists to prove the bijection so obviously, (u,v) maps to one and only one (x,y) for the above. But what about the other way around? What is the best approach
Do we need to calculate:
u=arcsin(xcosv), v=arccos(sinu/x), u=arccos(sinv/y), v=arcsin(ycosu) and then look at each individually? Or could we divide one by the other and obtain tan(u)tan(v)=xy so that u=arctan(xy/tan(v) and v=arctan(xy/tan(u))?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 13, 2010

### Staff: Mentor

I'm not sure that you can solve for u and v from your equations for x and y. Maybe it's possible, but I haven't come up with anything. Your attempts seem like good ideas at first, but you need u in terms of x and y alone, and the same for v. If you have learned about the Inverse Function Theorem (see http://en.wikipedia.org/wiki/Inverse_function_theorem), you can use it to show that there is an inverse mapping from a point (x, y) to a point (u, v).

3. Jun 13, 2010

### Gekko

Last edited by a moderator: Apr 25, 2017
4. Jun 13, 2010

### Office_Shredder

Staff Emeritus
The u and v one paragraph above is the inverse transformation

5. Jun 14, 2010

### Gekko

Of course! Thanks