# Writing x^2 + y^2 = 1 + sin^2(xy) in polar form

1. Oct 10, 2012

### Rasmus

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

$$x^2 + y^2 = 1 + sin^2(xy)$$

in polar form assuming

$$x = rcos(\phi)$$
$$y = rsin(\phi)$$

$$0<r, 0<= \phi < 2pi$$

solve for r as a function of $\phi$

3. The attempt at a solution

$$(rcos(\phi))^2 + (rsin(\phi))^2 = 1 + sin^2(r^2cos(\phi)sin(\phi))$$

$$r^2(cos^2(\phi) + sin^2(\phi)) = 1 + sin^2(r^2cos(\phi)sin(\phi))$$

$$r^2 = 1 + sin^2(r^2cos(\phi)sin(\phi))$$

At this point I'm feeling pretty lost, since I have no idea how to get the all r:s alone on one side of the equation. More specifically I don't understand how to get them out of the trig function.

2. Oct 10, 2012

### sweet springs

Hello.
I continued your calculation. it should be
φ = 1/2 * arcsin [2 arcsin{sqrt(r^2 - 1)} / r^2 ]
Regards.

3. Oct 10, 2012

### Rasmus

Thanks for the help, but I'm afraid I don't really understand exactly what you have done.

It looks like you applied arcsin twice as well as some other operations to solve for $\phi$ rather than r.

Wouldn't arcsin be a problem since that isn't defined for [0 to 2pi]?

4. Oct 10, 2012

### sweet springs

Hi.
The function stands for both φ　and -φ, i.e. even function or symmetric for x axis.
The function stands for both φ　and π-φ, i.e. even function or symmetric for y axis.
The function stands for both φ　and π/２-φ, i.e. symmetric for y=x
So you can narrow the range of φ to [0,π/4] in your estimation.
Regards.