Goniometric relationships

1. Jul 6, 2012

The Alchemist

1. The problem statement, all variables and given/known data
Given the following relation for θ:
$\frac{I_{\pi}}{I_{\sigma}} = \frac{\sin^2{\theta}}{1 + \cos^2{\theta}}$
solve for θ

2. Relevant equations
$\cos^2 x + \sin^2 x = 1$

3. The attempt at a solution
If I solve this I get: $\theta = \arcsin{\left(\pm\frac{2 I_{\pi}}{I_{\sigma}+I_{\pi}}\right)}$
But the paper where this equation is from says: Consequently θ becomes:
$\theta = \arctan{\left(\pm\frac{2 I_{\pi}}{I_{\sigma}-I_{\pi}}\right)}$

Is there a way to come to the arctan expression? Or is the paper wrong? I'm quite stuck.

2. Jul 6, 2012

Curious3141

Are you missing a square root sign on both your expressions?

Because I'm getting $\theta = \arcsin{\left(\pm \sqrt{\frac{2I_{\pi}}{I_{\sigma} + I_{\pi}}}\right)} = \arctan{\left(\pm \sqrt{\frac{2I_{\pi}}{I_{\sigma} - I_{\pi}}}\right)}$

To get the arctan expression, find cos θ then divide sin θ by cos θ to get tan θ, then take the arctangent.

3. Jul 6, 2012

The Alchemist

You're right I missed the square root in the arcsin, the arctan though doesn't have one!

When I solve for cosine θ $\theta = \arccos{ \left( \pm \sqrt{ \frac{I_{\sigma} - I_{\pi}}{I_{\sigma} + I_{\pi}}}\right)}$
How can I then come to an arctan expression?

4. Jul 6, 2012

Curious3141

Just divide the sine by the cosine. The denominators cancel out.

There should be a sqrt on the arctan expression as well.

5. Jul 6, 2012

The Alchemist

Oh, I was too dazzled! thanks for the help.
So the equation in the paper misses the sqrt...

6. Jul 6, 2012

Curious3141

Mistakes in print are nothing new.