How Do You Solve for Theta in Goniometric Equations?

[The Alchemist]

Homework Statement

Given the following relation for θ:
\frac{I_{\pi}}{I_{\sigma}} = \frac{\sin^2{\theta}}{1 + \cos^2{\theta}}
solve for θ

Homework Equations

\cos^2 x + \sin^2 x = 1

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.

 

[Curious3141]

Homework Statement

Given the following relation for θ:
\frac{I_{\pi}}{I_{\sigma}} = \frac{\sin^2{\theta}}{1 + \cos^2{\theta}}
solve for θ

Homework Equations

\cos^2 x + \sin^2 x = 1

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.

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.

 

[The Alchemist]

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.

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?

 

[Curious3141]

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?

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

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

 

[The Alchemist]

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

 

[Curious3141]

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

Mistakes in print are nothing new.