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Goniometric relationships

  1. Jul 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Given the following relation for θ:
    [itex]\frac{I_{\pi}}{I_{\sigma}} = \frac{\sin^2{\theta}}{1 + \cos^2{\theta}}[/itex]
    solve for θ


    2. Relevant equations
    [itex]\cos^2 x + \sin^2 x = 1[/itex]


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

    Is there a way to come to the arctan expression? Or is the paper wrong? I'm quite stuck.
     
  2. jcsd
  3. Jul 6, 2012 #2

    Curious3141

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

    Because I'm getting [itex]\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)}[/itex]

    To get the arctan expression, find cos θ then divide sin θ by cos θ to get tan θ, then take the arctangent.
     
  4. Jul 6, 2012 #3
    You're right I missed the square root in the arcsin, the arctan though doesn't have one!

    When I solve for cosine θ [itex]\theta = \arccos{ \left( \pm \sqrt{ \frac{I_{\sigma} - I_{\pi}}{I_{\sigma} + I_{\pi}}}\right)}[/itex]
    How can I then come to an arctan expression?
     
  5. Jul 6, 2012 #4

    Curious3141

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

    There should be a sqrt on the arctan expression as well.
     
  6. Jul 6, 2012 #5
    Oh, I was too dazzled! thanks for the help.
    So the equation in the paper misses the sqrt...
     
  7. Jul 6, 2012 #6

    Curious3141

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    Mistakes in print are nothing new.
     
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