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Math Help - ellipse, unit circle, angles

  1. #1
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    ellipse, unit circle, angles

    Please see attached image.

    Knowing only theta and x, find theta_2 and theta_3

    (the green curve is an ellipse inscribed within the reddish unit circle)
    Attached Thumbnails Attached Thumbnails ellipse, unit circle, angles-math-problem.jpg  
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  2. #2
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    Spoiler:

    \tan\theta_2 = \frac{\sin\theta - x}{\cos\theta}

    \tan\theta_3 = \frac{\sin^2\theta}{\cos\theta(\sin\theta - x)}
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  3. #3
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    In general , we have


     \tan(\theta_n) = (\frac{1}{b})^n \tan(\theta_0)


     x_n = (\frac{1}{b})^{n+1} \cdot \frac{ 1 - b }{ \sqrt{ 1 + (\frac{1}{b})^{2n} \tan^2(\theta_0) }}

    Once the value of  b is comfirmed ,

    we can find out the angle . ( if we keep drawing the lines similarly , a sequence is formed ! )
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  4. #4
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    These are very unexpected, interesting responses (to me), which is precisely what I hoped to get by posting this problem.

    Opalg, I'm hesitant to say you're wrong, but your answer doesn't work when I graph it. Either it's wrong or I'm making some mistake when I graph it.

    Simplependulum, I'm still trying to wrap my head around your response. I had never thought of expressing things in that way.

    Here is my solution (obviously not the only one):

    First, use the given info-- x, \sin{\theta}--to find the semi-minor axis of the ellipse, b:

    x=\sin{\theta}-b\sin{\theta}

    so, b=\frac{\sin{\theta}-x}{\sin{\theta}}

    Then, the equations for \theta_2 and \theta_3 are, respectively:

    \tan{\theta_2}=\sqrt{b^2(\sec^2{\theta}-1)}

    and

    \tan{\theta_3}=\arccos{\frac{b}{\sqrt{\tan^2{\thet  a}+b^2}}}

    You derive these equations by fooling around with (x, y) intercepts and pythagorus' theorem. I won't go into the details unless you really want.

    Happily, these equations turn out to be reducible to:

    \tan{\theta_2}=b\tan{\theta}

    and

    \tan{\theta_3}=\frac{\tan{\theta}}{b}
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