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Thread: Solution to trigonometric problem...

  1. #1
    Newbie
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    Aug 2010
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    Solution to trigonometric problem...

    Hi all!

    I have a problem which I need to solve, but I can't be sure that there is a simple solution.

    I have:

    $\displaystyle \frac{cos^2(\phi(x))}{(1-cos(\phi(x)))} = f(x)$

    and I want to write $\displaystyle \phi(x)$ as:

    $\displaystyle \phi(x) = g(x)$.

    The problem looks quite nice, and slightly familiar, but I can't for the life of me work it out. My trigonometry has deserted me.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    uadractic

    Hello, Guffmeister!

    I'll simplify the wording of the problem . . .


    $\displaystyle \text{Given: }\:\dfrac{\cos^2\!\phi}{1-\cos\phi} \:=\: f$

    $\displaystyle \text{Solve for }\phi$

    Cross-multiply:. $\displaystyle \cos^2\!\phi \;=\;f\!\cdot\!(1-\cos\phi)
    $

    . . . . . . . . . . . .$\displaystyle \cos^2\!\phi \;=\;f - f\!\cdot\!\cos\phi$

    . .$\displaystyle \cos^2\!\phi + f\!\cdot\!\cos\phi - f \;=\;0 $


    Quadratic Formula: .$\displaystyle \cos\phi \;=\;\dfrac{-f \pm \sqrt{f^2 + 4f}}{2}$

    Therefore: .$\displaystyle \phi \;=\;\cos^{-1}\!\bigg(\dfrac{-f \pm \sqrt{f^2 + 4f}}{2}\,\bigg)$

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  3. #3
    Newbie
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    Aug 2010
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    Waaaaagh!

    Great! That was very simple, and not even a trigonometric identity in sight. I thought it looked like a nice equation to solve.

    Thanks very much.
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