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Math Help - Find theta as a function of phi

  1. #1
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    Find theta as a function of phi

    I have found the equation 2\cos \theta \sin \theta \tan \phi +2\sin \theta ^2 -1=0 and want to solve out \theta as a function of \phi .

    The history of this equation is from a physics problem where a ball is thrown at angle \theta down a slope of angle \phi . The question is what angle \theta to throw the ball so that it has the greatest range. I have checked the equation numerically but cannot solve it.
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  2. #2
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    Hello, fysikbengt!

    2\cos\theta\sin\theta\tan\phi + 2\sin^2\!\theta-1\:=\:0

    \text{Solve for }\theta\text{ as a function of }\phi.

    \text{We have: }\;\underbrace{2\cos\theta\sin\theta}_{\sin2\theta  }\tan\phi + \underbrace{2\sin^2\!\theta - 1}_{-\cos2\theta} \:=\:0

    . . \sin2\theta\tan\phi - \cos2\theta \:=\:0 \quad\Rightarrow\quad \sin2\theta\tan\phi \:=\:\cos2\theta

    . . \frac{\sin2\theta}{\cos2\theta} \:=\:\frac{1}{\tan\phi} \quad\Rightarrow\quad \tan2\theta \:=\:\cot\phi

    . . 2\theta \:=\:\tan^{-1}(\cot\phi) \:=\: \left(\tfrac{\pi}{2} - \phi\right)

    . . \theta \:=\:\tfrac{1}{2}\left(\tfrac{\pi}{2}-\phi\right)

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  3. #3
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    Thanks

    Quote Originally Posted by Soroban View Post

    . . 2\theta \:=\:\tan^{-1}(\cot\phi) \:=\: \left(\tfrac{\pi}{2} - \phi\right)


    So easy!

    I really had to think about that last step but now it is obvious to me.

    Thanks for help.
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