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Thread: any chance of solving for theta_g?

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    any chance of solving for theta_g?

    Any chance of solving for $\displaystyle \theta_\gamma$?

    $\displaystyle x=\sqrt{\frac{b^2\sin^2\theta_\gamma+\cos^2\theta_ \gamma}{\tan^2\theta_\gamma+1}}$
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    Quote Originally Posted by rainer View Post
    Any chance of solving for $\displaystyle \theta_\gamma$?

    $\displaystyle x=\sqrt{\frac{b^2\sin^2\theta_\gamma+\cos^2\theta_ \gamma}{\tan^2\theta_\gamma+1}}$
    Start with the fact that $\displaystyle \tan^2\theta_\gamma+1 = \sec^2\theta_\gamma = 1/\cos^2\theta_\gamma$. So $\displaystyle x = \sqrt{(b^2\sin^2\theta_\gamma+\cos^2\theta_\gamma) \cos^2\theta_\gamma}$. Square both sides, replace $\displaystyle \sin^2\theta_\gamma$ by $\displaystyle 1-\cos^2\theta_\gamma$, and you have a quadratic equation for $\displaystyle \cos^2\theta_\gamma$.
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