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Math Help - Proving that a trig equation follows from another given trig equation

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    Proving that a trig equation follows from another given trig equation

    \tan({\frac{\theta}{2}})=\sqrt{\frac{1-e}{1+e}}\tan({\frac{\phi}{2}})

    Then prove that \cos\phi=\frac{\cos\theta-e}{1-e\cos\theta}
    Last edited by mr fantastic; July 24th 2009 at 04:36 AM.
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    Hello,

    Let t=\tan\tfrac \theta2 and p=\tan\tfrac\phi2
    We have t=\sqrt{\frac{1-e}{1+e}} \cdot p

    Square it : t^2=\frac{1-e}{1+e} \cdot p^2


    We know that \cos(x)=\frac{1-\tan^2\tfrac x2}{1+\tan^2\tfrac x2} (that's used in Weierstrass substitution - go google)

    So :

    \begin{aligned}\cos \phi &=\frac{1-p^2}{1+p^2} \\<br />
&=\frac{1-\frac{1+e}{1-e}\cdot t^2}{1+\frac{1+e}{1-e}\cdot t^2} \\<br />
&=\frac{1-e-(1+e) t^2}{1-e+(1+e) t^2} \\<br />
&=\frac{1-t^2-e(1+t^2)}{1+t^2-e(1-t^2)}\end{aligned}

    ...=\frac{\frac{1-t^2}{1+t^2}-e}{1-e \cdot\frac{1-t^2}{1+t^2}}

    ...=\frac{\cos\theta-e}{1-e\cos\theta}
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