# Thread: Proving that a trig equation follows from another given trig equation

1. ## Proving that a trig equation follows from another given trig equation

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

Then prove that $\displaystyle \cos\phi=\frac{\cos\theta-e}{1-e\cos\theta}$

2. Hello,

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

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

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

So :

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

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

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