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

• July 24th 2009, 03:14 AM
jashansinghal
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 $\cos\phi=\frac{\cos\theta-e}{1-e\cos\theta}$
• July 24th 2009, 04:56 AM
Moo
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} \\
&=\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}

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

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