Changing ##\tan\frac{x}{2}## to ##\cos x## for a proof

[brotherbobby]

Homework Statement

Given $\tan\dfrac{\theta}{2}=\sqrt{\dfrac{1-e}{1+e}}\tan\dfrac{\phi}{2}$, prove $\boxed{\cos\phi=\dfrac{\cos\theta-e}{1-e\cos\theta}}$

Relevant Equations

(1) Componendo and dividendo : If $\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow \dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}$.

(2) $\cos x=\dfrac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$

Problem Statement : The statement of the problem again. If $\tan\dfrac{\theta}{2}=\sqrt{\dfrac{1-e}{1+e}}\tan\dfrac{\phi}{2}$, prove $\boldsymbol{\cos\phi=\dfrac{\cos\theta-e}{1-e\cos\theta}}$.

Attempt : Transposing and squaring what's given, we get $\tan^2\dfrac{\phi}{2}=\dfrac{1+e}{1-e}\tan^2\dfrac{\theta}{2}$.
Upon applying the componendo and dividendo to both sides of the equation, $\small{\dfrac{1-\tan^2\phi/2}{1+\tan^2\phi/2}=\dfrac{1-e-(1+e)\tan^2\theta/2}{1-e+(1+e)\tan^2\theta/2}\Rightarrow \cos\phi=\dfrac{1-\tan^2\theta/2-e(1+\tan^2\theta/2)}{1+\tan^2\theta/2-e(1-\tan^2\theta/2)}}{\large{\color{red}{\overset{\mathbf ?}{\mathbf \Rightarrow}}}} \cos\phi=\dfrac{\cos\theta-e}{1-e\cos\theta}$.

Request : As you can see on the right side of the above calculation, I cannot show how my calculated value $\cos\phi=\dfrac{1-\tan^2\theta/2-e(1+\tan^2\theta/2)}{1+\tan^2\theta/2-e(1-\tan^2\theta/2)}=\dfrac{\cos\theta-e}{1-e\cos\theta}?$

I cannot show how the second expression reduces to the third, which is what's required.

A hint would be welcome.

 

[fresh_42]

I haven't checked but it looks as if the Weierstraß substitution $\tan\left(\dfrac{x}{2}\right)=t$ could make calculations a lot easier.

 

[brotherbobby]

Sorry I got it. My bad. Just divide the numerator and denominator of the last expression I obtained by $1+\tan^2\theta/2$.

To conclude in a good way, I complete the solution using writing ink and $\text{Autodesk Sketchbook}^{\circledR}$, assuming am not violating anything.

 

[fresh_42]

Here is my solution by the Weierstraß substitution:

Set $\tan\left(\dfrac{\phi}{2}\right)=t\, , \,\tan\left(\dfrac{\theta}{2}\right)=s\,.$ Then $s^2(1+e)=t^2(1-e)$ and
\begin{align*}
\dfrac{\cos(\theta)-e}{1-e\cos(\theta)}&=\dfrac{\dfrac{1-s^2}{1+s^2}-e}{1-e\dfrac{1-s^2}{1+s^2}}=\dfrac{1-s^2-e-es^2}{1+s^2-e+es^2}=\dfrac{(1-e)-t^2(1-e)}{(1-e)+t^2(1-e)}=\dfrac{1-t^2}{1+t^2}=\cos\left(\dfrac{\phi}{2}\right)\,.
\end{align*}

It is basically the same but with far less to write.

 

[brotherbobby]

I didn't follow your last but one step @fresh_42 . I paste it below.

There are terms containing $e$ in the numerator and denominator, viz. $\dfrac{1-t^2-e(1-t^2)}{1+t^2-e(1+t^2)}$. Why do these terms vanish?

 

[renormalize]

I didn't follow your last but one step @fresh_42 . I paste it below.

View attachment 357484

There are terms containing $e$ in the numerator and denominator, viz. $\dfrac{1-t^2-e(1-t^2)}{1+t^2-e(1+t^2)}$. Why do these terms vanish?

Just factor the numerator and denominator and divide-out the common factor:$$\dfrac{(1-e)-t^{2}(1-e)}{(1-e)+t^{2}(1-e)}=\dfrac{(1-e)(1-t^{2})}{(1-e)(1+t^{2})}=\dfrac{(1-t^{2})}{(1+t^{2})}$$

 

[brotherbobby]

Just factor the numerator and denominator and divide-out the common factor:$$\dfrac{(1-e)-t^{2}(1-e)}{(1-e)+t^{2}(1-e)}=\dfrac{(1-e)(1-t^{2})}{(1-e)(1+t^{2})}=\dfrac{(1-t^{2})}{(1+t^{2})}$$

Many thanks and apologies. So there's nothing special about the variable $e$ except $e\ne 1$.

 

[fresh_42]

Many thanks and apologies. So there's nothing special about the variable $e$ except $e\ne 1$.

If $e=1$ then $\tan \left(\dfrac{\theta}{2}\right)=0$ and $\theta=0.$ But then, the formula for $\cos(\phi)$ isn't defined anymore so that we can exclude the case $e=1.$

 

[D H]

Many thanks and apologies. So there's nothing special about the variable $e$ except $e\ne 1$.

The equation $\tan\frac\theta 2 = \sqrt{\frac{1-e}{1+e}} \tan\frac\phi 2$ pertains only if $0\le e<1$. Here, $\theta$ and $\phi$ are the eccentric and true anomalies.

In the case that $e=1$, one of two (or both) of total energy and total angular momentum is zero. Angular momentum being zero means that true anomaly is undefined, as is eccentric anomaly. This is a radial trajectory (aka a degenerate trajectory) and needs special treatment. Total energy being zero (which also results in $e=1$) is a parabolic trajectory. This needs separate treatment. Finally $e>1$ means a non-degenerate hyperbolic trajectory. It too needs special treatment.