1
$\begingroup$

Hi so I have this equation

$$\frac{1-e^2}{1+e\cos(\theta -\theta_{0})} = 1-e\cos\eta$$

and using the identity

$$\cos\theta = \frac{1-\tan^2 \frac\theta2}{1+\tan^2 \frac\theta2}$$

this becomes

$$\sqrt{1-e} \tan\left(\frac{\theta - \theta_{0}}2\right) = \sqrt{1+e} \tan\frac\eta2.$$

However, I'm unsure about the steps in between. Can someone help me out please?

$\endgroup$
  • $\begingroup$ @BarryCipra apologies, supposed to be a + on the bottom. edited it now! $\endgroup$ – SFL Apr 23 '18 at 20:45
2
$\begingroup$

Set $u=\tan((\theta-\theta_0)/2)$ and $v=\tan(\eta/2)$ for simplicity.

The left hand side becomes $$ \frac{1-e^2}{1+e\dfrac{1-u^2}{1+u^2}} =\frac{(1-e^2)(1+u^2)}{(1+e)+(1-e)u^2} $$ The right hand side becomes $$ 1-e\frac{1-v^2}{1+v^2}=\frac{(1-e)+(1+e)v^2}{1+v^2} $$ It's better to set $M=1-e$ and $P=1+e$, so we get a simpler identity $$ \frac{MP(1+u^2)}{P+Mu^2}=\frac{M+Pv^2}{1+v^2} $$ that can be rearranged to $$ MP+MPv^2+MP(1+v^2)u^2=MP+P^2v^2+M^2u^2+MPu^2v^2 $$ Pull all terms with $u^2$ on the left hand side $$ (MP+MPv^2-M^2-MPv^2)u^2=P^2v^2-MPv^2 $$ that simplifies to $$ M(P-M)u^2=P(P-M)v^2 $$ Assuming $P-M=1+e-1+e=2e\ne0$, this becomes $$ (1-e)\tan^2\frac{\theta-\theta_0}{2}=(1+e)\tan^2\frac{\eta}{2} $$

$\endgroup$
  • $\begingroup$ Beat me to it, so +1. $\endgroup$ – David Quinn Apr 23 '18 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.