Explanation of derivation made at wikipedia.

Question

in this wikipedia article A deriviation to convert true and eccentric anomaly. I am however quite stunned by a single line - trying to reproduce but after half a dozen sheets of paper I can't find how it is done.

The problem lies in the following:

$$\tan(E) = \frac{\sqrt{1-e^2} \sin(\theta)}{e + \cos(\theta)}$$

$$\tan \left( \frac{\theta}{2}\right) = \sqrt{\frac{1+e}{1-e}} \cdot \tan \left( \frac{E}{2}\right)$$

I am really stuck in how to get from the first line to the second line (which is at wikipedia explained by a single word "also").

Answer 1

I doubt that the second equation can be derived from nothing but the first, and I don't see how the Wikipedia article implies that.

With $\cos(2\alpha)=\cos^2\alpha-\sin^2\alpha$, thus $1+\cos(2\alpha)=2\cos^2\alpha$, and with $\sin(2\alpha)=2\sin\alpha\cos\alpha$, we have $\tan\alpha=\sin(2\alpha)/(1+\cos(2\alpha))$, and thus

$$\tan\left(\frac\theta2\right)=\frac{\sin\theta}{1+\cos\theta}$$

and likewise

$$\tan\left(\frac E2\right)=\frac{\sin E}{1+\cos E}\;.$$

Above the equations you quote, the article has

$$\cos E = \frac{ e + \cos \theta }{1 + e \cos \theta }$$

and

$$\sin E = \frac{ \sqrt{1 - e^2} \, \sin \theta }{1 + e \cos \theta }\;.$$

Substituting those in the formula for $\tan(E/2)$ and then using the one for $\tan(\theta/2)$ yields your second equation.

Answer 2

Going backwards,

$$\tan(E)=\tan(2\frac E2)=\frac{2\tan(\frac E2)}{1-\tan^2(\frac E2)}\\ =\frac{2\sqrt{\dfrac{1-e}{1+e}}\tan(\frac \theta2)}{1-\dfrac{1-e}{1+e}\tan^2(\frac\theta2)}\\ =\frac{2\sqrt{1-e^2}\sin(\frac \theta2)\cos(\frac \theta2)}{(1+e)\cos^2(\frac\theta2)-(1-e)\sin^2(\frac\theta2)}\\ =\frac{\sqrt{1-e^2}\sin(\theta)}{e+\cos(\theta)}.$$