Solving the inverse of cos^2

Question

The following equation provides the inclination ($i$) of a galaxy, using the ratio of its two axes:

$$ \cos^2 i = {(b/a)^2 − (b/a)^2_{eos} \over 1 − (b/a)^2_{eos}} $$

All I need however is to determine the value of $i$. Can someone walk me through solving this for both a normal $\cos\theta$ (using $\arccos\theta$ I assume), and then $\cos^2\theta$?

Update

Taking the basic trig provided by DonAntonio, I get this:

$$ \frac{\cos 2i+1}{2} = {(b/a)^2 − (b/a)^2_{eos} \over 1 − (b/a)^2_{eos}} $$

Then ... (poorly formatted I know) ... $$ i = \frac{\arccos\Bigg(\bigg(2\big({(b/a)^2 − (b/a)^2_{eos} \over 1 − (b/a)^2_{eos}}\big)\bigg)-1\Bigg)}{2} $$

Thanks.

Answer 1

Hints:

Don't struggle with that squared cosine. Better, remember some basic trigonometric identities:

$$\cos 2x=\cos^2x-\sin^2x=2\cos^2x-1\Longrightarrow$$

$$\Longrightarrow \color{red}{\cos^2x=\frac{\cos 2x+1}{2}}$$