# Solving the inverse of cos^2

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.

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Take square root, then arccos. What's so difficult here ? – Jean-Claude Arbaut Mar 22 '13 at 16:23
I'm just not familiar with the notation (was daydreaming during my trig classes unfortunately). So cos^2(i) is the same as [cos(i)]^2? – Carl Mar 22 '13 at 16:34
Yes, it's the same – Jean-Claude Arbaut Mar 22 '13 at 17:32

$$\cos 2x=\cos^2x-\sin^2x=2\cos^2x-1\Longrightarrow$$
$$\Longrightarrow \color{red}{\cos^2x=\frac{\cos 2x+1}{2}}$$
Good idea, Carl...and don't forget to divided by two at the end to get $\,x\,$ and not only $\,2x\,$... – DonAntonio Mar 22 '13 at 16:42