# 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.

• Take square root, then arccos. What's so difficult here ? Mar 22, 2013 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, 2013 at 16:34
• Yes, it's the same Mar 22, 2013 at 17:32

Hints:

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

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

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

• Very helpful. Thanks for that Don. So after substituting the alternative, I can solve for i. I might update the question to include this as it's easier to do formulas in there.
– Carl
Mar 22, 2013 at 16:37
• Good idea, Carl...and don't forget to divided by two at the end to get $\,x\,$ and not only $\,2x\,$... Mar 22, 2013 at 16:42
• Done. Keen for an edit on that if I have anything wrong. Thanks again for the guidance, it really helps to break these things down for me.
– Carl
Mar 22, 2013 at 16:46