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 $\cos^2\theta$?


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


  • 1
    $\begingroup$ Take square root, then arccos. What's so difficult here ? $\endgroup$ Mar 22, 2013 at 16:23
  • $\begingroup$ 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? $\endgroup$
    – Carl
    Mar 22, 2013 at 16:34
  • 2
    $\begingroup$ Yes, it's the same $\endgroup$ Mar 22, 2013 at 17:32

1 Answer 1



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

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

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

  • $\begingroup$ 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. $\endgroup$
    – Carl
    Mar 22, 2013 at 16:37
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    $\begingroup$ Good idea, Carl...and don't forget to divided by two at the end to get $\,x\,$ and not only $\,2x\,$... $\endgroup$
    – DonAntonio
    Mar 22, 2013 at 16:42
  • $\begingroup$ 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. $\endgroup$
    – Carl
    Mar 22, 2013 at 16:46

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