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I've seen examples where it's just $\cos(\theta)$ and the distribution is $$ \frac{1}{\pi \sqrt{1 - y^2}} $$ for $-1 < y < 1$ but I've never came across a squared cosine. I'm assuming we can use trig identities and go from there?

Sammy Black
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Khaled Q
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    Show your work. How is the distribution derived? Which trig identity? How does the distribution transform when you transform $\cos \theta \mapsto \cos^2 \theta$? – Sammy Black Oct 15 '22 at 23:46
  • You know the PDF of $\theta$. How does one compute, in general, the PDF of $Y=g(\theta)$ for $g$ nice enough (differentiable, bijective somewhere or monotone or…)? If you can remember this process or the resulting formula, then you can tackle problems like this always with at least something to try. – Nap D. Lover Oct 16 '22 at 02:38

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