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2. ## Re: Finding pdf question

you can note that

$P[\cos(X) < y] = 2 P[X < \cos^{-1}(y)]$

When you look at this carefully you find that

$2 P[X < \cos^{-1}(y)] = 1- \dfrac 1 \pi \cos^{-1}(y)$

with a resulting PDF of

$\dfrac{1}{\pi \sqrt{1-y^2}},~-1 \leq y \leq 1$