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Last edited by JaguarXJS; Dec 15th 2016 at 06:59 PM.
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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 $\dfrac{1}{\pi \sqrt{1-y^2}},~-1 \leq y \leq 1$