Sine of Uniformly Distributed Random Variable

snipez90
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Homework Statement


Suppose U follows a uniform distribution on the interval (0, 2pi). Find the density of sin(U)


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The Attempt at a Solution


Well if U ~ (0, 2pi), then sin(U) should follow a distribution on [-1, 1]. I know one way to do tackle such problems is to let a be some element in [-1, 1] and then try to find

P(sin(U) \leq a).

The big problem is that if I use this last expression, the next step seems to be to take the arcsin to get U is less than or equal to arcsin(a), but this seems ridiculous since we don't have monotonicity when dealing with the interval (0, 2pi). Is there another way to approach this problem?
 
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Why not draw a picture of sin(u) on [0, 2\pi]. Then look at cases. If 0 < a < 1 it isn't hard to see which u give sin(u) ≤ a. Use the arcsin plus some common sense looking at the graph and break it into cases on the value of a.
 
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