# Sine of Uniformly Distributed Random Variable

1. Oct 19, 2009

### snipez90

1. The problem statement, all variables and given/known data
Suppose U follows a uniform distribution on the interval (0, 2pi). Find the density of sin(U)

2. Relevant equations

3. 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?

2. Oct 19, 2009

### LCKurtz

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.