# Distribution of sine of uniform random variable on $[0, 2\pi]$

Let $X$ be a continuous random variable having uniform distribution on $[0, 2\pi]$. What distribution has the random variable $Y=\sin X$ ? I think, it is also uniform. Am I right?

• No, it is not. Loosely speking: Since the sine is "slow" near its extremes, if is more likely that $Y\approx 1$ than that $Y\approx 0$. Nov 17, 2014 at 18:14
• math.stackexchange.com/q/118108/321264 Oct 10, 2020 at 15:00

By the Jacobian formula, $$f_Y(y)=\frac{\mathbf 1_{|y|\lt1}}{\pi\sqrt{1-y^2}}.$$ This is the so-called Arcsine distribution, which famously appears in probability theory and in number theory, as shown by two mathematical giants from the 20th century:
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The support of the distribution is of course $[-1,1]$. For $y\in[0,1]$, we have \begin{align} \Pr(Y\le y) & = \Pr(0\le X\le\arcsin y\text{ or }2\pi\ge X\ge \pi-\arcsin y) \\[10pt] & = \frac{\arcsin y + (2\pi-(\pi-\arcsin y))}{2\pi}. \end{align} The density is the derivative of that, and that is not a constant function, so it's not uniformly distributed.
• OK, here it is: $\dfrac1{\pi\sqrt{1-y^2}}$. ${}\qquad{}$ Nov 17, 2014 at 18:39