Find the PDF of $Y= \sin{(\pi X)}$, where $X \sim U[0,1]$

Question

Let $X\sim U_{(0,1)}$ and lets define $Y= \sin{(\pi X)}$. I want to get the pdf of $Y$.
My attempt:

Clearly, $y\in(-1,1)\Rightarrow 1-y^2\ge0$, so

$$ F_Y(y)=\Bbb P(Y\le y)=\Bbb P\big(\sin{(\pi X)}\le y\big)=\Bbb P\big(\sin^2{(\pi X)}\le y^2\big)=\Bbb P\big(1\le y^2+\cos^2{(\pi X)}\big)=\Bbb P\Big(\big|\cos{(\pi X)}\big|\ge \sqrt{1-y^2}\Big) $$

but,

$$ \big|\cos{(\pi X)}\big|\ge\sqrt{1-y^2}\; \Leftrightarrow\;\cos{(\pi X)}\le -\sqrt{1-y^2}\;\text{ or }\; \cos{(\pi X)}\ge\sqrt{1-y^2} $$

so, I'm not sure if the following follows:

$$ F_Y(y)=\Bbb P\Big( \cos{(\pi X)}\le-\sqrt{1-y^2} \Big)+\Bbb P\Big( \cos{(\pi X)}\ge\sqrt{1-y^2} \Big) $$

and, since $X\in(0,1),\;\; \cos{(\pi X)}$ is invertible, where $\arccos$ is decreasing, so

$$ F_Y(y)=1-F_X\Big(\frac{1}{\pi}\arccos{-\sqrt{1-y^2}}\Big)+F_X\Big(\frac{1}{\pi}\arccos{\sqrt{1-y^2}}\Big) $$

so

$$ f_Y(y)=\frac{1}{\pi\sqrt{1-y^2}}\Big(f_X\big(\frac{1}{\pi}\arccos{-\sqrt{1-y^2}}\big)+f_X\big(\frac{1}{\pi}\arccos{\sqrt{1-y^2}}\big)\Big)=\frac{1}{\pi\sqrt{1-y^2}}\Big(\mathbb{1}_{(0,1)}\big(\frac{1}{\pi}\arccos{-\sqrt{1-y^2}}\big)+\mathbb{1}_{(0,1)}\big(\frac{1}{\pi}\arccos{\sqrt{1-y^2}}\big)\Big) $$

and got stucked here, assuming all of the above follows. Would appreciate any help.

Answer 1

The CDF of $Y= \sin(\pi X)$ is

$$F_Y (y) = \mathbb{P} (Y \leq y) = \mathbb{P} (\sin(\pi X) \leq y) = 2 \, \mathbb{P} \left(0 \leq X \leq \frac{1}{\pi} \, \arcsin(y)\right) = \frac{2}{\pi} \, \arcsin(y)$$

Hence, the PDF is

$$f_Y (y) = \frac{2}{\pi \sqrt{1-y^2}}$$