Finding marginal distribution of uniform distribution on unit circle

Question

Consider the map $f: [0,1)\to \mathbb S^1, t \mapsto (\cos 2\pi t,\sin2\pi t)$

and let $\lambda^{1}$ be the one-dimensional Lebesgue measure. Let $P$ be the uniform distribution on $\mathbb S^1:=\{ (x,y): x^2+y^2=1\}$. That is $P(B)=\lambda^1(f^{-1}(B))$. Find the first marginal distribution $P_1$ which represents the measure of given by the projection $\pi_1: \mathbb S^1 \to [-1,1],(x,y) \mapsto x$.

My idea:

consider $x \in [-1,1]$ arbitrary. Then $P_{1}([-1,x])=P([-1,1]\times [-1,x])=\lambda^{1}(f^{-1}([-1,1]\times [-1,x]))$

Then I want to use the transformation formula, however, I am unsure since I am unable to take the determinant of $Df = (-\sin(2\pi t), \cos(2\pi t))$

Any ideas?

Answer 1

\begin{align} x & = \cos(2\pi t) \\[8pt] dx & = - 2\pi\sin(2\pi t)\, dt \\[8pt] \frac{|dx|}{2\pi} & = \sin(2\pi t)\, dt = y\, dt \\[8pt] dt & = \left| \frac{dx}{\sin(2\pi t)} \right| = \left|\frac{dx}{2\pi y}\right| = \frac{|dx|}{\sqrt{1-x^2}} \end{align}

Or, alternatively:

Let $(X,Y)= (\cos(2\pi T),\sin(2\pi T))$ where $T$ is uniformly distributed in $[0,1].$ \begin{align} f_X(x) = {} & \frac d {dx} \Pr(X\le x) \\[8pt] = {} & \frac d {dx} \Pr(\arccos x \le 2\pi T \le 2\pi-\arccos x) \\[8pt] = {} & \frac d {dx} \,\frac{2\pi - 2\arccos x}{2\pi} = \frac 1 {\sqrt{1-x^2}}. \end{align}