If $k$ is a positive integer and $U \sim \text{Unif}(0,1)$, why does it hold that $\sin(2\pi kU) \sim \sin(2\pi U)$? If $k$ is a positive integer and $U \sim \text{Unif}(0,1)$ is the standard uniform distributed, why does it hold that
$$
\sin(2\pi kU) \sim \sin(2\pi U)
$$
Is there is an intuition behind this or a proof?
 A: We know that $$\varphi_U(u) = \frac{e^{iu} - 1}{iu}.$$ So $$\varphi(2l\pi) = \begin{cases}1 & \text{if $l=0$}\\ 0 & \text{if $l\neq 0$}\end{cases}$$ and if $p\in \mathbb N$, $p \neq 0$, $$\varphi(2\pi pl) = \varphi(2\pi l)$$ Let $Y_k = \sin\left(2\pi k U\right)$
\begin{align}
\mathbb E\left[Y_k^n\right] &= \mathbb E\left[\left(\frac{e^{i(2\pi kU)} - e^{-i(2\pi kU)}}{2i}\right)^n\right]\\
&= \frac{1}{(2i)^n}\sum_{m=0}^{n}\binom{n}{m} \mathbb E\left[e^{i(2\pi kU)m}e^{-i(2\pi kU)(n-m)}\right]\\
&= \frac{1}{(2i)^n}\sum_{m=0}^{n}\binom{n}{m} \varphi_{U}(2\pi k(2m-n))\\
&= \frac{1}{(2i)^n}\sum_{m=0}^{n}\binom{n}{m} \varphi_{U}(2\pi (2m-n))\\
&= \frac{1}{(2i)^n}\sum_{m=0}^{n}\binom{n}{m} \mathbb E\left[e^{i(2\pi U)m}e^{-i(2\pi U)(n-m)}\right]\\
&= \mathbb E\left[\left(\frac{e^{i(2\pi U)} - e^{-i(2\pi U)}}{2i}\right)^n\right]\\
&= \mathbb E\left[Y_1^n\right]
\end{align}
This proves that $Y_k$ and $Y_1$ have the same moments and so the same distribution.

Another simple approach of proving it: Let $X$ be uniform in the set of integers $\{0,\ldots,k-1\}$ and $V\sim Uni(0,1)$, So $\frac{X+V}{k}\sim U$.
And $$\sin(2\pi kU)\sim\sin \left(2\pi (X+V)\right) = \sin (2\pi U)$$
