How to prove it is a strictly stationary process?

I am going to write a demonstration that the process is stationary, but I am aware this is not what the original poster asked since it is not based on calculating the characteristic function. This post is in response to a comment I made.

Strict sense. It suffices to show that
$$\mathbb{P}(z\sin(\omega t+\theta)\le\xi)$$ is independent of $t$. But this is clear since $s=\sin(\omega t+\theta)$ distributes in $(-1,1)$ with density $f_s=1/(\pi\sqrt{1-s^2})$ independent of $t$.

Wide sense. This becomes a bit unnecessary, but anyways. We need to show that the autocorrelation depends only on the temporal displacement, not the coordinate, that is, $\mathbb{E}[\xi(t)\xi(t+\tau)]$ is function of $\tau$ and not $t$. We have that $$ \begin{multline} \mathbb{E}[\xi(t)\xi(t+\tau)]=\cos(\omega t)\cos(\omega (t+\tau))\mathbb{E}[z^2\sin^2\theta]+\sin(\omega t)\sin(\omega (t+\tau))\mathbb{E}[z^2\cos^2\theta]+\\ \sin(\omega(2t+\tau))\mathbb{E}[z^2\sin\theta\cos\theta]~~.\\ \end{multline}$$

But $\mathbb{E}[z^2\sin^2\theta]=\mathbb{E}[z^2\cos^2\theta]=\mathbb{E}[z^2]/2=\mu_z^2/2$ (assuming it exists). We can separate the variables in the expectation since thet are independent. Also $\mathbb{E}[z^2\sin\theta\cos\theta]=0$. We conclude $$\mathbb{E}(\xi(t)\xi(t+\tau))=\frac{\mu_z^2}{2}\cos(\omega\tau)~~,$$ independent of t. Therefore the process is wide sense stationary.