Analytic solution to $\alpha , \beta \in \mathbb{R}$ such that $\cos \alpha \cdot \sin \beta = \cos \bigl(\sin (\alpha \cdot \beta)\bigr)$? Are there any $\alpha , \beta \in \mathbb{R}$ such that $$\cos \alpha \cdot \sin \beta = \cos \bigl(\sin (\alpha \cdot \beta)\bigr)?$$
The trivial solutions are $(\alpha, \beta)=(0, \dfrac{\pi}{2})$. But are there more? I dont see an obvious way of tackling this problem but I keep coming back to it because it seems interesting to me (look at the graph of the solutions in desmos: https://www.desmos.com/calculator/ms8ad8cxqt).
I tried simplifying matters by trying out the case where $\alpha = \beta$ where we get that $$\cos \alpha \cdot \sin \alpha = \cos \bigl(\sin (\alpha ^{2})\bigr)$$ and using the fact that $2\sin \alpha \cos \alpha = \sin 2\alpha$ we really want to find $\alpha \in \mathbb{R}$ such that $$\dfrac{1}{2}\sin 2\alpha = \cos \bigl(\sin (\alpha ^{2}) \bigr),$$ though I have to admit that this does not make the problem easier (it seems to me at least).
Has anyone looked at this problem before and have a solution or maybe some hints or suggestions on how to go further in solving the problem?

Here's one thing i did with this problem. Please let me know if it is legitimate or maybe trivial, hehe.
Suppose $\alpha , \beta \in \mathbb{R}$ where $\alpha \approx 0$ and $\beta \approx 0$, then it immediately follows that $\alpha \cdot \beta \approx 0$.
Now what I was thinking is that we can use the small angle approximations of $\cos \theta \approx 1-\dfrac{\theta ^{2}}{2}$ and $\sin \theta \approx \theta$ and get that $$\cos \alpha \cdot \sin \beta \approx \Bigl(1-\dfrac{\alpha ^{2}}{2}  \Bigr)\cdot \beta = \beta - \dfrac{\alpha ^{2}\beta}{2}$$ and $$\cos \Bigl(\sin \alpha \cdot \beta \Bigr)\approx \cos \alpha \cdot \beta \approx 1-\dfrac{\alpha^{2}\beta^{2}}{2}.$$ So now we may ask: for which $\alpha , \beta \in \mathbb{R}$ does $$\beta - \dfrac{\alpha ^{2}\beta}{2}=1-\dfrac{\alpha^{2}\beta^{2}}{2}?$$
After some algebraic manipluations we get that $$\dfrac{(\beta -1)(\beta \cdot \alpha^{2}+2)}{\alpha^{2}}= 0 \Leftrightarrow \boxed{\beta = 1 \: \text{or} \: \beta = -\dfrac{2}{\alpha^{2}}}.$$ Since it is assumed that $\beta \approx 0$ I dont know if the solution $\beta = -\dfrac{2}{\alpha^{2}}$ is even legitimate since if $\alpha \approx 0$, then $\Bigl|-\dfrac{2}{\alpha^{2}}\Bigr|>>0.$ Anyhow, if we let $\beta = 1$ (can this be considered $\beta \approx 0$?) and let $\alpha$ be something $\approx 0$, say $\alpha = 0.001$, then we get $$\cos \alpha \cdot \sin \beta =\cos 0.001 \cdot \sin 1 \approx 0.\bar{9}\cdot 0.84147\approx 0.84147$$ and $$\cos \Bigl( \sin \alpha \cdot \beta \Bigr)\approx \cos \Bigl( \sin 0.001\cdot 1 \Bigr)\approx \cos 0.\bar{9}\approx 0.\bar{9}.$$ Thus $$\cos \alpha \cdot \sin \beta \approx \cos \Bigl( \sin \alpha \cdot \beta \Bigr),$$ whenever $\beta =1$ and $\alpha \approx 0$.
