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?
 A: The solution set $S$  consists of disjoint  flowers $\gamma_{jk}$.  These flowers are more or less circular "distorted Lissajous figures" and become more complex further out in the plane. It will be difficult to analyze the individual $\gamma_{jk}$, but we can  put each of them in a nice small $(x,y)$-square with center $(0,0)$.

To this end we draw $\xi$- and $\eta$-axes with $\pm45^\circ$ slopes, intersecting at  the point ${\bf c}_{00}=\bigl(0,{\pi\over2}\bigr)$.  The units on these axes are chosen such that the centers ${\bf c}_{jk}$ of the flowers obtain new coordinates $${\bf c}_{jk}=[j\pi,k\pi]\qquad\bigl((j,k)\in{\mathbb Z}^2\bigr)\ ,$$ whereby we use square brackets for such coordinates. This setup leads to the transformation formulas
$$\alpha=\xi+\eta,\qquad \beta=-\xi+\eta+{\pi\over2}\ .\tag{1}$$
We now partition the plane into squares, aligned with the new axes, by putting
$$\xi=j\pi+x,\quad \eta=k\pi+y\qquad\bigl(\ [x,y]\in\left[-{\pi\over2},{\pi\over2}\right]^2\ \bigr)\ .$$
In this way each  ${\bf c}_{jk}$ is the center of its square of (new) side length $\pi$. The formulas  $(1)$ now appear as
$$\alpha=(j+k)\pi+x+y,\qquad\beta=(-j+k)\pi-x+y+{\pi\over2}\ .\tag{2}$$
This implies
$$\cos\alpha\sin\beta=(-1)^{k+j}\cos(x+y)(-1)^{k-j}\cos(y-x)
=1-\sin^2 x-\sin^2 y\ ,\tag{3}$$
where the RHS is independent of $j$ and $k$.
For points of  $S$ we want $$\cos\alpha\sin\beta=\cos\bigl(\sin(\alpha\cdot\beta)\bigr)\geq\cos1>{1\over2}\ .$$
From $(3)$ it then follows that, independently of $j$ and $k$, we must have $$\sin^2 x+\sin^2 y<{1\over2}\qquad\bigl(\> [j\pi+x,k\pi+y]\in S\>\bigr)\ ;\tag{4}$$
in particular $$|x|<{\pi\over4}, \quad |y|<{\pi\over4}\qquad \bigl([x,y]\in\gamma_{jk}\bigr)\ .$$ Formula $(4)$ is the decisive equation of the present answer: It shows that the formal procedure has lead to the right separation of the different flowers $\gamma_{jk}$.
To sum it up: When $(j,k)\in{\mathbb Z}^2$ is given and fixed, by $(3)$ and $(2)$  the flower $\gamma_{jk}$ has an $[x,y]$-equation of the form
$$1-\sin^2 x-\sin^2 y=
 \cos\circ\sin\left(\bigl((j+k)\pi+x+y\bigr)\bigl((-j+k)\pi-x+y+{\pi\over2}\bigr)\right)\ .$$
I don't see an obvious simplification of the $\bigl(\ldots\bigr)$ part here.
A: There are infinitely many solutions but it is hard to write them down.
Take $\alpha=2\pi t$,  $\beta=\frac{\pi}{2}t$ and look at the sign of
$$f(t)=\cos(2\pi t)\cdot \sin(\frac{\pi}{2}t) - \cos(\sin({\pi^{2}}t^{2}))$$
when $t$ runs over integers.
The second term is always between $(-1,1)$ since $\sin()$ is zero only at multiplicities of $\pi$ and $\pi$ is irrational. At the same time if $t = 4k+1$ the first term is $+1$ and if $t = 4k+3$ it is $-1$.
In words, $f(t)$ changes sign infinitely many times.
A: 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 $\alpha \approx 0$.
