For two given integers $k,l\in\mathbb{N}$ and any real values $A,B,C,D$ satisfying $A^2+B^2+C^2+D^2=1$, does the following system of equations has a solution, on $x,y\in\mathbb{R}$?
- $\cos(kx)\cos(ly)=A$
- $\cos(kx)\sin(ly)=B$
- $\sin(kx)\cos(ly)=C$
- $\sin(kx)\sin(ly)=D$
I think that in general the answer is no, but in that case, what extra conditions do I need to ask $A,B,C,D$ to have a solution? Thanks!
Indeed, generally, the system does not have a solution.
Let me assume that $k = l = 1$. (If $k = 0$ or $l = 0$, this leads to a degenerate case of your system which you can study apart easily. If $k > 1$ or $l> 1$, you can simply rescale the unknowns and look for $x' = kx$ and $y' = ly$.)
So you are trying to solve $$ \begin{cases} \cos x \cos y = A, \\ \cos x \sin y = B, \\ \sin x \cos y = C, \\ \sin x \sin y = D. \end{cases} $$ You can start by solving for $x$. You get $\cos^2 x = A^2+B^2$ and $\sin^2 x = C^2+D^2$. Thanks to you hypothesis, you can find such an $x \in \mathbb{R}$ (in fact, there are many solutions).
Then you can solve for $y$ using the first two lines or the last two lines. Here, you indeed have two many constraints. Since $\cos^2 y$ must be the same whether you compute it using the first two lines or the last two ones, this imposes $A^2/(A^2+B^2) = C^2/(C^2+D^2)$. Also, for the signs to agree, you must have $ABCD > 0$.
Under these two additional assumptions (which are anyway necessary), I believe your system has a solution (in fact, many).