I have a function $\mathbb{R}\to\mathbb{R}$, $y=f(x)$. I want to find its equation $y'=g(x')$ when the basis $(x,y)$ is rotated by $a$. $(x',y')= R_a[(x,y)]$, where $R_a$ is the rotation of angle $a$.
I assume that f is symétric (f(x)=f(-x)) and that $g$ is still a function after the rotation !
I thought naively that I could rotate by -a, compute f, and then rotate by a. But this is wrong when $f$ is linear and commutes with $R_a$ ! More precisely, I thought that $$g(x)= P_y \circ R_a (0,f \circ P_x \circ (R_{-a}[(x,0)]))$$, where $P_x$ and $P_y$ are the projectors $R^2 \rightarrow R$ on the two coordinates
Thank you for your help.
I know of no answer in general unless the function has a sort of symmetry about the origin. For instance, on the real numbers the only non-identity rotation is multiplication by $-1$. We can say something about $g(-x)$ if $g$ is odd or even, for example.