If $f(2\alpha-\theta) = f(\theta)$, then $\theta=\alpha$ is a line of symmetry of $r=f(\theta)$. How do you derive $f(2\alpha-\theta) = f(\theta)$?

Question

For Polar Coordinates I know that for x-axis symmetry $f(-\theta)=f(\theta)$,

for y-axis symmetry $f(\theta)=f(\pi-\theta)$,

and for symmetry about the origin $f(\theta)=f(\theta+\pi)$.

The big question however, is given the above information or otherwise, how do you get to $f(2\alpha-\theta) = f(\theta)$?

Thanks in advance. regards.

Answer 1

In order for a polar function $f$ to have radial line of symmetry $\theta = \alpha$, we must have $f(\alpha - \phi) = f(\alpha + \phi)$ for all $\phi$. Now, put $\phi = \theta - \alpha$ to obtain the equation $$ f(2\alpha - \theta) = f(\theta) $$ for all $\theta$.