For the polar curve
$$r = \cos(2\theta)$$
I have run through the tests for polar symmetry about the $y$-axis, the $x$-axis and about the origin. According to the replacements for $(r,\theta)$, this function appears to only have symmetry about the $x$-axis since replacing $(r,\theta)$ with $(-r,-\theta)$ and $(-r, \theta)$ yield the equation
$$
r = -\cos(2\theta)
,$$
which is a different equation from the original. However, a quick visual verification on Desmos indicates that this graph appears to have symmetry with regards to the $x,y$-axes and also the origin.
Since the resulting equation is not the same as $r = \cos(2\theta)$, for the y-axis and origin symmetry tests, would we still say that the polar function has symmetry about the $y$-axis and origin due to its graph?
Best Answer
A polar graph is symmetric with respect to $r=\frac{\pi}{2}$ [the $y$-axis] if replacing $(r,\theta)$ with either $(r,\pi-\theta)$ or $(-r,-\theta)$ gives an equivalent equation.
$r=\cos(2(\pi-\theta))=\cos(2\theta)$ so the polar graph is symmetric with respect to the $y$-axis.
A polar graph is symmetric with respect the pole [the origin] if replacing $(r,\theta)$ with either $(r,\pi+\theta)$ or $(-r,\theta)$ gives an equivalent equation.
$r=\cos(2(\pi+\theta))=\cos(2\theta)$ so the graph is symmetric with respect to the origin.