Suppose we have a function as $r=\cos \frac{\theta}{2}$.
I know the graph is symmetric about $x$ axis. Replacing $\theta$ by $-\theta$ shows it. I want to know how the graph of this function is symmetric about $y$ axis also. Please help.
I am really not getting this. I have tried using $(-r,-\theta)$ and $(r,\pi-\theta)$.
Some polar plots are symmetric across the $y$ axis because the radius is the same for $\theta = \frac\pi2 + \alpha$ as it is for $\theta = \frac\pi2 - \alpha,$ that is, because $r\left(\frac\pi2 + \alpha\right) = r\left(\frac\pi2 - \alpha\right).$
This one is not like that. Except for certain special values of $\theta,$ $$r\left(\frac\pi2 + \alpha\right) = \cos\left(\frac\pi4 + \frac\alpha2\right) \neq \cos\left(\frac\pi4 - \frac\alpha2\right) = r\left(\frac\pi2 - \alpha\right).$$
Instead, the $y$-axis symmetry comes from a set of points with negative radius values that are symmetric with the points with positive radius values. In fact for $0 \leq \theta \leq 2\pi$ you get only points on or above the $x$ axis: $r \geq 0$ when $0 \leq \theta \leq \pi$; and $r \leq 0$ when $\pi \leq \theta \leq 2\pi.$
What you might try is to compute $(x,y)$ for $\theta = \pi + \alpha$ and for $\theta = \pi - \alpha$ (angles measured symmetrically outward from the middle angle $\theta = \pi$):
\begin{align} && \theta &= \pi - \alpha & \theta &= \pi + \alpha \\ x &= r\cos\theta & x &= \cos\left(\frac{\pi - \alpha}2\right)\cos(\pi - \alpha) & x &= \cos\left(\frac{\pi + \alpha}2\right)\cos(\pi + \alpha) \\ y &= r\sin\theta & y &= \cos\left(\frac{\pi - \alpha}2\right)\sin(\pi - \alpha) & y &= \cos\left(\frac{\pi + \alpha}2\right)\sin(\pi + \alpha) \\ \end{align}
Alternatively, take angles measured symmetrically inward from the two ends of the range $[0,2\pi]$, that is, $\theta = \alpha$ and for $\theta = 2\pi - \alpha$:
\begin{align} && \theta &= \alpha & \theta &= 2\pi - \alpha \\ x &= r\cos\theta & x &= \cos\left(\frac{ \alpha}2\right)\cos( \alpha) & x &= \cos\left(\frac{ 2\pi - \alpha}2\right)\cos( 2\pi - \alpha) \\ y &= r\sin\theta & y &= \cos\left(\frac{ \alpha}2\right)\sin( \alpha) & y &= \cos\left(\frac{ 2\pi - \alpha}2\right)\sin( 2\pi - \alpha) \\ \end{align}