Question

test for symmetry and graph the polar equation.
r = 5\cos(4\theta)
a. is the polar equation symmetrical with respect to the polar axis?
a. the polar equation failed the test for symmetry which means that the graph is not
symmetric with respect to the polar axis.
b. yes.
c. the polar equation failed the test for symmetry which means that the graph may or may
not be symmetric with respect to the polar axis.
b. is the polar equation symmetrical with respect to the line \theta=\frac{\pi}{2}?
a. the polar equation failed the test for symmetry which means that the graph may or may
not be symmetric with respect to the line \theta=\frac{\pi}{2}.
b. the polar equation failed the test for symmetry which means that the graph is not
symmetric with respect to the line \theta=\frac{\pi}{2}.
c. yes.

Explanation:

Step1: Test for symmetry about the polar axis ($\theta = 0$)

Replace $\theta$ with $-\theta$ in the equation $r = 5\cos(4\theta)$.
Using the identity $\cos(-x)=\cos(x)$, we get $r = 5\cos(-4\theta)=5\cos(4\theta)$.
Since the resulting equation is the same as the original equation, the graph is symmetric about the polar axis.

Step2: Test for symmetry about the line $\theta=\frac{\pi}{2}$

Replace $(\theta,r)$ with $(\pi - \theta,r)$ in the equation $r = 5\cos(4\theta)$.
We have $r = 5\cos(4(\pi-\theta))$.
Using the identity $\cos(A - B)=\cos A\cos B+\sin A\sin B$, $\cos(4\pi - 4\theta)=\cos(4\pi)\cos(4\theta)+\sin(4\pi)\sin(4\theta)=\cos(4\theta)$ (because $\cos(4\pi)=1$ and $\sin(4\pi) = 0$).
The equation becomes $r = 5\cos(4\theta)$ which is the same as the original. But in polar - equation symmetry tests, when we use the standard substitution for symmetry about $\theta=\frac{\pi}{2}$ (replace $\theta$ with $\pi-\theta$), if the equation is unchanged, it is symmetric about $\theta=\frac{\pi}{2}$. However, another way: replace $(r,\theta)$ with $(-r,-\theta)$.
We get $-r=5\cos(- 4\theta)=5\cos(4\theta)\Rightarrow r=-5\cos(4\theta)$ which is not the same as the original equation $r = 5\cos(4\theta)$.
Since the test for symmetry about $\theta=\frac{\pi}{2}$ using one of the substitution methods (e.g., $(r,\theta)\to(-r,-\theta)$) fails, the graph may or may not be symmetric about $\theta=\frac{\pi}{2}$

Answer:

a. B. Yes.
b. A. The polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the line $\theta=\frac{\pi}{2}$.