Question

test for symmetry and graph the polar equation.
$r^{2}=64cos(2\theta)$
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 may or may
not be symmetric with respect to the polar axis.
b. the polar equation failed the test for symmetry which means that the graph is not
symmetric with respect to the polar axis.
c. yes.
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 is not
symmetric with respect to the line $\theta=\frac{pi}{2}$.
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 line $\theta=\frac{pi}{2}$.

Explanation:

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

Replace $\theta$ with $-\theta$.
We know that $\cos(2(-\theta))=\cos(2\theta)$ (since $\cos(-x)=\cos(x)$).
The equation $r^{2}=64\cos(2\theta)$ becomes $r^{2}=64\cos(2(-\theta)) = 64\cos(2\theta)$.
Since the equation remains the same after the substitution, the graph is symmetric about the polar axis.

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

Replace $(r,\theta)$ with $(-r,\pi - \theta)$.
First, $\cos(2(\pi-\theta))=\cos(2\pi - 2\theta)=\cos(2\theta)$ (using the identity $\cos(A - B)=\cos A\cos B+\sin A\sin B$ with $A = 2\pi$ and $B=2\theta$, and $\cos(2\pi)=1,\sin(2\pi)=0$).
The left - hand side: $(-r)^{2}=r^{2}$.
The equation $r^{2}=64\cos(2\theta)$ becomes $(-r)^{2}=64\cos(2(\pi - \theta))$, or $r^{2}=64\cos(2\theta)$.
But if we use the substitution $(r,\theta)\to(r,\pi - \theta)$ (another way to test symmetry about $\theta=\frac{\pi}{2}$), $r^{2}=64\cos(2(\pi - \theta))=64\cos(2\theta)$. However, when we use the substitution $(r,\theta)\to(-r,-\theta)$ (a more comprehensive test), we note that the standard tests for symmetry about $\theta=\frac{\pi}{2}$:
Replace $r$ with $-r$ and $\theta$ with $\pi-\theta$. The original equation $r^{2}=64\cos(2\theta)$ gives $(-r)^{2}=64\cos(2(\pi - \theta))$. But also, if we use the substitution $(r,\theta)\to(r,\pi - \theta)$:

$$
LATEXBLOCK0
$$

But the general rule is: A failure in a symmetry test does not necessarily mean the graph is not symmetric. For the line $\theta=\frac{\pi}{2}$, if we consider the fact that $r^{2}=64\cos(2\theta)$ is a lemniscate. The standard form of a lemniscate $r^{2}=a^{2}\cos(2\theta)$ is symmetric about the polar axis ($\theta = 0$) and the pole. When we test for symmetry about $\theta=\frac{\pi}{2}$ using the substitution $(r,\theta)\to(-r,\pi - \theta)$ (or other equivalent substitutions), we find that the equation may not pass the strict substitution test in the same straightforward way as for the polar axis. In fact, $r^{2}=64\cos(2\theta)$ is not symmetric about $\theta=\frac{\pi}{2}$. For example, when $\theta=\frac{\pi}{4}$, $r^{2}=64\cos(\frac{\pi}{2}) = 0$. When $\theta=\frac{3\pi}{4}$, $r^{2}=64\cos(\frac{3\pi}{2})=0$. But if we take a non - zero value, say $\theta = 0$, $r^{2}=64$, $r=\pm8$ and for $\theta=\pi$, $r^{2}=64$, $r = \pm 8$, but for $\theta=\frac{\pi}{2}$, $r^{2}=64\cos(\pi)=- 64$ (no real - valued $r$). And if we consider points in the upper - half and lower - half planes with respect to $\theta=\frac{\pi}{2}$, we can see the lack of symmetry.

Answer:

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