Answer
Horizontal asymptotes at $ (-\frac{1}{2}, 1)$ and $(\frac{1}{2}, -1) $, and no vertical asymptotes
Work Step by Step
$x= cos(θ)$, $y=cos(3θ)$
Boundaries of the parametric are $-\pi \leq θ \leq \pi$
To find horizontal asymptotes, we find the derivative of $y$ and set it to zero.
$\frac{dy}{dθ}= -3sin(3θ) $
$-3sin(3θ)=0$
$sin(3θ)=0$
$3θ= \frac{arcsin(0)}{3}=0$
$θ=0$
$sin(0) = 0$
This then also becomes:
$θ= \frac{k\pi}{3}$, where $k$ is an integer.
$sin(k\pi)=0$, so $sin(\frac{3k\pi}{3})=0$
Now to find the points where the asymptotes exist by plugging in the $θ$ values for x and y
$θ=0$
$X: cos(0)=1$
$Y: cos(0)=1$
There is no asymptote here because the function is discontinuous at this point.
$θ=\frac{\pi}{3}$
$X: cos(\frac{\pi}{3})= \frac{1}{2}$
$Y: cos(\frac{3\pi}{3})=-1$
Horizontal asymptote at ($\frac{1}{2}, -1)$
$θ=\frac{2\pi}{3}$
$X: cos(\frac{2\pi}{3})= -\frac{1}{2}$
$Y: cos(\frac{6\pi}{3})=1$
Horizontal asymptote at ($-\frac{1}{2}, 1)$
$θ=\pi$
$X: cos(0)=-1$
$Y: cos(0)=-1$
There is no asymptote here because the function is discontinuous at this point.
To find vertical asymptotes, we find the derivative of $x$ and set it to zero.
$\frac{dx}{dθ}= -sin(θ)$
$sin(θ)=0$
$θ = k\pi$, where $k$ is an integer
Now to find the points where the asymptotes exist by plugging in the $θ$ values for x and y
$θ=0$
$X: cos(0)=1$
$Y: cos(0)=1$
There is no asymptote here because the function is discontinuous at this point.
$θ=\pi$
$X: cos(0)=-1$
$Y: cos(0)=-1$
There is no asymptote here because the function is discontinuous at this point.
Horizontal asymptotes at $ (-\frac{1}{2}, 1)$ and $(\frac{1}{2}, -1) $, and no vertical asymptotes
