Jaslyn Gordon

2023-02-25

How to graph ${r}^{2}=\mathrm{cos}\left(2\theta \right)$?

Zachariah Patel

the first and fourth quadrants $|\theta |\le \frac{\pi }{4}$ In the 2nd and 3rd,
$|\pi -\theta |\le \frac{\pi }{4}$.
I used infinity symbol $\infty$ to depict the shape of this double-
loop, looking like a fallen 8. For getting 8-erect, the equation is
${r}^{2}=-\mathrm{cos}2\theta$
Use a table
$\left\{\left(r,\theta \right)\right\}.=\left\{\left(0,-\frac{\pi }{4}\right)\left(\frac{1}{\sqrt{2}},-\frac{\pi }{8}\right)\left(1,0\right)\left(\frac{1}{\sqrt{2}},\frac{\pi }{8}\right)\left(0,$
$\frac{\pi }{4}\right)\right\}$, for one loop. Its mirror image with respect to $\theta =\frac{\pi }{2}$
is the other loop.
Strictly, $r=\sqrt{{x}^{2}+{y}^{2}}\ge 0$.
Graphs of both ${r}^{2}=±\mathrm{cos}2\theta$ are combined.

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