Jaslyn Gordon

2023-02-25

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

Zachariah Patel

Beginner2023-02-26Added 7 answers

the first and fourth quadrants $\left|\theta \right|\le \frac{\pi}{4}$ 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

$\left\{(r,\theta )\right\}.=\{(0,-\frac{\pi}{4})(\frac{1}{\sqrt{2}},-\frac{\pi}{8})(1,0)(\frac{1}{\sqrt{2}},\frac{\pi}{8})(0,$

$\frac{\pi}{4}\left)\right\}$, 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 ${r}^{2}=\pm \mathrm{cos}2\theta$ are combined.

$|\pi -\theta |\le \frac{\pi}{4}$.

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

$\left\{(r,\theta )\right\}.=\{(0,-\frac{\pi}{4})(\frac{1}{\sqrt{2}},-\frac{\pi}{8})(1,0)(\frac{1}{\sqrt{2}},\frac{\pi}{8})(0,$

$\frac{\pi}{4}\left)\right\}$, 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 ${r}^{2}=\pm \mathrm{cos}2\theta$ are combined.