# How do you graph r^2=cos(2theta)?

Apr 20, 2016

Called Lemniscate, the graph looks like $\infty$, symmetrical about the initial line $\theta = 0$ and the pole r = 0.

#### Explanation:

In the 1st and 4th 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

${r}^{2} = - \cos 2 \theta$

Use a table

{(r, theta)}.= {(0, -pi/4) (1/sqrt2, -pi/8) (1, 0) (1/sqrt2, pi/8) (0,

pi/4)}, 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 ${r}^{2} = \pm \cos 2 \theta$ are combined.
graph{((x^2+y^2)^2-x^2+y^2)((x^2+y^2)^2+x^2-y^2)=0[-2 2 -1 1]}.

Interestingly, an easy rotation of this graph through $\frac{\pi}{4}$

produces a grand 8-petal flower.

graph{((x^2+y^2)^2-x^2+y^2)((x^2+y^2)^2+x^2-y^2)((x^2+y^2)^2-2xy)((x^2+y^2)^2+2xy)=0[-2 2 -1 1]}.