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

Sep 10, 2016

See graph and details.

#### Explanation:

The period of $\cos 2 \theta$ is $\pi$.

As r is a cosine function, the graph is symmetrical about the initial

line $\theta = 0$.

As $\cos 2 \theta = 1 - 2 {\sin}^{2} \theta$, It is also a sine function. So, it

is symmetrical about $\theta = \frac{\pi}{2}$...

Understanding r as the modulus of the position vector from pole to

the point $\left(r , \theta\right)$ on $r = \cos 2 \theta \ge 0$,

$2 \theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] .$ and this is one period for $\cos 2 \theta$..

So, $\theta \in \left[- \frac{\pi}{4} , \frac{\pi}{4}\right] ,$.

A short Table for making the graph in Q1:.

$\left(r , \theta\right) : \left(1 , 0\right) \left(\frac{\sqrt{3}}{2} , \frac{\pi}{12}\right) \left(\frac{1}{\sqrt{2}} , \frac{\pi}{8}\right) \left(\frac{1}{2} , \frac{\pi}{6}\right) \left(0 , \frac{\pi}{4}\right)$.

Using symmetry, the other three quarters are traced.

graph{(x^2 + y^2)^1.5 - x^2 + y^2=0[-2 2 -2 2]}

${r}^{m} = \cos 2 \theta \mathmr{and} {r}^{m} = \sin s 2 \theta , m = 1 , 2 , 3 , 4 , \ldots$ all

generate such bi-loops called lemniscate.

Combined graph for ${r}^{3} = \sin 2 \theta \mathmr{and} {r}^{3} = \cos 2 \theta$:

graph{((x^2+y^2)^2.5-2xy)( (x^2+y^2)^2.5-x^2+y^2)=0[-2 2 -2 2]}

To 1.6 K Socratic viewers of this answer, here is another from

combinations in infinitude, using this and its rotation.

graph{((x^2+y^2)^2.5-2xy)( (x^2+y^2)^2.5-x^2+y^2)((x^2+y^2)^2.5-1.414xy-0.707(x^2-y^2))( (x^2+y^2)^2.5-0.702(x^2-y^2)+1.414xy)=0[-2 2 -2 2]}