Question

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

Answer 1

See graph and details.

Explanation:

The period of `cos 2theta` is `pi`.

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

line `theta = 0`.

As `cos 2theta = 1 - 2sin^2 theta`, It is also a sine function. So, it

is symmetrical about `theta = pi/2`...

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

the point `(r, theta)` on `r = cos 2theta >=0`,

`2theta in [-pi/2, pi/2].` and this is one period for `cos 2theta`..

So, `theta in [-pi/4, pi/4],`.

A short Table for making the graph in Q1:.

`(r, theta): (1, 0) (sqrt3/2, pi/12) (1/sqrt2, pi/8) (1/2, pi/6) (0, pi/4)`.

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 2theta and r^m =sins 2theta, m = 1, 2, 3, 4, ...` all

generate such bi-loops called lemniscate.

Combined graph for `r^3 = sin 2theta and r^3 = cos 2theta`:

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]}