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

##### 1 Answer

See graph and details.

#### Explanation:

The period of

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

line

As

is symmetrical about

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

the point

So,

A short Table for making the graph in Q1:.

Using symmetry, the other three quarters are traced.

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

generate such bi-loops called lemniscate.

Combined graph for

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]}