How do you graph $r=2 cos 2theta$ ?

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Answer 1 · 2018-07-15T15:59:03

Here, the non-negative $r in [0. 2]$ .

The period of $cos 2theta$ is $(2pi)/2=pi$ .

In one period $theta in [ -pi/2, pi/2 }$ ,

$r >= 0$ for $theta in [ - pi/4, pi/4 ]$ only.

Using,

$( x, y ) = r ( cos theta, sin theta )$ and $r =sqrt (x^2 + y^2 ) >=0$ ,

the Cartesian form of

$r = 2 cos 2theta = 2 ( cos^2theta - sin^2theta )$ is

$( x^2 + y^2 )^1.5 = 2 (x^2 - y^2 )$ .

The 2-loop Socratic graph is immediate.
graph{(x^2+y^2)^1.5-2(x^2-y^2)=0[-6 6 -3 3]}

Invalid ( r < 0 ) loops are shown below.
graph{(x^2+y^2)^1.5+2(x^2-y^2)=0[-6 6. -3 3]}

Valid and invalid combined graph:
graph{(x^2+y^2)^3-4(x^2-y^2)^2=0[-6 6. -3 3]}

How do you graph r=2 cos 2thetar=2 cos 2theta?