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]}