How do you graph # 4cos(3theta)#?

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Answer 1 · 2018-08-14T13:01:09

See explanation and graph.

#0 <= 4 sqrt (x^2 + y^2 ) = 4 cos 3theta in [ 0, 4 ]#

The period for this cosine function # = ( 2 pi )/3#.

#r >= 0# for #theta in [ 0, pi/6 ], [ pi/2, 2/3pi ], [ 7/6pi, 3/2pi ] and#

# [ 11/6pi, 2pi ]#, in one revolution #theta in [ 0, 2pi ]#. The first

and the last are halves, from the same loop. For subsequent

revolutions, these three three loops are redrawn.

For Socratic graphic utility that keeps off r-negative pixels, the

Cartesian form.

#sqrt ( x^2 + y^2 )#

# = 4 = cos 3theta =4 ( cos^3theta - 3 cos theta sin^2theta )#

#= 4 ( x^3/r^3- (3xy^2)/r^3)#, giving

#( x^2 + y^2 )^2 = 4 ( x^3 - 3 xy^2 )#

is used. See graph.
graph{( x^2 + y^2 )^2 - 4 ( x^3 - 3 xy^2 )=0}