How do you graph #r=2cos3theta#?

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Answer 1 · 2017-02-06T16:56:58

See Socratic graph for the Cartesian form of the equation

#r=2cos3theta>=0,# giving #3theta in Q_1 and Q_4#.

So, for one loop of this 3-petal rose,

#theta in [-pi/6, pi/6]# and this range is

#1/2(2/3pi)# = 1/2(period of #cos 3theta)#)

For the other half of one period r is negative and, if you choose to

admit negative r, the third loop would be redrawn.

For #r = 2sin3theta#, r_ loops would add three more.

I reiterate that #r = sqrt(x^2+y^2) >= 0#

I have used the Cartesian form of the equation, using

#cos 3theta=4cos^3theta-3costheta#

graph{(x^2+y^2)^2+6x(x^2+y^2)=8x^3 [-4, 4, -2, 2]}

I introduce here a 2-cosine-combined 10-petal rose for

my viewers of this answer.

Observe that four of the petals are much smaller.

graph{0.01((x^2+y^2)^2.5-x^4+6x^2y^2-y^4)(0.25(x^2+y^2)^3.5-x^6+15x^4y^2-15x^2y^4+y^6)(x^2+y^2-.04)=0}