Question

How do you graph `r=2cos3theta`?

Answer 1

See Socratic graph for the Cartesian form of the equation

Explanation:

`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}