Having seen that there were more than 1 K viewers in a day, I now add more.
The 2-D polar coordinates P ( r, theta)P(r,θ), r = sqrt (x^2 + y^2 ) >= 0√x2+y2≥0. It represents length of the position vector < r, theta >. theta<r,θ>.θ determines the direction. It increases for anticlockwise motion of P about the pole O. For clockwise rotation, it decreases. Unlike r, thetaθ admit negative values.
r-negative tabular values can be used by artists only.
The polar equation of a rose curve is either r = a cos ntheta or r = a sin nthetar=acosnθorr=asinnθ.
n is at your choice. Integer values 2,, 3, 4.. are preferred for easy counting of the number of petals, in a period. n = 1 gives 1-petal circle.
To be called a rose, n has to be sufficiently large and integer + a fraction, for images looking like a rose. For integer values, the petals might be redrawn, when the drawing is repeated over successive periods.
.
The period of both sin ntheta and cos nthetasinnθandcosnθ is 2pi/n2πn.
The number of petals for the period [0, 2pi/n][0,2πn] will be n or 2n ( including r-negative n petals ) according as n is odd or even, for 0 <= theta <= 2pi0≤θ≤2π. Of course, I maintain that r is length >=0≥0, and so non-negative. For Quantum Physicists, r > 0.
Foe example, consider r = 2 sin 3thetar=2sin3θ. The period is 2pi/32π3 and the number of petals will be 3. In continuous drawing. r-positive and r-negative petals are drawn alternately. When n is odd, r-negative petals are same as r-positive ones. So, the total count here is 3.
Prepare a table for (r, theta)(r,θ), in one period [0, 2pi/3][0,2π3], for theta = 0, pi/12, 2pi/12, 3pi/12, ...8pi/12. Join the points by smooth curves, befittingly. You get one petal. You ought to get the three petals for 0 <= theta <= 2pi..
For r = cos 3theta, the petals rotate through half-petal angle = pi/6, in the clockwise sense.
A sample graph is made for r = 4 cos 6theta, using the Cartesian
equivalent. It is r-positive 6-petal rose, for 0 <=theta <=2pi.
graph{(x^2+y^2)^3.5-4(x^6-15x^2y^2(x^2-y^2)-y^6)=0}
