Question

What is the type of curve represented by the equation `r=sin6theta`?

Answer 1

`r >=0 to theta in [0, pi/6], [pi/3, pi/2], [2/3pi, 5/6pi], [pi, 7/6pi], [4/3pi, 3/2pi] and [5/3pi, 11/6pi]` Graph is in the Cartesian frame. See explanation.

Explanation:

graph{(x^2+y^2)^3.5-6xy(x^4+y^4)+20x^3y^3=0 [-2 2 -1 1]} `r = sin 6theta >=0 to theta in [0, pi/6]`

The period = `(2pi)/6=pi/3`. Yet, for `theta in (pi/6, pi/3), r<0.`

There is just one loop: `r in [0, 1]`, for `theta in [0, pi/6]`.

For this loop, `theta = pi/12` is the line of symmetry

For conversion, I have used

`r=sin 6theta`

`= 6cos^5 theta sintheta-20cos^3 theta`

`sin^3theta+6costhetasin^5theta`

`=(6xy(y^4+x^4)-20x^3y^3)/r^6`, where `r=sqrt(x^2+y^2)`.

Likewise, in the same Cartesian frame, the graph is created for the

seldom cared yet grand

`r = sin (theta/6)`, with `r>=0` half-period `6pi`.

The shape is on uniform scale. The intruded arc over the zenith is

not a part of the graph, for one period. Please ignore it. Sans this

arc, the graph is OK.
graph{y-(x^2+y^2)(1-x^2-y^2)^0.5(6-32(x^2+y^2)(1-x^2-y^2))=0[-3 3 -1.5 1.5]}