Mathematical roses with $4n+2$ petals In polar coordinates $(r, \theta)$, the equation $$r = \sin\left(a \theta\right)$$ gives a rose with $a$ petals if $a$ is odd, or $2a$ petals if $a$ is even.
Thus, the number of petals generated for some values of $a$ are
   a   | petals
=======+========
   1   |   1
   2   |   4
   3   |   3
   4   |   8
   5   |   5
   6   |   12
   7   |   7

Conspicuously missing from this table are roses with $4n+2$ petals. How can you generate a rose of the same "shape" with $2, 6, 10, \ldots$ petals? If you can't, why (intuitively) is it impossible?
 A: What is happening here is that $\sin n\theta$ has $n$  positive lobes and $n$ negative lobes.  When $n$ is odd, the negative lobes exactly overlap the positive lobes in the graph, so you only see $n$ petals.  When $n$ is even, the negative lobes appear separately, so you see $n$ positive and $n$ negative lobes, for a total of $2n$ petals.
To get $2n$ petals when $n$ is odd, you can use the absolute value function to separate the negative and positive petals.  The graph of $$r=\lvert\sin n\theta\rvert$$ has exactly $2n$ petals, even in the case  when $n$ is odd.  So for example $r=\lvert\sin 3\theta\rvert$ has this graph: 

In the 3-petal rose $r = \sin3\theta$, three of the leaves are reflected across the origin onto the other three leaves, which is why the rose appears to have only 3 leaves:

But actually all six petals are there; it's just that they coincide in three pairs, so you can only see three petals in the graph.
The suggestion of Tony Jacobs elsewhere in this thread, of using $r = \sin^2 3\theta$, is essentially the same; the squaring operation forces the formerly invisible negative lobes onto the opposite side of the origin.  But the squaring also attenuates the petals: the petals of $\sin^2 n\theta$ are only as wide as the petals of $\sin 2n\theta$, whereas the petals of $\lvert\sin n\theta\rvert$ are exactly the same size and shape as the petals of $\sin n\theta$.
A: I just got a nice, 6-petaled flower with $r=\sin^2(3\theta)$. Is that what you're looking for?
