# 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?

• Offhand it seems like $r = \lvert\sin(2n\theta)\rvert$ ought to have exactly $2n$ leaves.
– MJD
Mar 12, 2014 at 16:39
• @MJD Could you explain why? I don't seem to get that answer on my TI-84 or on Wolfram|Alpha. Mar 12, 2014 at 16:54
• If you don't mind the petals overlapping, look at the graphs of $r=\sin(a\theta/2)$. Mar 12, 2014 at 16:59
• Sorry, I had it backwards. Here is $r=\lvert\sin(3\theta)\rvert$, with six petals. The trick only works for odd $n$, to get $2n$ leaves, but that covers exactly the cases you were asking about.
– MJD
Mar 12, 2014 at 17:05

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:
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 not as wide as the petals of $$\sin n\theta$$, whereas the petals of $$\lvert\sin n\theta\rvert$$ are exactly the same size and shape as the petals of $$\sin n\theta$$.
I just got a nice, 6-petaled flower with $r=\sin^2(3\theta)$. Is that what you're looking for?