We are investigating why, in polar roses, when n is even there are 2n petals, but when n is odd, there are n petals.
Rose curves are planar curves defined by the polar equation r= cos(n) or r=a sin (b* theta), where b is greater than or equal to 2 and is an integer.
Cartesian Equivalent
To convert from polar to Cartesian coordinates, the equation becomes:
x=cos(n) (cos*theta), y=cos (n) (sin*theta)
Here are some examples of rose curves:
r=2sin5θ r=2cos2θ r=2sin10θ
Symmetry
We look for symmetry along the x-axis, the y-axis, and the origin.
We look for symmetry of graphs that use sin.
r= a sin (b*theta)
we multiply by negative theta, to get a a sin (-b*theta), which tells us that - a sin b*theta = -r, with coordinates of (-r, -theta), and so those polar roses will be symmetric along the y-axis.
We then look for symmetry of graphs that use cosine, and conclude that those polar roses will be symmetric along the x-axis, because of the same principle pertaining to the sine graph and the y-axis. Cosine is at its maximum at theta = 0 and 2pi, therefore at those points, the value will be furthest from the pole for cosine, while sin is at a displacement of zero.
We note that when the coordinates are reflexive, that the curve is symmetrical.
We observe that when n is even, we have 2n petals, and when n is odd, we have n petals.
We also observe that when n is odd, the petals overlap, because because cos(n *theta) can produce negative values, because of the constant.
WHY? Because when n is odd and theta = pi, cos(k) = -1. That resets the graph back to the same point as r=1. The rest of the graph will be a repetition.
On the other hand, if n is even, the cos (n theta) = 1 when theta equals pi, which is on the opposite side of the origin. The rest of the graph follows that pattern, but it is reflected, so it creates double the amount of petals.
That is the reason why when n is even, we have 2n petals in a polar rose, and when n is odd, we have n petals in a polar rose.
Important things to remember:
One Interesting fact that we noted: