Precalculus

Rose curves

What does a polar graph look like when the distance from the pole swings out and back several times on one trip around, and how can you tell from the equation how many petals it has and where they point?

You need

Graphing sine and cosinePrecalculus

Where you are 3 in Polar equations and their graphs

Trigonometry7

Polar coordinates and vectors10

Polar equations and their graphs3

Limacons and cardioids Polar circles and symmetry tests

Rose curves

Why this exists

In Polar circles and symmetry tests the equation r = a\cos\theta gives a distance that swings once from a to -a as the direction turns half way around, and the graph is a circle. Now make the swing faster. In r = a\cos 3\theta the distance reaches its largest size, falls to the pole and turns negative three times as often. Each time the distance leaves the pole and comes back, the point draws a loop, and the loops together look like a flower. The rectangular equations are hard to use (r = \cos 2\theta becomes (x^2 + y^2)^3 = (x^2 - y^2)^2), but the polar ones can be read at a glance: how many petals, how long, which way they point, and how much turning draws them all. The petal count depends on whether the multiplier of \theta is odd or even.

Definition

A rose curve is the graph of

r = a\cos n\theta \qquad \text{or} \qquad r = a\sin n\theta,

where a \ne 0 and n is a positive integer. As in Plotting polar points, r may be negative: (r, \theta) with r < 0 is the point |r| units from the pole in the direction \theta + \pi.

A petal is the piece of the curve drawn while \theta runs between two neighboring zeros of r: the point leaves the pole and returns to it. The tip of a petal is its point farthest from the pole, and the petal length is the distance from the pole to the tip.

Intuition

Graph r against \theta on rectangular axes, as in Graphing sine and cosine. For r = 4\cos 3\theta this is a cosine wave of height 4 and period \displaystyle \frac{2\pi}{3} (Finding the period from the equation). Its zeros are \displaystyle \frac{\pi}{3} apart.

Figure 1Each arch of r = 4\cos 3\theta between two zeros becomes one petal. The half arches at the two ends join into one petal.

Read the wave as instructions for the polar point. Across one arch, from one zero to the next, the direction turns through \displaystyle \frac{\pi}{3} while the distance grows from 0 to 4 and shrinks back to 0: the point runs out from the pole and back, one petal in a wedge of angle \displaystyle \frac{\pi}{3}. An arch above the axis draws its petal in the directions \theta is pointing. An arch below the axis has negative r, so its petal is drawn on the opposite side of the pole.

So petals are counted by counting arches, as long as two arches do not draw the same petal. Whether they do is where odd and even part ways.

Results

Petal count. When n is odd, the rose has n petals. When n is even, it has 2n petals.

Petal length. Every petal has length |a|, and every point of the rose lies within |a| of the pole. Each petal fills a wedge of angle \displaystyle \frac{\pi}{n} at the pole.

Tips. The tips come from |\cos n\theta| = 1, that is n\theta = k\pi, or from |\sin n\theta| = 1, that is \displaystyle n\theta = \frac{\pi}{2} + k\pi, for an integer k. A tip where r = -|a| points in the direction \theta + \pi. For r = a\cos n\theta with a > 0 the tips point at \displaystyle \theta = \frac{2k\pi}{n} when n is odd, and at \displaystyle \theta = \frac{k\pi}{n} when n is even.

Sine roses are turned cosine roses. Since

\sin n\theta = \cos\left(n\left(\theta - \frac{\pi}{2n}\right)\right),

the graph of r = a\sin n\theta is the graph of r = a\cos n\theta turned \displaystyle \frac{\pi}{2n} counterclockwise about the pole.

One trip. The whole rose is drawn once as \theta runs over [0, \pi) when n is odd, and over [0, 2\pi) when n is even.

The case n = 1. r = a\cos\theta and r = a\sin\theta are the circles of Polar circles and symmetry tests: one “petal” of length |a|, drawn once on [0, \pi), as the odd rule says.

Symmetry. r = a\cos n\theta is symmetric about the polar axis, since \cos(-n\theta) = \cos n\theta. r = a\sin n\theta is symmetric about the line \displaystyle \theta = \frac{\pi}{2}, since (-r, -\theta) gives -r = a\sin(-n\theta), which is the original equation.

Why it is true

Take r = a\cos n\theta; the sine rose is a turned copy of it. The zeros of \cos n\theta are at \displaystyle \theta = \frac{\pi}{2n} + \frac{k\pi}{n}, spaced \displaystyle \frac{\pi}{n} apart. Between two neighbors, r keeps one sign and |r| rises to |a| once and falls back, so each such interval draws one petal filling a wedge of angle \displaystyle \frac{\pi}{n}. With a > 0, the petals where r > 0 have tips at \displaystyle \theta = \frac{2k\pi}{n}, and the petals where r < 0 have their tips at \displaystyle \theta = \frac{(2k + 1)\pi}{n}, drawn in the direction

\frac{(2k + 1)\pi}{n} + \pi = \frac{(2k + 1 + n)\pi}{n}.

Now compare \theta with \theta + \pi:

a\cos(n(\theta + \pi)) = a\cos(n\theta + n\pi) = (-1)^n a\cos n\theta.

n odd. Then r(\theta + \pi) = -r(\theta), and (-r, \theta + \pi) is the same point as (r, \theta). Every point drawn on [\pi, 2\pi) was already drawn on [0, \pi). Also 2k + 1 + n is even, so the negative petals point at even multiples of \displaystyle \frac{\pi}{n}, exactly where the positive ones do. The tips are at \displaystyle \frac{2k\pi}{n} for k = 0, 1, \ldots, n - 1: n petals, each \displaystyle \frac{\pi}{n} wide, with gaps \displaystyle \frac{\pi}{n} wide between them. The interval [0, \pi) holds n arches, one for each petal.

n even. Then r(\theta + \pi) = r(\theta), and (r, \theta + \pi) is the image of (r, \theta) through the pole, a different point. Now 2k + 1 + n is odd, so the negative petals point at odd multiples of \displaystyle \frac{\pi}{n}, between the positive ones. Tips sit at every multiple of \displaystyle \frac{\pi}{n}: 2n petals, each \displaystyle \frac{\pi}{n} wide, filling the whole turn with no gaps. The interval [0, 2\pi) holds 2n arches, one for each petal, so the full turn is needed.

Worked examples

Example 1

Three petals from a table

Graph r = 4\cos 3\theta. Give the number of petals, their length, the directions of the tips, and the order in which the petals are drawn.

Solution

Here n = 3 is odd, so there are 3 petals of length 4, drawn once for 0 \le \theta < \pi. A table across the first two arches uses exact values of cosine at 3\theta.

\theta	3\theta	r
0	0	4
\displaystyle \frac{\pi}{12}	\displaystyle \frac{\pi}{4}	\displaystyle 2\sqrt{2}
\displaystyle \frac{\pi}{9}	\displaystyle \frac{\pi}{3}	2
\displaystyle \frac{\pi}{6}	\displaystyle \frac{\pi}{2}	0
\displaystyle \frac{2\pi}{9}	\displaystyle \frac{2\pi}{3}	-2
\displaystyle \frac{\pi}{4}	\displaystyle \frac{3\pi}{4}	\displaystyle -2\sqrt{2}
\displaystyle \frac{\pi}{3}	\pi	-4

From \theta = 0 to \displaystyle \frac{\pi}{6} the point runs from (4, 0) back to the pole: the upper half of a petal on the polar axis. From \displaystyle \frac{\pi}{6} to \displaystyle \frac{\pi}{2}, r is negative, so the point is drawn in the directions \theta + \pi; its tip, \displaystyle (-4, \frac{\pi}{3}), points at \displaystyle \frac{4\pi}{3}. From \displaystyle \frac{\pi}{2} to \displaystyle \frac{5\pi}{6}, r is positive again, with tip \displaystyle \left(4, \frac{2\pi}{3}\right). From \displaystyle \frac{5\pi}{6} to \pi, r is negative and the directions \theta + \pi run from \displaystyle \frac{11\pi}{6} to 2\pi: the lower half of the first petal.

The tips solve 3\theta = k\pi, and they point at 0, \displaystyle \frac{2\pi}{3} and \displaystyle \frac{4\pi}{3}. Each petal is \displaystyle \frac{\pi}{3} wide, and a gap of \displaystyle \frac{\pi}{3} separates neighbors.

Figure 2Example 1. Three petals of length 4, with equal gaps between them, traced once for 0 \le \theta < \pi.

Example 2

Four petals and negative r

Graph r = 3\sin 2\theta, and say which values of \theta draw each petal.

Solution

Here n = 2 is even: 4 petals of length 3, and \theta must run over all of [0, 2\pi). The zeros of \sin 2\theta are at \displaystyle \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, so each quarter turn draws one petal. The tips solve \displaystyle 2\theta = \frac{\pi}{2} + k\pi, giving \displaystyle \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.

\displaystyle 0 < \theta < \frac{\pi}{2}: \sin 2\theta > 0, and the petal lies in quadrant I with tip \displaystyle \left(3, \frac{\pi}{4}\right).
\displaystyle \frac{\pi}{2} < \theta < \pi: r < 0, so the petal is drawn opposite, in quadrant IV. Its tip is \displaystyle \left(-3, \frac{3\pi}{4}\right), which points at \displaystyle \frac{7\pi}{4}.
\displaystyle \pi < \theta < \frac{3\pi}{2}: r > 0, a petal in quadrant III with tip at \displaystyle \frac{5\pi}{4}.
\displaystyle \frac{3\pi}{2} < \theta < 2\pi: r < 0, a petal in quadrant II; the tip \displaystyle \left(-3, \frac{7\pi}{4}\right) points at \displaystyle \frac{3\pi}{4}.

Figure 3Example 2. The petals of r = 3\sin 2\theta and the values of \theta that draw them. Negative r puts the second and fourth petals opposite their directions.

Stopping at \theta = \pi would leave out the petals in quadrants II and III.

Example 3

Counting before graphing

Nazira says that r = 2\sin 5\theta has more petals than r = 2\cos 4\theta, because 5 > 4. Count the petals of each and find where the tips point.

Solution

For r = 2\cos 4\theta, n = 4 is even, so there are 2 \cdot 4 = 8 petals of length 2. The tips solve 4\theta = k\pi: \displaystyle \theta = \frac{k\pi}{4}, every eighth of a turn, along both axes and both diagonals.

For r = 2\sin 5\theta, n = 5 is odd, so there are only 5 petals of length 2. The tips with r = 2 solve \displaystyle 5\theta = \frac{\pi}{2} + 2k\pi:

\theta = \frac{\pi}{10},\ \frac{\pi}{2},\ \frac{9\pi}{10},\ \frac{13\pi}{10},\ \frac{17\pi}{10}.

The tips with r = -2, at \displaystyle \theta = \frac{3\pi}{10} + \frac{2k\pi}{5}, turned by \pi, land on the same five directions. So Nazira has it backward: the even multiplier 4 gives 8 petals and the odd multiplier 5 gives 5. One tip of the five-petaled rose points straight up, at \displaystyle \frac{\pi}{2}.

Example 4

Petal counts that cannot happen

Olzhas wants a rose with 8 petals of length 5 and one tip pointing straight up. Then he wants one with 6 petals. Write an equation for each, or show none exists.

Solution

Eight is even, so it cannot come from an odd n (which gives an odd count). From an even n, 2n = 8 gives n = 4, and the length gives |a| = 5. For r = 5\cos 4\theta the tips are at \displaystyle \theta = \frac{k\pi}{4}, which includes \displaystyle \frac{\pi}{2}; indeed \displaystyle 5\cos\left(4 \cdot \frac{\pi}{2}\right) = 5\cos 2\pi = 5. So r = 5\cos 4\theta works. (r = 5\sin 4\theta does not: its tips are at \displaystyle \frac{\pi}{8} + \frac{k\pi}{4}.)

For 6 petals, an odd n would have to be 6, which is not odd, and an even n would have to satisfy 2n = 6, so n = 3, which is not even. With n a positive integer, no rose of either form has 6 petals. The possible counts are the odd numbers 1, 3, 5, \ldots and the multiples of 4: 4, 8, 12, \ldots

Example 5

Why half a turn is enough for one rose and not another

Raushan plots r = 6\sin 3\theta and r = 6\cos 2\theta at \displaystyle \theta = \frac{\pi}{6} and at \displaystyle \theta = \frac{7\pi}{6}. Compare the points, and say what this shows about the interval each rose needs.

Solution

For r = 6\sin 3\theta: at \displaystyle \frac{\pi}{6}, \displaystyle r = 6\sin\frac{\pi}{2} = 6. At \displaystyle \frac{7\pi}{6}, \displaystyle r = 6\sin\frac{7\pi}{2} = -6, and \displaystyle \left(-6, \frac{7\pi}{6}\right) is 6 units out in the direction \displaystyle \frac{7\pi}{6} + \pi, which is \displaystyle \frac{\pi}{6} after a full turn. Both are the point \displaystyle (3\sqrt{3}, 3). Since \sin(3\theta + 3\pi) = -\sin 3\theta, this happens at every \theta: [0, \pi) is enough.

For r = 6\cos 2\theta: at \displaystyle \frac{\pi}{6}, \displaystyle r = 6\cos\frac{\pi}{3} = 3, the point \displaystyle \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right). At \displaystyle \frac{7\pi}{6}, \displaystyle r = 6\cos\frac{7\pi}{3} = 3 again, but in the opposite direction: the point \displaystyle \left(-\frac{3\sqrt{3}}{2}, -\frac{3}{2}\right). Since \cos(2\theta + 2\pi) = \cos 2\theta, every point of the second half turn is the image through the pole of a point of the first, and those images are new petals. This rose needs all of [0, 2\pi).

Example 6

A rose crossing a circle

Find every point where the rose r = 4\cos 2\theta meets the circle r = 2.

Solution

Set the right sides equal: 4\cos 2\theta = 2, so \displaystyle \cos 2\theta = \frac{1}{2}. With \theta in [0, 2\pi), 2\theta runs over [0, 4\pi), so \displaystyle 2\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3} and

\theta = \frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{7\pi}{6},\ \frac{11\pi}{6}, \quad r = 2.

That is only half the answer. Every point of the circle also has names with r = -2, since (-2, \theta) is 2 units from the pole as well. So a point of the rose with r = -2 is also on the circle. Solve 4\cos 2\theta = -2: \displaystyle \cos 2\theta = -\frac{1}{2}, so \displaystyle 2\theta = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3} and \displaystyle \theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} with r = -2. Turning each by \pi gives the points

\left(2, \tfrac{4\pi}{3}\right),\ \left(2, \tfrac{5\pi}{3}\right),\ \left(2, \tfrac{\pi}{3}\right),\ \left(2, \tfrac{2\pi}{3}\right).

The pole is on the rose but not on the circle. There are 8 points in all, two on each petal, at every multiple of \displaystyle \frac{\pi}{6} except the multiples of \displaystyle \frac{\pi}{2}.

Figure 4Example 6. Each petal crosses the circle twice. Half of the eight points come from 4\cos 2\theta = 2, the other half from 4\cos 2\theta = -2.

Boundaries

n must be a positive integer. For a non-integer n, such as \displaystyle r = \cos\frac{3\theta}{2}, the curve does not close up after one turn, petals overlap, and the counting rule does not apply.

No added constant. r = b + a\cos n\theta with b \ne 0 is not a rose; for n = 1 it is a limaçon (Limacons and cardioids).

Petals are counted on the graph, not by trips. Over [0, 2\pi) an odd rose is drawn twice.

Setting equations equal misses points. A point can lie on two curves under different names (Example 6), so solving f(\theta) = g(\theta) alone can miss intersections.

Where this goes next

Roses, circles and limaçons (Limacons and cardioids) make up most of Polar equations and their graphs. In calculus the petal intervals return: the area of a petal comes from the interval of \theta that draws it (Area in polar coordinates), and the curve’s direction from Slope of a polar curve.

Exercises

Petals, length and tips

For each rose, give the number of petals, the petal length, the directions of the tips in [0, 2\pi), and an interval of \theta that draws the whole graph once.

r = 7\cos 5\theta
r = 2\sin 4\theta
r = -3\cos 3\theta
r = 8\sin\theta
r = 6\sin 6\theta
r = -4\sin 2\theta

Points and tracing

For r = 5\cos 3\theta, find the rectangular coordinates of the points at \displaystyle \theta = \frac{\pi}{4} and at \displaystyle \theta = \frac{5\pi}{4}. What do you notice?
For r = 2\cos 2\theta, find the rectangular coordinates of the points at \displaystyle \theta = \frac{\pi}{3} and at \displaystyle \theta = \frac{4\pi}{3}. What do you notice?
Find every \theta in [0, 2\pi) at which r = 3\cos 2\theta passes through the pole.
Which interval of \theta, containing 0, draws the petal of r = 6\cos 5\theta whose tip lies on the polar axis?
As \theta runs over \displaystyle \left[\frac{\pi}{3}, \frac{2\pi}{3}\right], the rose r = 8\sin 3\theta draws one petal. Where is its tip, in rectangular coordinates?

Writing the equation

A rose with 5 petals of length 3 and one tip on the positive polar axis.
A rose with 4 petals of length 7 whose tips point at \displaystyle \frac{\pi}{4}, \displaystyle \frac{3\pi}{4}, \displaystyle \frac{5\pi}{4} and \displaystyle \frac{7\pi}{4}.
A rose with 12 petals of length 1 and one tip on the positive polar axis.
A rose of the form r = a\sin n\theta with a > 0, with 3 petals of length 2, one of them pointing straight down.

A rose and a circle

Find every point where r = 6\cos 3\theta meets the circle r = 3. Give each as (3, \theta) with \theta in [0, 2\pi).

Reasoning

Explain why no rose r = a\cos n\theta or r = a\sin n\theta has 2, 6 or 10 petals.
Show that the graph of r = a\sin n\theta is the graph of r = a\cos n\theta turned \displaystyle \frac{\pi}{2n} counterclockwise about the pole.
Show that every rose r = a\cos n\theta is symmetric about the polar axis, and that when n is even it is also symmetric about the pole.
Raushan graphs r = 5\cos 3\theta for 0 \le \theta \le 2\pi and counts six loops as her pencil moves, so she says the rose has six petals. Is she right?
Explain why no point of r = a\cos n\theta is farther than |a| from the pole, and which points are exactly |a| away.
Show that r = -a\cos n\theta has the same graph as r = a\cos n\theta when n is even, but not when n is odd.

Answers

5 petals of length 7; tips at \displaystyle 0, \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \frac{8\pi}{5}; one trip on [0, \pi). Odd n = 5, and \cos 5\theta = 1 at \displaystyle \theta = \frac{2k\pi}{5}.
8 petals of length 2; tips at the odd multiples of \displaystyle \frac{\pi}{8}, from \displaystyle \frac{\pi}{8} to \displaystyle \frac{15\pi}{8}; one trip on [0, 2\pi). Even n = 4, and \displaystyle 4\theta = \frac{\pi}{2} + k\pi.
3 petals of length 3; tips at \displaystyle \frac{\pi}{3}, \pi, \frac{5\pi}{3}; one trip on [0, \pi). r = 3 where \cos 3\theta = -1, that is 3\theta = \pi + 2k\pi; the tips with r = -3 turn onto the same three directions.
One petal of length 8: the circle with center (0, 4) and radius 4, tip (0, 8) at \displaystyle \theta = \frac{\pi}{2}; one trip on [0, \pi). This is n = 1.
12 petals of length 6; tips at \displaystyle \theta = \frac{\pi}{12} + \frac{k\pi}{6} for k = 0, 1, \ldots, 11; one trip on [0, 2\pi). Even n = 6.
4 petals of length 4; tips at \displaystyle \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}; one trip on [0, 2\pi). r = -4 at \displaystyle \frac{\pi}{4} and \displaystyle \frac{5\pi}{4} turns onto the other diagonals. It is the same graph as r = 4\sin 2\theta (exercise 22).
Both are \displaystyle \left(-\frac{5}{2}, -\frac{5}{2}\right). At \displaystyle \frac{\pi}{4}, \displaystyle r = 5\cos\frac{3\pi}{4} = -\frac{5\sqrt{2}}{2}; at \displaystyle \frac{5\pi}{4}, \displaystyle r = 5\cos\frac{15\pi}{4} = \frac{5\sqrt{2}}{2}. With n odd, the second half turn retraces the first.
\displaystyle \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) and \displaystyle \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right). Both have \displaystyle r = 2\cos\frac{2\pi}{3} = 2\cos\frac{8\pi}{3} = -1, in opposite directions, so the points are images through the pole. With n even, the second half turn draws new petals.
\displaystyle \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, where \cos 2\theta = 0.
\displaystyle -\frac{\pi}{10} \le \theta \le \frac{\pi}{10}. The zeros of \cos 5\theta nearest 0 are at \displaystyle 5\theta = \pm\frac{\pi}{2}; the petal is \displaystyle \frac{\pi}{5} wide.
(0, -8). On this interval 3\theta runs over [\pi, 2\pi], so r \le 0, with r = -8 at \displaystyle \theta = \frac{\pi}{2}; that point is 8 units out in the direction \displaystyle \frac{3\pi}{2}, straight down.
r = 3\cos 5\theta. Five petals need odd n = 5, and \cos 0 = 1 puts a tip at \theta = 0.
r = 7\sin 2\theta. Four petals need n = 2, and \sin 2\theta = \pm 1 at the four diagonal directions. (r = -7\sin 2\theta has the same graph.)
r = \cos 6\theta. Twelve petals need 2n = 12, so n = 6, and |a| = 1.
r = 2\sin 3\theta. Three petals need n = 3; the tips with r = 2 solve \displaystyle 3\theta = \frac{\pi}{2} + 2k\pi, giving \displaystyle \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}, and \displaystyle \frac{3\pi}{2} points straight down.
Six points: \displaystyle \left(3, \frac{\pi}{9}\right), \displaystyle \left(3, \frac{5\pi}{9}\right), \displaystyle \left(3, \frac{7\pi}{9}\right), \displaystyle \left(3, \frac{11\pi}{9}\right), \displaystyle \left(3, \frac{13\pi}{9}\right), \displaystyle \left(3, \frac{17\pi}{9}\right). On [0, \pi), \displaystyle \cos 3\theta = \frac{1}{2} gives \displaystyle \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9} with r = 3; \displaystyle \cos 3\theta = -\frac{1}{2} gives \displaystyle \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9} with r = -3, which turn by \pi.
An odd n gives n petals, an odd number. An even n gives 2n petals, and 2n is a multiple of 4 because n is even. The numbers 2, 6 and 10 are even but not multiples of 4. A sine rose is a turned cosine rose, so it has the same counts.
\displaystyle a\sin\left(n\theta + \frac{\pi}{2}\right) = a\cos n\theta, and \displaystyle n\theta + \frac{\pi}{2} is \displaystyle n\left(\theta + \frac{\pi}{2n}\right). So if (r, \theta) is on r = a\cos n\theta, then \displaystyle \left(r, \theta + \frac{\pi}{2n}\right) is on r = a\sin n\theta, and that point is (r, \theta) turned \displaystyle \frac{\pi}{2n} counterclockwise. Every point of the sine rose arises this way, by reading the same identity backward.
Replacing \theta by -\theta gives r = a\cos(-n\theta) = a\cos n\theta, the original, so the graph is symmetric about the polar axis. Replacing \theta by \theta + \pi gives a\cos(n\theta + n\pi) = (-1)^n a\cos n\theta, the original when n is even, so then the graph is symmetric about the pole.
No, it has 3. Since 5\cos(3\theta + 3\pi) = -5\cos 3\theta, the point at \theta + \pi is the point at \theta, so on [\pi, 2\pi] she retraced the same three petals.
|\cos n\theta| \le 1, so the distance |r| = |a|\,|\cos n\theta| \le |a|. Equality holds only where |\cos n\theta| = 1, at \displaystyle \theta = \frac{k\pi}{n}: those points are the tips.
-a\cos n\theta = a\cos(n\theta - \pi), and n\theta - \pi is \displaystyle n\left(\theta - \frac{\pi}{n}\right), so r = -a\cos n\theta is r = a\cos n\theta turned \displaystyle \frac{\pi}{n}. For even n the tips sit at every multiple of \displaystyle \frac{\pi}{n}, so the turn maps the rose onto itself. For odd n (take a > 0) the tips sit at the even multiples, and the turn moves them into the gaps: a different graph.