The equation exhibits symmetry with respect to the line [latex]\theta =\frac{\pi }{2}[/latex], the polar axis, and the pole.

Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution [latex]u=2\theta[/latex].

[latex]\begin{align}&0=4\cos 2\theta \\ &0=4\cos u \\ &0=\cos u \\ &{\cos }^{-1}0=\frac{\pi }{2} \\ &u=\frac{\pi }{2}&& \text{Substitute }2\theta \text{ back in for }u. \\ &2\theta =\frac{\pi }{2} \\ &\theta =\frac{\pi }{4} \end{align}[/latex]

So, the point [latex]\left(0,\frac{\pi }{4}\right)[/latex] is a zero of the equation.

Now let’s find the maximum value. Since the maximum of [latex]\cos u=1[/latex] when [latex]u=0[/latex], the maximum [latex]\cos 2\theta =1[/latex] when [latex]2\theta =0[/latex]. Thus,

[latex]\begin{gathered}{r}^{2}=4\cos \left(0\right)\\ {r}^{2}=4\left(1\right)=4\\ r=\pm \sqrt{4}=2\end{gathered}[/latex]

We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line [latex]\theta =\frac{\pi }{2}[/latex], and the polar axis, we only need to plot points in the first quadrant.

Make a table similar to the table below.

[latex]\theta[/latex]	0	[latex]\frac{\pi }{6}[/latex]	[latex]\frac{\pi }{4}[/latex]	[latex]\frac{\pi }{3}[/latex]	[latex]\frac{\pi }{2}[/latex]
[latex]r[/latex]	2	[latex]\sqrt{2}[/latex]	0	[latex]\sqrt{2}[/latex]	0

Plot the points on the graph.

Analysis of the Solution

Making a substitution such as [latex]u=2\theta[/latex] is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown.

Some of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points of [latex]r[/latex]. This is because there are no real square roots for these values of [latex]\theta[/latex]. In other words, the corresponding r-values of [latex]\sqrt{4\cos \left(2\theta \right)}[/latex]
are complex numbers because there is a negative number under the radical.

Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line [latex]\theta =\frac{\pi }{2}[/latex] and the pole.

Now we will find the zeros. First make the substitution [latex]u=4\theta[/latex].

[latex]\begin{gathered}0=2\cos 4\theta \\ 0=\cos 4\theta \\ 0=\cos u\\ {\cos }^{-1}0=u\\ u=\frac{\pi }{2}\\ 4\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{8}\end{gathered}[/latex]

The zero is [latex]\theta =\frac{\pi }{8}[/latex]. The point [latex]\left(0,\frac{\pi }{8}\right)[/latex] is on the curve.

Next, we find the maximum [latex]|r|[/latex]. We know that the maximum value of [latex]\cos u=1[/latex] when [latex]\theta =0[/latex]. Thus,

[latex]\begin{gathered}r=2\cos \left(4\cdot 0\right) \\ r=2\cos \left(0\right) \\ r=2\left(1\right)=2 \end{gathered}[/latex]

The point [latex]\left(2,0\right)[/latex] is on the curve.

The graph of the rose curve has unique properties, which are revealed in the table below.

[latex]\theta[/latex]	0	[latex]\frac{\pi }{8}[/latex]	[latex]\frac{\pi }{4}[/latex]	[latex]\frac{3\pi }{8}[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\frac{5\pi }{8}[/latex]	[latex]\frac{3\pi }{4}[/latex]
[latex]r[/latex]	2	0	−2	0	2	0	−2

As [latex]r=0[/latex] when [latex]\theta =\frac{\pi }{8}[/latex], it makes sense to divide values in the table by [latex]\frac{\pi }{8}[/latex] units. A definite pattern emerges. Look at the range of r-values: 2, 0, −2, 0, 2, 0, −2, and so on. This represents the development of the curve one petal at a time. Starting at [latex]r=0[/latex], each petal extends out a distance of [latex]r=2[/latex], and then turns back to zero [latex]2n[/latex] times for a total of eight petals.

Analysis of the Solution

When these curves are drawn, it is best to plot the points in order, as in the table of Example 8’s solution. This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn.

The graph of the equation shows symmetry with respect to the line [latex]\theta =\frac{\pi }{2}[/latex]. Next, find the zeros and maximum. We will want to make the substitution [latex]u=5\theta[/latex].

[latex]\begin{gathered}0=2\sin \left(5\theta \right)\\ 0=\sin u\\ {\sin }^{-1}0=0\\ u=0\\ 5\theta =0\\ \theta =0\end{gathered}[/latex]

The maximum value is calculated at the angle where [latex]\sin \theta[/latex] is a maximum. Therefore,

[latex]\begin{gathered} r=2\sin \left(5\cdot \frac{\pi }{2}\right) \\ r=2\left(1\right)=2 \end{gathered}[/latex]

Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for [latex]n[/latex] odd yields the same number of petals as [latex]n[/latex], there will be five petals on the graph.

Create a table of values similar to the table below.

[latex]\theta[/latex]	0	[latex]\frac{\pi }{6}[/latex]	[latex]\frac{\pi }{3}[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\frac{2\pi }{3}[/latex]	[latex]\frac{5\pi }{6}[/latex]	[latex]\pi[/latex]
[latex]r[/latex]	0	1	−1.73	2	−1.73	1	0