91Ó°ÊÓ

Observe that the sine function has a range of [-1, 1]. Therefore, as \(\theta\) varies from 0 to \(2\pi\), the polar curve will have a range of \(r = [-1, 1]\).

We need to find key points on the curve to assist in sketching. These key points are when \(r=0\) or when \(\sin(2\theta) = 0\). Solve the equation for \(\theta\): \[\sin(2\theta) = 0\] The solutions for \(\theta\) are the integer multiples of \(\frac{\pi}{2}\): \[\theta = n \frac{\pi}{2} \quad \text{for } n = 0, 1, 2, ..., 4\] These are the angles where the curve intersects the origin, which are the angles \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\).

As \(\theta\) varies from 0 to \(2\pi\), the polar curve starts at the origin and sweeps outwards when \(r\) is positive and reflects about the pole when \(r\) is negative. As we identified earlier, the curve intersects the origin at angles \(\theta = n\frac{\pi}{2}\). 1. Between \(\theta = 0\) and \(\theta = \frac{\pi}{2}\), the curve starts at the origin, sweeps outwards because \(r\) is positive, and returns to the origin. 2. Between \(\theta = \frac{\pi}{2}\) and \(\theta = \pi\), the curve reflects across the pole (as \(r\) is negative), and returns to the origin. 3. Between \(\theta = \pi\) and \(\theta = \frac{3\pi}{2}\), the curve repeats its pattern in the first quadrant, now in the third quadrant, starting at the origin, sweeping outwards, and returning to the origin. 4. Finally, between \(\theta = \frac{3\pi}{2}\) and \(\theta = 2\pi\), the curve reflects across the pole once again and mimics the pattern in the second quadrant, starting at the origin and returning to the origin at \(\theta = 2\pi\). Connecting these patterns, the curve consists of 2 pairs of two overlapping petals. One pair is in quadrants 1 and 3, and another pair is reflected across the pole, covering quadrants 2 and 4.