91Ó°ÊÓ

There are three types of symmetry in polar coordinates: symmetry about the polar axis, the pole, and the line \(\theta = \pi / 2\). The tests for these types of symmetry are as follows: if the equation remains unchanged when \(\theta\) is replaced with \(- \theta\), \(r\) with \(-r\), or \(\theta\) with \(\theta + \pi\), respectively. When testing \(r = 1 - \sin \theta\) for symmetry, it is obvious that the equation doesn't have symmetry with respect to the polar axis or the pole, as replacing \(\theta\) or \(r\) with their negatives changes the equation. Now, we’ll replace \(\theta\) with \(\theta + \pi\) and simplify the equation \(r = 1 - \sin (\theta + \pi)\) to observe that it does not remain the same. Hence, the given equation doesn’t have any type of symmetry.

To sketch the graph of the polar equation \(r = 1 - \sin \theta\), you can convert the polar to Cartesian coordinates, which can be done by substitifying \(x = r\cos \theta\) and \(y = r\sin \theta\). The resulting equation is \(x = (1 - y)\cos \theta\) and \(y = (1 - y)\sin \theta\). Squaring and adding both equations lead to \((x^2 + y^2) = (1 - y)^2\), which simplifies to \(x^2 + y^2 = 1 - 2y + y^2\). Further simplifying it gives a recognizable form of a circle equation, \(x^2 + (y-1/2)^2 = (1/2)^2\), a circle of radius 1/2 and center at (0,1/2). Plot this circle for the graph of the given polar equation.