91Ó°ÊÓ

Test for symmetry about x-axis, y-axis and origin. These are the three general types of symmetries for polar curves: \n- Symmetry about the x-axis can be tested by replacing \(\theta\) with \(-\theta\) in the equation and simplifying. If we get the original equation, then the graph is symmetric about x-axis.\n- Symmetry about the y-axis can be tested by replacing \(\theta\) with \(\(\pi -\theta\)\) in the equation. If we get the original equation, then the graph is symmetric about y-axis.\n- Symmetry about the origin can be tested by replacing \(\theta\) with \(\(\theta + \pi\)\) or by replacing \(r\) with \(-r\) in the equation. If we get the original equation, then the graph is symmetric about the origin. \nFor \(r =2 \sin \theta\), if we replace \(\theta\) by \(-\theta\), the equation becomes \(r = 2\sin(-\theta) = -2 \sin\theta\). Since the resulting equation is not the same as the original, so the graph of the equation is not symmetric about the x-axis. But, if we replace \(\theta\) by \(\pi-\theta\), the equation becomes \(r = 2\sin(\pi-\theta) = 2\sin\theta\). So, the graph is symmetric about the y-axis. Therefore, we only need to plot for \(\theta\) from 0 to \pi, since the graph will just reflect about the y-axis for \(\theta\) from \(-\pi\) to 0.

Plot points for various values of \(\theta\) from 0 to \pi. Calculate \(r\) for each \(\theta\). For instance choose, \(\theta=0, \pi/6, \pi/4, \pi/3, \pi/2, 2\pi/3, 3\pi/4, 5\pi/6\) and \(\pi\). We find that \(r= 0, 1, \sqrt{2}, \sqrt{3}, 2, \sqrt{3}, \sqrt{2}, 1\) and 0 respectively.

Draw these points on the polar graph. For each point, start at the origin, move along the \(\theta\) angle, then move along the radius \(r\). This set of points will form a circle. Remember to reflect the graph across the y-axis because it has y-axis symmetry. Note that as \(r\) can take on negative values, a negative radius means to move in the opposite direction of the angle.