91Ó°ÊÓ

Apply the three types of symmetry rules to the equation \(r = 1+ \sin \theta\). - For the symmetry about the x-axis (or the polar axis), if replacing \( \theta \) by \( -\theta \) yields an equivalent equation, then the graph is symmetric about the x-axis. If we replace \( \theta \) with \( -\theta \) in our equation, \(r = 1+ \sin(-\theta)\), simplifies to \(r = 1- \sin(\theta) \) which is not the same as the original equation, so there's no symmetry about the x-axis. - For symmetry about the y-axis, if replacing \( r \) and \( \theta \) by \( -r \) and \( \theta+ \pi \) respectively yields an equivalent equation, the graph is symmetric about the y-axis. If we replace \( r \) with \( -r \) and \( \theta \) with \( \theta + \pi \) we get \(-r = 1+ \sin (\theta + \pi)\), and simplified this gives \(-r =1-\sin \theta\), which is also not the same as the original equation, so there's no symmetry about the y-axis.- Finally, for symmetry about the origin, if replacing \( r \) by \( -r \) and \( \theta \) by \( \theta+ \pi \) yields an equivalent equation, the graph is symmetric about the origin. In this case we again get \(-r =1-\sin \theta\), which is not the same equation, so there's also no symmetry about the origin.

To sketch the polar curve, we can start by transforming polar equation to Cartesian coordinates. As the equation involves only \( r \) and \( \sin \theta \), we know that \( y = r \sin \theta \), we can multiply both sides of \( r = 1+ \sin \theta \) by \( r \) to get \( r^2 = r +r\sin \theta \) or \( r^2 = r +y \). The polar curve is a limaçon that is centered at (0.5, 0) on x-y Cartesian system and extends from -0.5 to 1.5 along the x-axis.