## Precalculus (6th Edition) Blitzer

To determine the graph, at first we will check the symmetry across the polar axis, the line $\theta =\frac{\pi }{2}$ and across the pole. To find symmetry across the polar axis, we will replace $\theta$ with $-\theta$, Then, \begin{align} & r=1+\sin \left( -\theta \right) \\ & =1-\sin \theta \end{align} So it is not symmetrical across the polar axis because $r\ne 1+\sin \theta$. Now, we will check across the line $\theta =\frac{\pi }{2}$; then we will replace $\left( r,\theta \right)=\left( -r,-\theta \right)$. \begin{align} & -r=1+\sin \left( -\theta \right) \\ & -r=1-\sin \theta \\ & r=-1+\sin \theta \end{align} So it is not symmetrical across the line $\theta =\frac{\pi }{2}$ because $r\ne 1+\sin \theta$. Now we will check symmetry across the pole by replacing $r\,\,\text{with}\,\,-r$, \begin{align} & -r=1+\sin \theta \\ & r=-1-\sin \theta \\ \end{align} So it is not symmetrical across the pole because $r\ne 1+\sin \theta$. So from the above assumption, it has not satisfied any kind of symmetry. Now we will plot the graph with the range taken as the period of $\sin \theta$, $\left[ 0,2\pi \right]$; thus the value of $\theta$ will vary from $0\,\,\text{to}\,\,2\pi$. Put $\theta =0{}^\circ$ in $r=1+\sin \theta$, Then, \begin{align} & r=1+\sin 0{}^\circ \\ & =1+0 \\ & =1 \end{align} So at $\theta =0$, $r=1$. Similarly, we put in values between $0\,\,\text{to}\,\,2\pi$ and find the values of r seen below, $\theta$ 0 $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi$ $\frac{7\pi }{6}$ $\frac{4\pi }{3}$ $\frac{3\pi }{2}$ $\frac{5\pi }{3}$ $\frac{11\pi }{6}$ $2\pi$ $r$ 1 1.5 1.87 2 1.87 1.5 1 0.5 0.13 0 0.13 0.5 1