Chapter 6.5, Problem 16E

Solution

To find: Whether the graph is symmetric about the $x$ -axis, $y$ -axis, or origin using the polar symmetry test. Also, use a graph to support the answer.

The graph is symmetric about $y$ -axis only and supported by the figure (1).

Given information:

The given equation is $r=1-3\sin \theta$ .

Formula used:

Odd-even identities: $\sin \left(-\theta \right)=-\sin \theta$ .

Sine of a difference identity: $\sin \left(A-B\right)=\sin A\cos B-\cos A\sin B$ .

Sine of a sum identity: $\sin (A+B)=\sin A\cos B+\cos A\sin B$ .

Calculation:

Symmetry about the $x$ -axis:

Case 1: Replace $\left(r,\theta \right)$ by $\left(r,-\theta \right)$ .

  $\begin{array}{l}r=1-3\sin \left(-\theta \right)\\ r=1+3\sin \theta \end{array}$

Case 2: Replace $\left(r,\theta \right)$ by $\left(-r,\pi -\theta \right)$ .

  $-r=3+3\sin \left(\pi -\theta \right)$

  $\begin{array}{c}-r=1-3\left(\sin \pi \cos \theta -\cos \pi \sin \theta \right)\\ -r=1-3\left(0+\sin \theta \right)\\ -r=1-3\sin \theta \\ r=-1+3\sin \theta \end{array}$

Both substitution do not give the same original polar equation.

Thus, $r=1-3\sin \theta$ is not symmetric about the $x$ -axis.

Symmetry about the $y$ -axis.

Case 1: Replace $\left(r,\theta \right)$ by $\left(-r,-\theta \right)$ .

  $\begin{array}{c}-r=1-3\sin \left(-\theta \right)\\ r=-1-3\sin \left(\theta \right)\end{array}$

Case 2: Replace $\left(r,\theta \right)$ by $\left(r,\pi -\theta \right)$ .

  $\begin{array}{c}r=1-3\sin \left(\pi -\theta \right)\\ r=1-3\left(\sin \pi \cos \theta -\cos \pi \sin \theta \right)\\ r=1-3\left(0+\sin \theta \right)\\ r=1-3\sin \theta \end{array}$

The second substitution gives the the original polar equation so it is symmetric about the $y$ -axis.

Symmetry about the origin:

Case 1: Replace $\left(r,\theta \right)$ by $\left(-r,\theta \right)$ .

  $\begin{array}{c}-r=1-3\sin \left(\theta \right)\\ r=-1+3\sin \left(\theta \right)\end{array}$

Case 2: Replace $\left(r,\theta \right)$ by $\left(r,\pi +\theta \right)$ .

  $\begin{array}{l}r=1-3\sin \left(\theta +\pi \right)\\ r=1-3\left(\sin \theta \cos \pi +\cos \theta \sin \pi \right)\\ r=1-3\left(-\sin \theta +0\right)\\ r=1+3\sin \theta \end{array}$

Both substitution do not give the same as the original polar equation so it is not symmetric about the origin.

Therefore, the graph is symmetric about $y$ -axis only.

  

Figure (1)

From the figure (1), it is validated that the curve has symmetry about $y$ -axis.