Chapter 6.5, Problem 19E

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 $r=\frac{3}{1+\sin \theta }$ .

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: $\sin \left(A+B\right)=\sin A\cos B-\cos A\sin B$ .

Calculation:

Symmetry about the $x$ -axis:

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

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

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

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

Both substitution do not give the same original polar equation $r=\frac{3}{1+\sin \theta }$ .

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

Symmetry about the $y$ -axis.

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

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

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

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

Second substitution gives the original polar equation $r=\frac{3}{1+\sin \theta }$ .

Thus, $r=\frac{3}{1+\sin \theta }$ is symmetric about the $y$ -axis.

Symmetry about the origin:

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

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

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

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

Both substitution do not give the same original polar equation $r=\frac{3}{1+\sin \theta }$ .

Thus, $r=\frac{3}{1+\sin \theta }$ is not symmetric about the origin.

Therefore, graph of $r=\frac{3}{1+\sin \theta }$ is symmetric about $y$ -axis only.

Plot the graph for $r=\frac{3}{1+\sin \theta }$ .

  

Figure (1)

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