Chapter 6.5, Problem 18E

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 $r=7\sin \left(3\theta \right)$ .

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 $\left(r,-\theta \right)$ .

  $\begin{array}{l}r=7\sin \left(-3\theta \right)\\ r=-7\sin \left(3\theta \right)\end{array}$

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

  $\begin{array}{c}-r=7\sin \left[3\left(\pi -\theta \right)\right]\\ -r=7\sin \left(3\pi -3\theta \right)\\ -r=7\left(\sin 3\pi \cos 3\theta -\cos 3\pi \sin 3\theta \right)\\ -r=7\left(\sin 3\theta +0\right)\\ r=-7\sin 3\theta \end{array}$

Both substitution do not give the same original polar equation $r=7\sin \left(3\theta \right)$ .

Thus, $r=7\sin \left(3\theta \right)$ is not symmetric about the $x$ -axis.

Symmetry about the $y$ -axis.

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

  $\begin{array}{c}-r=7\sin \left(-3\theta \right)\\ r=7\sin \left(3\theta \right)\end{array}$

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

  $r=7\sin \left[3\left(\pi -\theta \right)\right]$

  $\begin{array}{c}r=7\sin \left(3\pi -3\theta \right)\\ r=7\left(\sin 3\pi \cos 3\theta -\cos 3\pi \sin 3\theta \right)\\ r=7\left(\sin 3\theta +0\right)\\ r=7\sin 3\theta \end{array}$

Second substitution gives the original polar equation $r=7\sin \left(3\theta \right)$ .

Thus, $r=7\sin \left(3\theta \right)$ is symmetric about the $y$ -axis.

Symmetry about the origin:

Case1: Replace $\left(r,\theta \right)$ by $\left(-r,\theta \right)$ .

  $\begin{array}{c}-r=7\sin \left(3\theta \right)\\ r=-7\sin \left(3\theta \right)\end{array}$

Case2: Replace $\left(r,\theta \right)$ by $\left(r,\pi +\theta \right)$ .

  $r=7\sin \left[3\left(\pi +\theta \right)\right]$

  $\begin{array}{c}r=7\sin \left(3\pi +3\theta \right)\\ r=7\left(\sin 3\pi \cos 3\theta +\cos 3\pi \sin 3\theta \right)\\ r=7\left(-\sin 3\theta +0\right)\\ r=-7\sin 3\theta \end{array}$

Both substitution do not give the same original polar equation $r=7\sin \left(3\theta \right)$ .

Thus, $r=7\sin \left(3\theta \right)$ is not symmetric about the origin.

Therefore, graph of $r=7\sin \left(3\theta \right)$ is symmetric about $y$ -axis only.

  

Figure (1)

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