Chapter 6.5, Problem 58E

Solution

To prove: The graph of $r=3\sin 4\theta$ is symmetric about the $y$ -axis, or origin using polar symmetry test.

It is prove that the graph of $r=3\sin 4\theta$ is symmetric about $y$ -axis and the origin.

Given information:

The given equation is $r=3\sin 4\theta$ .

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 $y$ -axis.

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

  $\begin{array}{c}-r=3\sin 4\left(-\theta \right)\\ r=3\sin 4\theta \end{array}$

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

  $\begin{array}{c}r=3\sin \left[4\left(\pi -\theta \right)\right]\\ r=3\left(\sin 4\pi \cos 4\theta -\cos 4\pi \sin 4\theta \right)\\ r=3\left(0-\sin 4\theta \right)\\ r=-3\sin 4\theta \end{array}$

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

Symmetry about the origin:

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

  $\begin{array}{c}-r=3\sin 4\left(\theta \right)\\ r=-3\sin 4\theta \end{array}$

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

  $\begin{array}{c}r=3\sin \left[4\left(\pi +\theta \right)\right]\\ r=3\left(\sin 4\pi \cos 4\theta +\cos 4\pi \sin 4\theta \right)\\ r=3\left(0+\sin 4\theta \right)\\ r=3\sin 4\theta \end{array}$

The second substitution gives the original polar equation so it is symmetric about the origin.

Therefore, it is prove that the graph of $r=3\sin 4\theta$ is symmetric about $y$ -axis and the origin.