Chapter 6.5, Problem 17E

Solution

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

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

Given information:

The given equation is $r=5\cos \left(2\theta \right)$ .

Formula used:

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

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

Cosine of a sum identity: $\cos \left(A+B\right)=\cos A\cos B-\sin A\sin B$ .

Calculation:

Symmetry about the $x$ -axis:

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

  $\begin{array}{l}r=5\cos \left(-2\theta \right)\\ r=5\cos \left(2\theta \right)\end{array}$

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

  $\begin{array}{c}-r=5\cos \left[2\left(\pi -\theta \right)\right]\\ -r=5\cos \left(2\pi -2\theta \right)\\ -r=5\left(\cos 2\pi \cos 2\theta +\sin 2\pi \sin 2\theta \right)\\ -r=5\left(\cos 2\theta +0\right)\\ r=-5\cos 2\theta \end{array}$

First substitution gives the same original polar equation $r=5\cos \left(2\theta \right)$ .

Thus, $r=5\cos \left(2\theta \right)$ is symmetric about the $x$ -axis.

Symmetry about the $y$ -axis.

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

  $\begin{array}{c}-r=5\cos \left(-2\theta \right)\\ r=-5\cos \left(2\theta \right)\end{array}$

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

  $\begin{array}{c}r=5\cos \left(2\pi -2\theta \right)\\ r=5\left(\cos 2\pi \cos 2\theta +\sin 2\pi \sin 2\theta \right)\\ r=5\left(\cos 2\theta +0\right)\\ r=5\cos 2\theta \end{array}$

Second substitution gives the original polar equation $r=5\cos \left(2\theta \right)$ .

Thus, $r=5\cos \left(2\theta \right)$ is symmetric about the $y$ -axis.

Symmetry about the origin:

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

  $\begin{array}{c}-r=5\cos 2\theta \\ r=-5\cos 2\theta \end{array}$

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

  $\begin{array}{c}r=5\cos \left(2\theta +2\pi \right)\\ r=5\left(\cos 2\pi \cos 2\theta -\sin 2\pi \sin 2\theta \right)\\ r=5\left(\cos 2\theta +0\right)\\ r=5\cos 2\theta \end{array}$

Second substitution gives the original polar equation $r=5\cos \left(2\theta \right)$ .

Thus, $r=5\cos \left(2\theta \right)$ is symmetric about the origin.

Therefore, graph is symmetric about $x-$ axis, $y$ -axis, and origin.

  

Figure (1)

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