Chapter 6.5, Problem 14E

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 $y$ -axis only and supported by the figure (1).

Given information:

The given equation is $r=1+2\cos \theta$ .

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:

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

  $\begin{array}{l}r=1+2\cos \left(-\theta \right)\\ r=1+2\cos \theta \end{array}$

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

  $-r=1+2\cos \left(\pi -\theta \right)$

  $\begin{array}{c}-r=1+2\left(\cos \pi \cos \theta +\sin \pi \sin \theta \right)\\ -r=1+2\left(-\cos \theta +0\right)\\ r=-1+2\cos \theta \end{array}$

First substitution gives the same original polar equation.

Thus, $r=1+2\cos \theta$ is symmetric about the $x$ -axis.

Symmetry about the $y$ -axis.

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

  $\begin{array}{c}-r=1+2\cos \left(-\theta \right)\\ -r=1+2\cos \left(\theta \right)\end{array}$

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

  $\begin{array}{c}r=1+2\cos \left(\pi -\theta \right)\\ r=1+2\left(\cos \pi \cos \theta +\sin \pi \sin \theta \right)\\ r=1+2\left(-\cos \theta +0\right)\\ r=1-2\cos \theta \end{array}$

Both substitution do not give the original polar equation

Thus, $r=1+2\cos \theta$ is not symmetric about the $y$ -axis.

Symmetry about the origin:

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

  $\begin{array}{c}-r=1+2\cos \theta \\ r=-1-2\cos \theta \end{array}$

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

  $\begin{array}{c}r=1+2\cos \left(\theta +\pi \right)\\ r=1+2\left(\cos \pi \cos \theta -\sin \pi \sin \theta \right)\\ r=1+2\left(-\cos \theta +0\right)\\ r=1-2\cos \theta \end{array}$

Both substitution do not give the original polar equation

Thus, $r=1+2\cos \theta$ is not symmetric about the origin.

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

Plot the graph for $r=1+2\cos \theta$ .

  

Figure (1)

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