Question

Determine if the graph is symmetric about the $x-$axis, the $y-$axis, or the origin. $r=2\cos 5\theta$?

Answer 1

Explanation for the correct answer:

Step 1: Finding symmetry of given equation about $x-$axis.

The given equation is

$r=2\cos 5\theta$

A graph is symmetric about $x-$axis if $r\left(\theta \right)=r(-\theta )$

$$\begin{gathered} \therefore r\left(-\theta \right)=2\cos \left\{5\left(-\theta \right)\right\} \\ =2\cos 5\theta \left[\because \cos \left(-\theta \right)=\cos \theta \right] \\ =r\left(\theta \right) \end{gathered}$$

$\therefore r\left(\theta \right)=r(-\theta )$. So, the graph is symmetric about $x-$axis.

Step 2: Finding symmetry of given equation about $y-$axis.

Now a graph is said to be symmetric about $y-$axis if $r\left(\theta \right)=r(\mathrm{\pi }-\theta )$

$$\begin{gathered} \because r\left(\mathrm{\pi }-\mathrm{\theta }\right)=2\cos \left[5\left(\mathrm{\pi }-\mathrm{\theta }\right)\right] \\ =2\cos \left[5\mathrm{\pi }-5\mathrm{\theta }\right] \end{gathered}$$

Since, $5\mathrm{\pi }-5\mathrm{\theta }$in second quadrant, where cos is negative, so

$2\cos \left[5\mathrm{\pi }-5\mathrm{\theta }\right]=-2\cos \left(5\theta \right)$

$\because r\left(\theta \right)\ne r\left(\mathrm{\pi }+\mathrm{\theta }\right)$, So, the graph is not symmetric about $y-$axis.

Step 3: Finding symmetry of equation about origin.

Now, a graph is said to be symmetric about origin if $r\left(\theta \right)=r\left(\mathrm{\pi }+\mathrm{\theta }\right)$

$$\begin{gathered} r\left(\mathrm{\pi }+\mathrm{\theta }\right)=2\cos \left[5\left(\mathrm{\pi }+\mathrm{\theta }\right)\right] \\ =2\cos \left[5\mathrm{\pi }+5\mathrm{\theta }\right] \\ =-2\cos 5\theta \end{gathered}$$

$\because r\left(\theta \right)\ne r\left(\mathrm{\pi }+\mathrm{\theta }\right)\Rightarrow$The graph is symmetric about origin.

Hence, the graph is symmetric about $x-$axis.