Question

Prove that, for every even integer n, the graph of $r=\sin n\theta$ is symmetrical with respect to the $x-$axis.

Short answer

The polar equation $r=\sin n\theta$ is symmetric with respect to $x-$axis.

Hence it is proved.

Step-by-step solution

Given information

$r=\sin n\theta$

Calculation

Consider the polar equation where $r=\sin n\theta ,n$is any integer.

The goal is to show that the equation is symmetric around the$x-$axis.

Let $(r,\theta )$be any point on the graph and $f\left(\theta \right)=\sin n\theta$where $n$is even.

By the definition of symmetry,

If a curve is symmetric with regard to the $x-$axis, then every point $(r,\theta )$on the graph is symmetrical about the $x-$axis if$(r,-\theta )$ is also on the graph.

That means the point $(r,\theta )$satisfies the relationship $r=f\left(\theta \right)$then some point of the form $(r,-\theta +2n\pi )$or $(-r,\pi -\theta +2n\pi )$satisfies the relationship for some even integer $n$

That is $r=f(-\theta +2n\pi )$ for some $n$ then the function is symmetric about $x-$axis.

Calculation

Take the equation $f\left(\theta \right)=\sin n\theta$

$$\begin{gathered} f(-\theta )=\sin n(-\theta ) \\ f(-\theta )=\sin (-n\theta ) \\ f(-\theta )=\sin (2\pi -n\theta ) \end{gathered}$$

Then,

$f(-\theta )=\sin n\theta$

As a result, every point on the graph$(r,\theta ),(r,-\theta )$ exists.

Therefore, the polar equation $r=\sin n\theta$ is symmetric with respect to $x-$axis.

Hence it is proved.