91Ó°ÊÓ

$r=\sin n\mathrm{θ}$

Consider the polar equation where $r=\sin n\mathrm{θ},n$is any integer.

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

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

By the definition of symmetry,

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

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

That is $r=f(-\mathrm{θ}+2n\mathrm{π})$ for some $n$ then the function is symmetric about $x-$axis.

Take the equation $f\left(\mathrm{θ}\right)=\sin n\mathrm{θ}$

$$\begin{gathered} f(-\mathrm{θ})=\sin n(-\mathrm{θ}) \\ f(-\mathrm{θ})=\sin (-n\mathrm{θ}) \\ f(-\mathrm{θ})=\sin (2\mathrm{π}-n\mathrm{θ}) \end{gathered}$$

Then,

$f(-\mathrm{θ})=\sin n\mathrm{θ}$

As a result, every point on the graph$(r,\mathrm{θ}),(r,-\mathrm{θ})$ exists.

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

Hence it is proved.