Prove that when n is a positive odd integer, the polar rose $r=\cos n\theta orr=\sin n\theta$ is traced twice on the interval $[0,2\pi ]$ and thus has exactly $n$ petals.

The polar roses $r=\cos n\theta$ and $r=\sin n\theta$ where $n$ is a odd positive integer.

A polar curve is a shape constructed using the polar coordinate system.

The goal is to show that $r=\cos n\theta$and$r=\sin n\theta$ have the same number of petals.

Let $(r,\theta )$by any point on the graph,

$f\left(\theta \right)=\cos n\theta$

Thus, $r=\cos n\theta$where $n$is a odd positive integer.

If $n$an any odd positive integer then by angle sum identity of cos

$$\begin{gathered} \cos n\theta =\cos n(\pi +\theta ) \\ =-\cos n\theta \end{gathered}$$

Therefore, the point $(-r,\cos n(\pi +\theta \left)\right)=(-r,-\cos n\theta )$has the same location as $(r,\theta )$

As a result, the graph drawn on the interval $[\pi ,2\pi ]$is identical to the graph shown on the interval $[0,\pi ]$

As a result, the graph has exactly $n$ petals and is traced twice in the interval.

Let $(r,\theta )$by any point on the graph,

$f\left(\theta \right)=\sin n\theta$

Thus, $f\left(\theta \right)=\sin n\theta$where $n$is a odd positive integer.

If $n$is any odd positive integer then by angle sum identity of cos

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

Therefore, for the point $(-r,\sin n(\pi +\theta \left)\right)=(-r,-\sin n\theta )$has the same location as $(r, \theta) .$

As a result, the graph is traced the $[\pi ,2\pi ]$corresponds to the graph drawn on the interval $[0,\pi ]$

Therefore, the graph is traced twice in the interval and has exactly $n$petals.