How can I show that the following two polar equations are the same, without graphing them? $r=8-\sin(2x)$ and $r=\sin(2x)-8$?

Question

How can I show in what sense the following two polar equations are “the same”: $$r=8-\sin (2x)$$ and $$r=\sin (2x)-8$$ They trace out the same graph. They are polar equations - consider the x to be a theta.

Answer 1

I can see that you're obviously allowing negative $r$.

Now, the transformation $$z\mapsto -z$$can be expressed in two ways, the first way, (1), is

$$r\mapsto -r$$

or, another way, (2):

$$\theta \mapsto \theta + (2n+1)\pi$$

Let's start with your first equation,

$$r=8-\sin 2x$$

Perform (1) to get

$$r=\sin 2x - 8$$

Then perform (2). Note that $\sin 2\theta$ is $\pi$- periodic, so the equation is unchanged.

Overall we have composed $z\mapsto -z$ with itself. So, we have done nothing at all (identity map overall).

Answer 2

They are not the same because $r>0$. In the first case $r=8-\sin 2x >0$ but in the second case$ r=\sin 2x-8<0$. Hence the second equation represents a null set: no curve in $(r,\theta)$ space.

Answer 3

Polar coordinate equations are usually defined only for $r\ge0$. But, as a matter of fact, graphing softwares (such as Desmos or GeoGebra) and even some books (see for instance this answer about an exercise in Spivak's Calculus book) allow for negative values of $r$.

To give a meaning to negative values of $r$, they make the most natural choice: they assume that $(,\ )$ corresponds, when $<0$, to the point $(−,\ +180°)$ (i.e. a negative radius means that the point lies in the opposite direction with respect to $θ$).

I don't know if this extension of polar coordinates to $<0$ is widely accepted, but it is enforced in graphing softwares because to graph a polar equation like $r=(\theta)$ they transform it into the parametric curve $$ x=f(\theta)\cos⁡\theta \quad y=f(\theta)\sin⁡\theta $$ and a negative value of $(\theta)$ amounts at taking the opposite vector, as described above.

In the case of $=\sin2\theta-8$, for example, we can then define $′=−$ and $′=+180°$, which inserted into the equation give: $-r'=\sin(2\theta'-360°)-8$, that is: $r'=-\sin2\theta'+8$, which is the "right" equation. Hence a graphing software will show in both cases the same graph.