Today, I am doing practice for SAT.
In a textbook example, I see $$r=\frac{1}{\sin\theta}$$
My textbook is telling me that this particular function looks the same whether it's graphed on a polar or a rectangular system.
I checked it on demos calculator, and confirmed this, that in both cases the graph is simply $y=1$.
This confuses me because functions like $r=\sin\theta$ in polar looks no where like it does in rectangular. Why?
Also addressed in my textbook is that because $\sin\theta$ is in the denominator. It cannot be zero. Thus, there are holes on the line $y=1$ at multiples of $\pi$ 1 unit above the $x$-axis.(yeah...the two are not really the same. But mostly)
(Does this mean that for the rectangular graph to be exactly the same as its polar counterpart, $$\sin^{-1}\frac{y}{\sqrt{x^2+y^2}}$$ cannot be equal to multiple of $\pi$?)
(Also is there distinction between a simple number and a number that is in radian?)
$y=1$ in rectangular and $r = 1/\sin \theta$ in polar should plot the same curve because $y = r\sin \theta$ is the fundamental transformation between these coordinate systems.
Similarly, $r = \sin \theta$ would transform very differently, you have to use $r^2 = r \sin \theta$ instead, getting the equation $$x^2 + y^2 = y$$ which is very different. The reason for that is that the fundamental transformations are $x = r \cos \theta$ and $y = r \sin \theta$.
if this doesn't help answer your question, please comment and explain what it is that is troubling you exactly.