# What are the characteristics of functions that look the same in both polar and rectangular graph?

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?)

• wouldn't any curve be the same if graphed on a polar and rectangular system? after all, they are equivalent representations – gt6989b Oct 26 '15 at 22:16
• I don't understand what your two cases are. If you have a $\theta$-axis that is a straight line and an $r$-axis the is a line orthogonal to it, then the graph of $r = 1/\sin\theta$ does not look like the graph you get if you take $(r,\theta)$ to be polar coordinates. The latter is a straight line. ${}\qquad{}$ – Michael Hardy Oct 26 '15 at 22:22
• @gt6989b, i am confused why then are sin function different? – most venerable sir Oct 26 '15 at 22:33
• @MichaelHardy, I don't really understand it either. Are you saying the opposite of @gt6989b? – most venerable sir Oct 26 '15 at 22:35

$y=1$ 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 $y = r \sin \theta$.
• @Doeser it seems like $x^2 + y^2 = y$ is equivalent to $x^2 + (y-1/2)^2 = (1/2)^2$, so it is a circle with the center at $(0,1/2)$ and radius $1/2$ – gt6989b Oct 28 '15 at 4:16