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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?)

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  • $\begingroup$ wouldn't any curve be the same if graphed on a polar and rectangular system? after all, they are equivalent representations $\endgroup$ – gt6989b Oct 26 '15 at 22:16
  • $\begingroup$ 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. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 26 '15 at 22:22
  • $\begingroup$ @gt6989b, i am confused why then are sin function different? $\endgroup$ – most venerable sir Oct 26 '15 at 22:33
  • $\begingroup$ @MichaelHardy, I don't really understand it either. Are you saying the opposite of @gt6989b? $\endgroup$ – most venerable sir Oct 26 '15 at 22:35
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$y=1$ 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.

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  • $\begingroup$ Oh, so each function is the same only if you plot it in their corresponding system? What kinda makes confused is that polar system is basically some concentric circles and diagonal lines with different slope, all graphed on a rectangular system (to help you plot the point). But I just realized it is the way you plot it that makes a difference. $\endgroup$ – most venerable sir Oct 27 '15 at 19:19
  • $\begingroup$ @Doeser i think this is correct... $\endgroup$ – gt6989b Oct 27 '15 at 19:22
  • $\begingroup$ When plotting in rectangular, where do you put holes in the curve? $\endgroup$ – most venerable sir Oct 27 '15 at 19:28
  • $\begingroup$ How do you plot x^2+y^2=y on a calculator, since you cannot isolate y? $\endgroup$ – most venerable sir Oct 27 '15 at 19:31
  • $\begingroup$ @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 $1/2$ $\endgroup$ – gt6989b Oct 28 '15 at 4:16

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