I am not sure I do it in a correct way, so I decided to ask for your opinion. For example, given the polar equation, which I have to plot on a polar graph, $$r = 2 \sin(2\theta)$$

if I have chosen subintervals $[0, π/4]$, $[π/4, π/2]$, $[3π/4, π]$, to determine value of $x$ and $y$ on the $xy$ - graph, do I take the value of $\theta$, or the double value of $\theta$ as in the original equation? Namely:

$$ x = r \cos(\theta)$$ or $$x = r \cos(2\theta)$$

for instance, for the angle π/4: $$ x = r \cos(π/4)$$ or $$x = r \cos(2π/4) = r \cos(π/2)$$

EDIT: I am adding more detailed description of the problem.

Initial expression: $$r = 2 \sin(2\theta)$$

To find $x$ and $y$ we use: $$ x = r \cos(\theta)$$ and $$ y = r \sin(\theta)$$

For example, I take the following values of $\theta$:

$\theta = \{0, π/4, π/2\}$

To compute $r$ in the initial expression I do this: $$r = 2 \sin(2x0)$$ => r = 0, therefore both $x$ and $y$ equal $0$, because $r=0$.

Next value of $r$: $$r = 2 \sin(2π/4)$$ => $$r = 2 \sin(π/2)$$ therefore, $r = 2$

And now which value of $\theta$ shall I use in both $ x = r \cos(\theta)$ and $ y = r \sin(\theta)$ - shall I use $π/4$, or shall I take $2\theta$ and hence use $π/2$? Which version is the correct one:

$$ x = 2 \cos(π/4)$$ $$ y = 2 \sin(π/4)$$ or $$ x = 2 \cos(π/2)$$ $$ y = 2 \sin(π/2)$$

I have visually checked the approach, in which I use the value of $\theta$ (not $2\theta$) for both $x$ and $y$, with the graph kindly suggested by Donald Splutterwit, and it seems that it is a correct way, because as I see in the graph,

when $\theta = π/6$, and thus $r = \sqrt{3}$, and then, using $\theta = π/6$, $x = 1.5$ and $y = 0.87$;

when $\theta = π/3$, and thus again $r = \sqrt{3}$, and then, using $\theta = π/3$, $x = 0.87$ and $y = 1.5$.

So I presume I shall use the exact value of $\theta$, and not the doubled one, correct?

Thank you very much!

  • $\begingroup$ I touched up some of the formatting you had...mainly the trigonometric functions and $\theta.$ $\endgroup$ – bjcolby15 May 29 '17 at 16:54
  • $\begingroup$ @bjcolby15 Thank you very much. It does look much better now. I will try to follow this formatting in future. $\endgroup$ – Vitale May 29 '17 at 17:06
  • $\begingroup$ You're welcome! When I saw $\Theta$ and $2 \partial$ and some of the trig functions in italics, I said, "hmm, that doesn't look right." (You can also use $\vartheta$ (\vartheta) if you want to get fancy.) $\endgroup$ – bjcolby15 May 29 '17 at 18:44

To change a polar equation into cartesian coordinates ... remember the equations \begin{eqnarray*} r^2=x^2+y^2 \\ x=r \cos(\theta) \\ y=r \sin ( \theta) \end{eqnarray*} Now use the double angle formula for sin ($\sin(2 \theta) = 2 \sin(\theta) \cos( \theta)$) & multiply the equation by $r^2$. The polar equation becomes \begin{eqnarray*} (x^2+y^2)^{\frac{3}{2}}=4 x y \end{eqnarray*} Now use the Desmos online graphing calculator https://www.desmos.com/calculator/zxgizn9onc enter image description here


Your equation is $r=2 \sin 2 \theta$ ... Multiplying by $r^2$ & using the double angle formula $\sin(2 \theta) = 2 \sin(\theta) \cos( \theta)$ ... We have \begin{eqnarray*} \color{green}{r^3}= 4 \color{red}{r\sin(\theta)} \color{blue}{r \cos( \theta)} \end{eqnarray*} Now recall that $\color{blue}{x=r \cos( \theta)} $ , $\color{red}{y=r\sin(\theta)}$ and $ \color{green}{r=(x^2+y^2)^{\frac{1}{2}}}$ So \begin{eqnarray*} \color{green}{(x^2+y^2)^{\frac{3}{2}}}= 4 \color{blue}{x} \color{red}{y}. \end{eqnarray*}

  • $\begingroup$ Thank you very much for your explanation, and for showing me the online graphing tool. I am not sure I understand how you got to this equation \begin{eqnarray*} (x^2+y^2)^{\frac{3}{2}}=4 x y \end{eqnarray*} I will try to deduce, and will be back in case I can't. $\endgroup$ – Vitale May 29 '17 at 17:11
  • $\begingroup$ One more question, if I may: please, share what language did you use to type this (I see how you did it, but I don't know where it comes from) \begin{eqnarray*} (x^2+y^2)^{\frac{3}{2}}=4 x y \end{eqnarray*} $\endgroup$ – Vitale May 29 '17 at 17:12
  • $\begingroup$ @Vitale ... You can press the edit button to see the LaTeX ... is that what you meant by "what language" ? $\endgroup$ – Donald Splutterwit May 29 '17 at 17:26
  • $\begingroup$ This is beautiful. Thank you very much. The only thing - I have asked a different question. I am graphing manually, and therefore given the equation $r=2 \sin 2 \theta$ and values of r and $\theta$ on particular intervals (for example, when $\theta$ = π/4), I want to find exact values of x and y, using the formulas $$x=r \cos(\theta)$$ and $$y=r \sin ( \theta)$$, as I showed in the original post. So my question is shall I use $2\theta$ in the last two formulas, or shall I use just $\theta$ - I have showed computations in the original post. $\endgroup$ – Vitale May 30 '17 at 6:33

Re: your edit - yes, use the exact value of $\theta$ in the equation if you're finding $x$ and $y$ coordinates in $x = r \cos \theta$ and $y = r \sin \theta$. You can also trace the graph below to verify those findings.


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