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How can I plot $\theta=\cos r$ in polar plane? Of course I know that it is different from $r=\cos \theta.$

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  • $\begingroup$ does this work: PolarPlot[ArcCos[t], {t, -Pi, Pi}]? $\endgroup$
    – Basheer Algohi
    Commented Dec 4, 2014 at 5:18
  • $\begingroup$ ContourPlot[θ == Cos[r], {r, 0, 50}, {θ, -π, π}, MaxRecursion -> 2, PlotPoints -> 50, PlotRange -> All, AspectRatio -> Automatic, DisplayFunction -> ReplaceAll[{r_Real, t_Real} :> {r*Cos[t], r*Sin[t]}]] mathematica.stackexchange.com/a/300018/72111 $\endgroup$
    – cvgmt
    Commented Mar 6 at 10:48

4 Answers 4

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  • If you can transform it to a parametric form:

    ParametricPlot[
  Evaluate @ CoordinateTransform["Polar" -> "Cartesian", {r, Cos[r]}], 
  {r, 0, 50}
]

    enter image description here

  • And if you have to use implicit form:

    ContourPlot[
    Evaluate[
        TransformedField[
            "Polar" -> "Cartesian", θ == Cos[r], {r, θ} -> {x, y}
        ] 
    ]
  , {x, 0, 50}
  , {y, -50, 50}
  , PlotPoints -> 25
  , AspectRatio -> Automatic
  , Frame -> False, Axes -> True
 ]

    enter image description here

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Does this work for you:

PolarPlot[{ArcCos[t], -ArcCos[t]}, {t, -1,1}]

enter image description here

Update: (Sjoerd C. de Vries comment)

For all r

r=(+/-)ArcCos[t]+2 n Pi

Taking few of r results:

Show[Table[
  PolarPlot[{ArcCos[t], -ArcCos[t]} + n 2 Pi, {t, -1, 1}, 
   PlotRange -> All], {n, -10, 10}]]

enter image description here

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  • $\begingroup$ Let $\theta$ is arbitrary, How is the graph? $\endgroup$
    – Nimbigli
    Commented Dec 4, 2014 at 5:45
  • $\begingroup$ I don't understand you. but in general for each theta you will find r using ArcCos[theta]. then {theta,r} will be a point in the plot. $\endgroup$
    – Basheer Algohi
    Commented Dec 4, 2014 at 5:52
  • 2
    $\begingroup$ I suppose you should take into account that r is unrestricted, so you have to work with Mod 2 pi. $\endgroup$
    – Sjoerd C. de Vries
    Commented Dec 4, 2014 at 7:28
  • $\begingroup$ Why is left side included? theta is Cos[r] so in <-1,1> interval, right? $\endgroup$
    – Kuba
    Commented Dec 4, 2014 at 9:40
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    $\begingroup$ That's probably kind of academic discussion but there is Cos[r] and r is > 0, if you assume that r can be negative, then ok. $\endgroup$
    – Kuba
    Commented Dec 4, 2014 at 9:54
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Another possible solution is to

  1. ContourPlot the equation directly;

  2. Make the coordinate transform on the coordinates of points inside the resulting graphic.

I've wrapped these steps in a function:

ClearAll@implicitPlot
implicitPlot[eq_, range__, coordSys_, opt : OptionsPattern[ContourPlot]] := 
 Module[{coord}, 
  With[{plot = ContourPlot[eq, range, opt, PlotRange -> All], 
    trans = #[coord, #2] &[Function, 
       CoordinateTransform[coordSys -> "Cartesian", {coord[1], coord[2]}]] /. 
      coord[i_] :> Part[coord, i]}, 
   plot /. GraphicsComplex[coord_, rest_] :> 
     GraphicsComplex[trans[coord\[Transpose]]\[Transpose], rest]]]

implicitPlot[Cos@r == theta, {r, 0, 40}, {theta, -8 Pi, 8 Pi}, "Polar", 
 PlotPoints -> 100]

Mathematica graphics

The advantage of this approach is, it allows us to directly set domain of definition under the interested coordinate system.

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for completeness this can be done with ContourPlot (solutions already given are faster and probably better quality)

ContourPlot[
 Cos[Norm[{x, y}]] - ArcTan[x, y] == 0, {x, 0, 100}, {y, -100, 100}, 
 PlotPoints -> 100]

enter image description here

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