I am not satisfied, I'll be wrong? I used this

Simplify[r*Cos[θ]^2 == 
Sqrt[2]*Sin[θ + π/4] /. {r -> Sqrt[x^2 + y^2], 
z -> ArcTan[y/x]}]

enter image description here

  • 1
    $\begingroup$ why would you expect the z-> rule to do anything when there is no z in your expression? $\endgroup$
    – george2079
    Commented Nov 1, 2016 at 23:54

2 Answers 2


Perhaps this?

Simplify@TransformedField["Polar" -> "Cartesian", 
  r*Cos[θ]^2 == Sqrt[2]*Sin[θ + Pi/4], {r, θ} -> {x, y}]

(Note Pi is spelled, like all Mathematica keywords, with an initial capital.)

  • $\begingroup$ **(Note Pi is spelled, like all Mathematica keywords, with an initial capital.)**The latter do not understand, ah one more thing, what would be the reverse order of Cartesian to polar thank you very much for your attention $\endgroup$ Commented Nov 1, 2016 at 23:29
  • $\begingroup$ @juanmuñoz The reverse would be TransformedField["Cartesian" -> "Polar",...]. You can find other transformations in the documentation for TransformedField, which explains how the function works. $\endgroup$
    – Michael E2
    Commented Nov 1, 2016 at 23:32
  • $\begingroup$ @juanmuñoz You wrote pi in your question. It should be Pi, with an initial capital P. Feyre has changed it to π for some unknown reason. (I would have thought it bad form to introduce fixes in someone else's question.) -- All Mathematica function names and other keywords use capital letters (Sin[], Cos[], E, etc.). $\endgroup$
    – Michael E2
    Commented Nov 2, 2016 at 2:13

Just a rule based way (I have voted for MichaelE2's answer which uses a built-in function):

eq = r Cos[t]^2 == Sqrt[2] Sin[t + Pi/4];
den = Sqrt[x^2 + y^2];
re = Simplify[
  TrigExpand[eq] /. {r -> den, Cos[t] -> x/den, Sin[t] -> y/den}];
Legended[Show[Plot[y /. Solve[re, y][[1]], {x, -3, 3}],
  PolarPlot[Sqrt[2] Sin[t + Pi/4]/Cos[t]^2, {t, -Pi/3, Pi/4}, 
   PlotStyle -> Red], Frame -> True], 
 LineLegend[{Blue, Red}, {"\!\(\*SuperscriptBox[\(x\), \(2\)]\)-x", 
   "r= \!\(\*SqrtBox[\(2\)]\)\!\(\*FractionBox[\(\(\\\ \)\(Sin[t + \
π/4]\)\), SuperscriptBox[\(Cos[t]\), \(2\)]]\)"}]]

enter image description here

Note: y /. Solve[re, y][[1]] yields :-x+x^2


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