Re: Convert function from polar to Cartesian

*To*: mathgroup at smc.vnet.net*Subject*: [mg112783] Re: Convert function from polar to Cartesian*From*: Peter Breitfeld <phbrf at t-online.de>*Date*: Thu, 30 Sep 2010 04:52:15 -0400 (EDT)*References*: <i7usgt$rdn$1@smc.vnet.net>

Sam Takoy wrote: > Hi, > > I've stumbled onto an unexpected problem. I have a function in polar > coordinate and I need to convert it to Cartesian coordinates: > > fPolar[r_, a_] = r^2 Cos[2 a]; > fCart[x_, y_] = fPolar[Sqrt[x^2 + y^2], ArcTan[x, y]]; > fCart[x, y] > > I get: (x^2 + y^2) Cos[2 ArcTan[x, y]]. > > But how to get rid of the ArcTan[x, y]? I've tried this: > > fCart[x, y] // > TrigExpand /. {Cos[ArcTan[x, y]] -> x/Sqrt[x^2 + y^2], > Sin[ArcTan[x, y]] -> y/Sqrt[x^2 + y^2]} > > but it does not do the substitution I need it to do. So what's the > proper way to get rid of the ArcTan[x, y] and for the simple example > above, to obtain x^2 - y^2? > > > Many thanks in advance, > > Sam > I think you can't always get rid of the ArcTan, but this works for your case: cartRule = {r -> Sqrt[x^2 + y^2], a -> ArcTan[x, y]}; (((fPolar[r, a] // TrigToExp) /. cartRule) // FullSimplify) // Expand Out= x^2 - y^2 -- _________________________________________________________________ Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de