Re: Convert function from polar to Cartesian

• To: mathgroup at smc.vnet.net
• Subject: [mg112793] Re: Convert function from polar to Cartesian
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Thu, 30 Sep 2010 04:54:06 -0400 (EDT)
• Reply-to: hanlonr at cox.net

```\$Version

"7.0 for Mac OS X x86 (64-bit) (February 19, 2009)"

fPolar[r_, a_] = r^2 Cos[2 a];

fCart[x_, y_] = fPolar[Sqrt[x^2 + y^2], ArcTan[x, y]];

fCart[x, y] // FullSimplify

(x - y) (x + y)

Bob Hanlon

---- Sam Takoy <sam.takoy at yahoo.com> 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

```

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