Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?

*To*: mathgroup at smc.vnet.net*Subject*: [mg91880] Re: [mg91799] How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 11 Sep 2008 06:16:39 -0400 (EDT)*References*: <200809080903.FAA25833@smc.vnet.net> <F29C8163-C0D9-41CE-9C78-D4A24DE414C7@mimuw.edu.pl> <366c6f340809101456o1170addeybfa514c3255fd439@mail.gmail.com>

On 11 Sep 2008, at 06:56, Peng Yu wrote: > On Wed, Sep 10, 2008 at 7:45 AM, Andrzej Kozlowski > <akoz at mimuw.edu.pl> wrote: >> I don't think there is a way to "simplify" one of these expressions >> into the >> other, but one can use Mathematica as an aid in proving that they >> are equal >> (for real x and y). One way to do this is: > > If this is case, I therefore might have to rely on hand calculation. > But I would think Mathematica should be improved so that mathematical > expressions can be manipulated in more versatile ways. > > Thanks, > Peng I always agree that Mathematica should be improved (although perhaps I put the limit the point where it would start putting human mathematicians out of their jobs). However, that is easy to say, the hard thing is to find the right algorithms and to implement them. Algorithmic simplification of expressions is a very complex problem and only for certain types of expressions (like polynomials, rational functions, exponentials, logs and a few others) sufficiently general methods exist (and this ignores the not exactly trivial problem of defining when one expression is to be considered "simpler" than another). But the situation is much worse when you wish to convert from one form to another an expression, when the conversion is only valid under certain assumptions. If you can suggest a concrete general algorithm that you think should be implemented in Mathematica and is not implemented now I am sure WRI will be happy to oblige but just saying "it should be improved" isn't very helpful. You can be sure that there are people who are trying to do that all the time. Andrzej Kozlowski

**References**:**How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?***From:*Peng Yu <PengYu.UT@gmail.com>

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**How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?**

**Re: Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?**