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*: [mg91866] 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 06:12:54 -0400 (EDT)*References*: <200809080903.FAA25833@smc.vnet.net>

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: expres = {ArcCos[x/Sqrt[x^2 + y^2]], Pi/2 - ArcTan[x/Abs[y]]}; Subtract @@ Assuming[Element[x | y, Reals], Simplify[Sin /@ expres]] 0 Subtract @@ Assuming[Element[x | y, Reals], Simplify[Cos /@ expres]] 0 So for real x and y, the two expressions have equal sines and cosines. That means that they must differ by an integer multiple of 2Pi. However, since the difference is a continuous function of x and y, it has to be constant. Now, putting (for example) x=0, y=1, we see that the constant must be zero, hence they are equal. Andrzej Kozlowski On 8 Sep 2008, at 18:03, Peng Yu wrote: > Hi, > > ArcCos[x/Sqrt[x^2+y^2]] > > and > > Pi/2-ArcTan[x/Abs[y]] > > are the same. > > But I can not get the first expression be simplified to the second > one. And the following command can not be simplified to zero in > mathematica as well. > > FullSimplify[ArcCos[x/Sqrt[x^2 + y^2]] - ( Pi/2 - ArcTan[x/Abs[y]]), > Element[x, Reals] && Element[y, Reals]] > > I'm wondering if there is any walkaround to do this simplification. > > Thanks, > Peng >

**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>