How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?
- To: mathgroup at smc.vnet.net
- Subject: [mg91799] How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?
- From: Peng Yu <PengYu.UT at gmail.com>
- Date: Mon, 8 Sep 2008 05:03:06 -0400 (EDT)
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
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- From: "Peng Yu" <pengyu.ut@gmail.com>
- Re: Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?
- From: "Peng Yu" <pengyu.ut@gmail.com>
- Re: Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?
- From: "Peng Yu" <pengyu.ut@gmail.com>
- Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Re: Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?