       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
>

```

• Prev by Date: Problems with Frame in ListPlots with Legends
• Next by Date: Re: How I can fit data with a parametric equation?
• Previous by thread: Re: Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?
• Next by thread: Re: Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]?