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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*: [mg91939] 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*: Sat, 13 Sep 2008 05:56:38 -0400 (EDT)
*References*: <200809080903.FAA25833@smc.vnet.net> <F29C8163-C0D9-41CE-9C78-D4A24DE414C7@mimuw.edu.pl>
In the earlier discussion of this problem I forgot that there is much
better approach. Simply use the two argument version of ArcTan to get:
Assuming[Element[x | y, Reals],
FullSimplify[ArcCos[x/Sqrt[x^2 + y^2]] - (Pi/2 - ArcTan[Abs[y], x])]]
0
Andrzej Kozlowski
On 10 Sep 2008, at 21:45, Andrzej Kozlowski 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:
>
> 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
>>
>
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