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MathGroup Archive 2007

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Re: Re: Inverse Tangent function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80054] Re: [mg80023] Re: [mg79970] Inverse Tangent function
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sat, 11 Aug 2007 02:15:51 -0400 (EDT)
  • References: <200708100538.BAA10433@smc.vnet.net> <32688575.1186748325330.JavaMail.root@m35> <op.twutlyhoqu6oor@monster.gateway.2wire.net>

Yes, of course. I guess I thought (for about 1 second that I devoted  
to this question) that Simplifying wasn't the main point but  
according to the OP my answer was "completely out of context".  I  
still don't know what that "context" was ???

Andrzej


On 10 Aug 2007, at 19:31, DrMajorBob wrote:

> Simplify greatly improves that result:
>
> TrigExpand[Tan[ArcTan[x] - ArcTan[y]]] // Simplify
>
> (x - y)/(1 + x y)
>
> Bobby
>
> On Fri, 10 Aug 2007 05:46:19 -0500, Andrzej Kozlowski  
> <akoz at mimuw.edu.pl> wrote:
>
>>
>> On 10 Aug 2007, at 07:38, Boen S. Liong wrote:
>>
>>> Please help:
>>>
>>> Let w1 = tan-1(x)  i.e. inverse tangent of x or tan(w1) = x
>>>
>>> w2 = tan-1(y)  i.e. tan(y) = w2
>>>
>>> and w3 = w2- w1
>>>
>>> problem: find tan(w3) in term of x and y.
>>>
>>> or better yet:
>>>
>>> let x= x1/z1
>>>
>>> and y= y1/z1
>>>
>>> problem:find tan(w3) in term of x and y.
>>>
>>> I know the formula for tan( u+v) or tan(u-v). but please help for  
>>> the
>>> above. thank you.
>>>
>>> Boen S. Liong
>>>
>>>
>>
>> TrigExpand[Tan[ArcTan[x] - ArcTan[y]]]
>>   x/(Sqrt[x^2 + 1]*Sqrt[y^2 + 1]*((x*y)/(Sqrt[x^2 + 1]*Sqrt[y^2 +  
>> 1]) +
>>             1/(Sqrt[x^2 + 1]*Sqrt[y^2 + 1]))) -
>>     y/(Sqrt[x^2 + 1]*Sqrt[y^2 + 1]*((x*y)/(Sqrt[x^2 + 1]*Sqrt[y^2 +
>> 1]) +
>>             1/(Sqrt[x^2 + 1]*Sqrt[y^2 + 1])))
>>
>> P.S. I don't see the point of the "or better yet" part of your  
>> question.
>>
>> Andrzej Kozlowski
>>
>>
>
>
>
> -- 
> DrMajorBob at bigfoot.com



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