Re: Re: Inverse Tangent function

• To: mathgroup at smc.vnet.net
• Subject: [mg80051] Re: [mg80023] Re: [mg79970] Inverse Tangent function
• From: DrMajorBob <drmajorbob at bigfoot.com>
• Date: Sat, 11 Aug 2007 02:14:17 -0400 (EDT)
• References: <200708100538.BAA10433@smc.vnet.net> <32688575.1186748325330.JavaMail.root@m35>
• Reply-to: drmajorbob at bigfoot.com

```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:
>
>>
>> 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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