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MathGroup Archive 2007

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Re: Inverse Tangent function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80050] Re: [mg79970] Inverse Tangent function
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Sat, 11 Aug 2007 02:13:46 -0400 (EDT)
  • References: <15063506.1186727312121.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

Something like...

Off[Solve::"ifun"]
Solve[{Tan[w1] == x, Tan[w2] == y,
   z == TrigExpand@Tan[w2 - w1]}, z, {w1, w2}]

{{z -> (-x + y)/(1 + x y)}}

or

Off[Solve::"ifun"]
Solve[TrigExpand /@ {Tan[w1] == x, Tan[w2] == y,
    z == Tan[w2 - w1]}, z, {w1, w2}]

{{z -> (-x + y)/(1 + x y)}}

Watch out for "artifact" solutions and branching, of course.

Bobby

On Fri, 10 Aug 2007 00:38:13 -0500, Boen S. Liong  
<mr_bean_curdy at yahoo.com> 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
>
>
>



-- 

DrMajorBob at bigfoot.com


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