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: > >> 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
- References:
- Inverse Tangent function
- From: "Boen S. Liong" <mr_bean_curdy@yahoo.com>
- Inverse Tangent function