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