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