Re: How to solve a simple Trig cofunction?

*To*: mathgroup at smc.vnet.net*Subject*: [mg50547] Re: [mg50500] How to solve a simple Trig cofunction?*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Thu, 9 Sep 2004 05:19:25 -0400 (EDT)*Reply-to*: hanlonr at cox.net*Sender*: owner-wri-mathgroup at wolfram.com

soln = Reduce[ {Cos[x+4]==Sin[3x+2], 0<=x<=Pi/2},x] x == (1/2)*(-3 + 2*Pi - 2*ArcTan[1 + Sqrt[2]]) To convert to degrees: soln /. a_?NumericQ :> a/Degree // FunctionExpand x == (90*(-3 + 2*Pi - 2*ArcTan[1 + Sqrt[2]]))/Pi %//N x == 26.556330730376516 Select[NSolve[Cos[x+4]==Sin[3x+2], x], 0<=(x/.#)<=Pi/2&] /. a_?NumericQ :> a/Degree {{x -> 26.55633073037652}} Bob Hanlon > > From: bfeeny at mac.com (Brian Feeny) To: mathgroup at smc.vnet.net > Date: 2004/09/08 Wed AM 05:14:59 EDT > To: mathgroup at smc.vnet.net > Subject: [mg50547] [mg50500] How to solve a simple Trig cofunction? > > Lets say I have: > > cos[X+4] = sin[3X+2] > > What would be the proper way to solve the above, for X, in > Mathematica, with the answer in degrees? > > I did this, but something tells me their has to be a better way: > > NSolve[Cos[X Degree +4 Degree] = Sin[3 X Degree+2 Degree], X] > > I mean, I wish i didn't have to enter "Degree"so many times, perhaps > their is a simple way to just tell it the whole thing is in Degree. > > Additionally, is their a way to constrain the above to 90 > X > 0? I > tried doing like: > > NSolve[Cos[X Degree +4 Degree] = Sin[3 X Degree+2 Degree], X /; 90 > >X>0] > > but that obviously is not the right way since that doesn't work. > Appreciate all the help. > > Brian > >