Re: solving for 2 angles

*To*: mathgroup at smc.vnet.net*Subject*: [mg89821] Re: solving for 2 angles*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Sat, 21 Jun 2008 05:32:25 -0400 (EDT)*Organization*: The Open University, Milton Keynes, UK*References*: <g3g00h$kvd$1@smc.vnet.net>

Steve wrote: > I have two equations with unknown angles az and el: > > J*cos(az)*sin(el)+K*sin(az)*sin(el)+L*cos(el)=0; > M*cos(az)*sin(el)+N*sin(az)*sin(el)+P*cos(el)=0; > > Is it a simple matter to solve for the angles? > > I tried taking the first equation, and re-writing as: > > sin(el)*[J*cos(az)+K*sin(az) ] = -L*cos(el) > > which leads to > > el= atan(-L / [J*cos(az)+K*sin(az)]) > > I then subsituted this into the second equation, but was not able to > isolate the az variable. Steve, This newsgroup is dedicated to Mathematica, software made and published by Wolfram Research, Inc. Thus, I assume that you are using it and I shall show you the syntax and command to use to get the complete set of solutions. eqns = {J*Cos[az]*Sin[el] + K*Sin[az]*Sin[el] + L*Cos[el] == 0, M*Cos[az]*Sin[el] + N*Sin[az]*Sin[el] + P*Cos[el] == 0}; Reduce[eqns, {az, el}] ((C[1] | C[2]) \[Element] Integers && ((L == 0 && P == 0 && az == Pi + 2 Pi C[1] && el == Pi + 2 Pi C[2]) || 2 2 2 2 L M (J != 0 && ((L (J + L - J Sqrt[J + L ]) != 0 && P == --- && J az == Pi + 2 Pi C[1] && 2 2 -J + Sqrt[J + L ] el == 2 (ArcTan[------------------] + Pi C[2])) || L 2 2 2 2 L M (L (J + L + J Sqrt[J + L ]) != 0 && P == --- && J <...> (* the rest of this *very* long output has been deleted for the reader's sake. Just evaluate the original expressions within Mathematica on your system to get it again. *) Regards, -- Jean-Marc