On 5/1/2013 2:24 PM, Nasser M. Abbasi wrote: > On 5/1/2013 11:16 AM, Jung wrote: >> Hello, >> >> I am trying to solve for: >> y >> r >> p >> >> in the following equations where I, J, K are known. >> >> I = siny * sinr + cosy * sinp * cosr >> J = -siny * cosr + cosy * sinp * sinr >> K = cosy * cosp >> >> The code I have used is: >> >> syms r p y i j k >> >> S = solve(i == sin(y)*sin(r)+cos(y)*sin(p)*cos(r), j == >> -sin(y)*cos(r)+cos(y)*sin(p)*sin(r), k == cos(y)*cos(p)) >> S.r >> S.p >> S.y >> >> but I cannot seem to get a result. >> > > > it is normally very hard to obtain analytical solutions for > trig equations since these are nonlinear and involves inverse functions > with branch cuts as well. So to solve for 'y','p', and 'r' > about, you might want to try a numerical approach. > > But I am not an expert in this. > > --Nasser >
Your problem interested me. I believe that I have a proof that, at least among real values for p, r, and y, there is no solution.
Look at the third equation cosy * cosp = 0
Suppose that cos(y) = 0. Then the first equation becomes (+-1)*sin(r) + 0 = 0 So sin(r) = 0.
The second equation is (+-1) * (+-1) + 0 = 0. This is impossible.
So in the third equation we have cos(p) = 0. so sin(p) = (+-1) Suppose sin(p) = 1. Use the addition formulas for cos(a+b) and sin(a+b). Then the first and second equations become cos(r-y) = 0 sin(r-y) = 0 But this is impossible.
So suppose sin(p) = -1. Then the first and second equations become -cos(r+y) = 0 -sin(r+y) = 0 This is impossible as well.
So there is no solution to the equations, at least over the reals, and I think even over complex, too.
Alan Weiss MATLAB mathematical toolbox documentation