Solving Trigonometric Equations with More Than One Variables

Hi.

My problem is that I have a 3x1 matrix with trigonometric expressions such that:

x = [cos(theta)*sin(phi)*sin(psi) + cos(phi)*sin(theta); -cos(psi)*sin(theta); cos(phi)*sin(psi)*cos(theta) - sin(phi)*cos(psi)*sin(theta)]

And it should be equal to another matrix

with real numbers like:

y = [0.6,-0.76,0]

My code is:

syms theta phi psi

eqn_1 = cos(theta)*sin(phi)*sin(psi) + cos(phi)*sin(theta) == 0.6;
eqn_2 = -cos(psi)*sin(theta) == -0.76 ;
eqn_3 = cos(phi)*sin(psi)*cos(theta) - sin(phi)*cos(psi)*sin(theta) == 0 ;
eqn_4 = theta == -30 * pi / 180 ;
 s = solve(eqn_1,eqn_2,eqn_3,eqn_4,theta,phi,psi,'ReturnConditions',true)  ;

s.theta

 
ans =
 
Empty sym: 0-by-1
 

s.phi

 
ans =
 
Empty sym: 0-by-1
 

s.psi

 
ans =
 
Empty sym: 0-by-1
 

But as you can see, it does not work. Can you help me,please?

Thanks.

Answer 1

[John D'Errico]

You need to understand that first, the use of solve like this will probably fail. You have 4 equations, but only 3 unknowns. Solve will try to find an exact solution, but the problem is over-determined, so there will be no exact solution.

As bad is the problem that this is equivalent to a polynomial problem in the unknowns, where the degree of the polynomial may be too high for an algebraic solution to be found.

syms theta phi psi
eqn_1 = cos(theta)*sin(phi)*sin(psi) + cos(phi)*sin(theta) == 0.6;
eqn_2 = -cos(psi)*sin(theta) == -0.76 ;
eqn_3 = cos(phi)*sin(psi)*cos(theta) - sin(phi)*cos(psi)*sin(theta) == 0 ;

First, ignore the constraint on theta. We will now get two primary solutions, although infinitely many solutions will exist.

sol = solve(eqn_1,eqn_2,eqn_3)

sol = struct with fields:
      phi: [2×1 sym]
      psi: [2×1 sym]
    theta: [2×1 sym]

vpa(sol.phi)

ans = 

vpa(sol.theta)

ans = 

vpa(sol.psi)

ans = 

eqn_4 = theta == -30 * pi / 180 ;

Note that it gets nastier looking if I use 'returnconditions' on the call. But do either of the solutions yield theta = pi/6? It is close, but not going to happen. Just for kicks though, lets try it.

sol = solve(eqn_1,eqn_2,eqn_3,'returnconditions',true); % nasty looking, so leave it hidden

Next, do any of those solutions ever yield theta = pi/6?

solve(sol.theta == pi/6)

ans = struct with fields:
     m: -1/3
    z1: 3^(1/2) + 2

So we learn that m must be -1/3. But I thought m had to be an integer.

sol.conditions(1)

ans = 

So no solutions exist.

[Yagmur Savkay Öztok]

Thanks for your huge effort, this is so helpful!

Wish you a nice day, John.

Answer 2

[Alan Stevens]

You have theta = 30degrees, so sin(theta) is 1/2.

Therefore, in equation 2, you have cos(psi)/2 = 0.76, so cos(psi) > 1. If you are dealing with real numbers this is invalid!

[Yagmur Savkay Öztok]

Yeah, but the main problem in here is that this code does not show any valid results even if there are real solutions for theta, phi and psi. I will be using a set of data and y matrix will change accordingly.

I need this code to solve for these angle values but... I'm stuck. I couldn't find any similar examples.

Also,thank you for the response ! Have a nice day.

[Sam Chak]

If no restriction on , then there are complex-valued solutions on Wolfram.

[Yagmur Savkay Öztok]

Thanks!

Have a nice day.