Solving trigonometric simultaneous equations

Is it possible to solve when the equations are little complex? For example lets's take an inverse kinematics problem. Where we get,

-0.2595    0.5536    0.7913   0.3004        r11   r12   r13   o1
-0.7781   -0.6052    0.1682   0.0639    =   r21   r22   r23   o2
 0.5721   -0.5721    0.5878   0.6167        r31   r32   r33   o3
      0         0         0        1          0     0     0    0

each of these symbols represents a trigonometric function such as below. Need to find θ1,θ2,θ3,θ4,θ5

[   sin(theta1)*sin(theta5) - cos(theta5)*(cos(theta4)*(cos(theta1)*sin(theta2)*sin(theta3) - cos(theta1)*cos(theta2)*cos(theta3)) + sin(theta4)*(cos(theta1)*cos(theta2)*sin(theta3) + cos(theta1)*cos(theta3)*sin(theta2))), cos(theta5)*sin(theta1) + sin(theta5)*(cos(theta4)*(cos(theta1)*sin(theta2)*sin(theta3) - cos(theta1)*cos(theta2)*cos(theta3)) + sin(theta4)*(cos(theta1)*cos(theta2)*sin(theta3) + cos(theta1)*cos(theta3)*sin(theta2))), cos(theta4)*(cos(theta1)*cos(theta2)*sin(theta3) + cos(theta1)*cos(theta3)*sin(theta2)) - sin(theta4)*(cos(theta1)*sin(theta2)*sin(theta3) - cos(theta1)*cos(theta2)*cos(theta3)), (1143*cos(theta1)*cos(theta2))/5000 + (127*cos(theta4)*(cos(theta1)*cos(theta2)*sin(theta3) + cos(theta1)*cos(theta3)*sin(theta2)))/1600 - (127*sin(theta4)*(cos(theta1)*sin(theta2)*sin(theta3) - cos(theta1)*cos(theta2)*cos(theta3)))/1600 - (1143*cos(theta1)*sin(theta2)*sin(theta3))/5000 + (1143*cos(theta1)*cos(theta2)*cos(theta3))/5000]
[ - cos(theta1)*sin(theta5) - cos(theta5)*(cos(theta4)*(sin(theta1)*sin(theta2)*sin(theta3) - cos(theta2)*cos(theta3)*sin(theta1)) + sin(theta4)*(cos(theta2)*sin(theta1)*sin(theta3) + cos(theta3)*sin(theta1)*sin(theta2))), sin(theta5)*(cos(theta4)*(sin(theta1)*sin(theta2)*sin(theta3) - cos(theta2)*cos(theta3)*sin(theta1)) + sin(theta4)*(cos(theta2)*sin(theta1)*sin(theta3) + cos(theta3)*sin(theta1)*sin(theta2))) - cos(theta1)*cos(theta5), cos(theta4)*(cos(theta2)*sin(theta1)*sin(theta3) + cos(theta3)*sin(theta1)*sin(theta2)) - sin(theta4)*(sin(theta1)*sin(theta2)*sin(theta3) - cos(theta2)*cos(theta3)*sin(theta1)), (1143*cos(theta2)*sin(theta1))/5000 + (127*cos(theta4)*(cos(theta2)*sin(theta1)*sin(theta3) + cos(theta3)*sin(theta1)*sin(theta2)))/1600 - (127*sin(theta4)*(sin(theta1)*sin(theta2)*sin(theta3) - cos(theta2)*cos(theta3)*sin(theta1)))/1600 - (1143*sin(theta1)*sin(theta2)*sin(theta3))/5000 + (1143*cos(theta2)*cos(theta3)*sin(theta1))/5000]
[                                                                             cos(theta5)*(cos(theta4)*(cos(theta2)*sin(theta3) + cos(theta3)*sin(theta2)) + sin(theta4)*(cos(theta2)*cos(theta3) - sin(theta2)*sin(theta3))),                                                                          -sin(theta5)*(cos(theta4)*(cos(theta2)*sin(theta3) + cos(theta3)*sin(theta2)) + sin(theta4)*(cos(theta2)*cos(theta3) - sin(theta2)*sin(theta3))),                                                 sin(theta4)*(cos(theta2)*sin(theta3) + cos(theta3)*sin(theta2)) - cos(theta4)*(cos(theta2)*cos(theta3) - sin(theta2)*sin(theta3)),                                                                        (1143*sin(theta2))/5000 + (1143*cos(theta2)*sin(theta3))/5000 + (1143*cos(theta3)*sin(theta2))/5000 - (127*cos(theta4)*(cos(theta2)*cos(theta3) - sin(theta2)*sin(theta3)))/1600 + (127*sin(theta4)*(cos(theta2)*sin(theta3) + cos(theta3)*sin(theta2)))/1600 + 4953/20000]
[                                                                                                                                                                                                                           0,                                                                                                                                                                                                                         0,                                                                                                                                                                                 0,                                                                                                                                                                                                                                                                                                                                                 1]

Is matlab capable of solving such equations?

Thank you.

Answer 1

[Anurag Ojha]

Hello Jupin

Yes, MATLAB is capable of solving complex equations, including inverse kinematics problems. MATLAB provides powerful numerical computation capabilities and has built-in functions for solving systems of equations. In this case, you can use the fsolve function to find the values of θ1, θ2, θ3, θ4, and θ5 that satisfy the given equations.

I have taken a simple example to demostrate. Kindly make changes as per your use case:

% Define the function to solve
function F = trigEquations(theta)
    % Unpack the angles
    theta1 = theta(1);
    theta2 = theta(2);
    
    % Equations to solve
    F(1) = sin(theta1) + cos(theta2) - 1.5;
    F(2) = cos(theta1) + sin(theta2) - 0.5;
end

% Initial guess for the angles
initial_guess = [0, 0];

% Use fsolve to solve the equations
options = optimoptions('fsolve', 'Display', 'iter');
[theta_solution, fval] = fsolve(@trigEquations, initial_guess, options);

                                             Norm of      First-order   Trust-region
 Iteration  Func-count     ||f(x)||^2           step       optimality         radius
     0          3                 0.5                             0.5              1
     1          6           0.0308187       0.707107            0.158              1
     2          9         0.000169659       0.175552           0.0122           1.77
     3         12         8.38458e-09      0.0140895         8.57e-05           1.77
     4         15         2.10737e-17    9.97403e-05         4.28e-09           1.77

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

% Display the results
fprintf('Solution for theta1: %.4f radians\n', theta_solution(1));

Solution for theta1: 0.5900 radians

fprintf('Solution for theta2: %.4f radians\n', theta_solution(2));

Solution for theta2: -0.3373 radians

Adding the documentation to get better understanding:

• https://www.mathworks.com/help/optim/ug/fsolve.html