Solving trigonometric non-linear equations in MATLAB

Hi there, I'm trying to solve some non-linear simultaneous equations with trigonometric functions. They are of the form:

    297.5*cos(x) + 489*sin(y) - 740.78 = 0;
    297.5*sin(x) - 489*cos(y) + 197 = 0;

I have tried to follow this format and so currently have this:

 %Mapping b(1) = x, b(2) = y
f = @(b) [297.5*cos(b(1)) + 489.5*sin(b(2)) -740.78; 297.5*sin(b(1)) - 489*cos(b(2)) +197];
B0 = rand(2,1)*2*pi;
[B,fv,xf,ops] = fsolve(f, B0);
ps = ['theta'; 'beta'];
fprintf(1, '\n\tParameters:\n')
for k1 = 1:length(B)
fprintf(1, '\t\t%s = % .4f\n', ps(k1,:), B(k1))
end

However, I am not getting any results. MATLAB says that the "last step was ineffective". Does this mean that my system is unsolveable or have I made a mistake in my code?

Thank you in advance!

[Walter Roberson]

There are (at least) two x solutions per cycle of 2 * Pi, and (at least) two y solutions per cycle of 2 * Pi. If you have the symbolic toolbox then you should be able to get exact solutions (to within the accuracy that "740.78 represents)

[AVoyage]

Thank you Walter!

I think I may have set MATLAB an unsolvable problem here. Sorry! I'll continue to fiddle.

[John D'Errico]

Walter told you exactly how to solve it, and to do so trivially.

[AVoyage]

Yes, sorry, I realise that. I meant fiddle with my underlying model that led to these equations because I have other systems of 4 and 5 non-linear simultaneous equations to solve later - just wanted to make sure that I got everything to work for the basic case first. To that end though I did use Walter's solution as a check on my answer, so thanks again Walter.

Answer 1

[John D'Errico]

This is not impossible to solve. The symbolic toolbox does it trivially. You can convert the result to a numeric one as I show.

syms x y
E1 = 297.5*cos(x) + 489*sin(y) - 740.78 == 0;
   E2 = 297.5*sin(x) - 489*cos(y) + 197 == 0;

[x,y] = solve(E1,E2,x,y)
x =
  2*atan(6674880334960949^(1/2)/548570849 - 73259375/548570849)
 -2*atan(6674880334960949^(1/2)/548570849 + 73259375/548570849)
y =
 -2*atan(6674880334960949^(1/2)/581777974 - 452801775/581777974)
  2*atan(6674880334960949^(1/2)/581777974 + 452801775/581777974)

vpa(x)
ans =
 0.030770500758327955535681743283777
 -0.55061054417726048251968900912215

vpa(y)
ans =
 1.1356089436618227709005164864045
 1.4861436665090379405781196310366

Or you can solve it numerically using a tool like fsolve. First, why did you have a problem?

First, I'll write f as you did:

f = @(b) [297.5*cos(b(1)) + 489.5*sin(b(2)) -740.78; 297.5*sin(b(1)) - 489*cos(b(2)) +197];

f([1 2])
ans =
     605.84      -740.78
     453.83          197

See that when I Passed f a vector of length 2, it created a MATRIX result.

What happened when you wrote the sub-expression:

297.5*cos(b(1)) + 489.5*sin(b(2)) -740.78

MATLAB parsed that as a vector of length 2. See the differrnce between the way Ill write it:

f = @(b) [297.5*cos(b(1)) + 489.5*sin(b(2)) - 740.78; 297.5*sin(b(1)) - 489*cos(b(2)) + 197];

f([1 2])
ans =
    -134.94
     650.83

The only difference was I put a space between the operator - and the number 740.78. Then I did the same for the constant term in the second part.

Using the second form for f, as I defined it, fsolve now works trivially too.

b = fsolve(f,[1 2])

Equation solved.

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

<stopping criteria details>

b =
     0.033827       1.1336

It finds only one solution of course, based on the starting values provided.

[John D'Errico]

Thinking about this question, I suppose the answer is somewhat non-obvious. It hinges on the difference between the fragments

[2 -1]
ans =
     2    -1

and

[2 - 1]
ans =
     1

I put them both in square brackets, but they are different animals. In the first case, MATLAB parses -1 as the application of unary minus to 1, then sees these two numbers are separated by a space, the equivalent to a comma. So it creates a vector of length 2.

In the second case, MATLAB sees the - operator between two numbers, so it applies the binary - operator between them. The result is now a scalar.

As I said, a subtle appearing difference, but an important one.

Answer 2

[Sumera Yamin]

hi john, can you comment on why my code (similar as the original question is not giving me any solution? I will really appreciate any help.

Ld= 0.8194  %constant
calib = @(K,L) [cos(K*L).^2 + Ld*K*sin(K*L).*cos(K*L) - 2.2;  cos(K*L).^2 - 0.299;  Ld*cos(K*L).^2 + (sin(K*L).*cos(K*L))/K - 0.262];
fun = @(b) calib(b(1),b(2));
initial_val=[2.73, 0.6];  % initial value of K and L respectively
[x,fval,exitflag,output] = fsolve(fun,initial_val)

[Walter Roberson]

You have three equations in two unknowns. You can solve the first two equations together; when you then substitute the values into the third equation, you will get 0.07334537675 instead of 0, indicating that the three equations are inconsistent with each other.

[Sumera Yamin]

It means if i only used two equations instead of 3, then the solution will be possible?

[Torsten]

A solution for the two equations would be possible, but not for all three.

Answer 3

[sanjay kolape]

% write code

l1 = 10; % length of first arm
l2 = 10; % length of second arm

X_coordinate = input("Enter value of X coordinate : "); % theta1 values
Y_coordinate = input("Enter value of Y coordinate : ");  % theta2 values

x0 = 0; y0 = 0; %coordinate of fixed link

syms theta1 theta2

eqn1 = l1*cos(theta1) + l2*cos(theta1-theta2) ==  X_coordinate;
eqn2 =  l1*sin(theta1) + l2*sin(theta1-theta2) ==  Y_coordinate;

[theta1, theta2]= solve(eqn1,eqn2,theta1,theta2);

theta1 = theta1(2);
theta2 = theta2(2);

x1 = l1*cos(theta1);
y2=l1*sin(theta1);

X1 = [x0 x1];                      Y1= [y0 y1];
X2 = [x1 X_coordinate];   Y2 = [y1 Y_coordinate];

comet(X1,Y1);
hold on
comet(X2,Y2);

[Walter Roberson]

It is not clear how this code answers the original Question ?