what's the relation between A , B and M,G for this Nonlinear system of equation ?

A*sin(3*Phi)-B*sin(Phi) = G ;

A*cos(3*Phi)-B*cos(Phi) = M ;

Answer 1

[Torsten]

A^2 + B^2 - 2*A*B*cos(2*Phi) = G^2 + M^2

[YOUSSEF El MOUSSATI]

Ok, thanks for you , but I want the relation between A,B and C,D without sin (Phi) and code(Phi)

[Torsten]

As you can see, there is a relation between A,B,G and M in one equation that contains the variable "phi". Thus there cannot be a relation between them without "phi".

[YOUSSEF El MOUSSATI]

Yes , but we can use the relation between sin(phi)^2+cos^2(phi)=1

[Torsten]

I already used it two times :

A*sin(3*Phi) - B*sin(Phi) = G -> (square both sides)
A^2*sin^2(3*Phi) + B^2*sin^2(Phi) - 2*A*B*sin(3*Phi)*sin(Phi) = G^2  (1)
A*cos(3*Phi) - B*cos(Phi) = M -> (square both sides)
A^2*cos^2(3*Phi) + B^2*cos^2(Phi) - 2*A*B*cos(3*Phi)*cos(Phi) = M^2  (2)

Adding (1) and (2) gives

A^2*(sin^2(3+Phi)+cos^2(3*Phi)) + B^2*(sin^2(Phi)+cos^2(Phi)) - 2*A*B*(sin(3*Phi)*sin(Phi)+cos(3*Phi)*cos(Phi)) = G^2 + M^2 ->
A^2 + B^2 - 2*A*B*(sin(3*Phi)*sin(Phi)+cos(3*Phi)*cos(Phi)) = G^2 + M^2

Using

sin(x)*sin(y)+cos(x)*cos(y) = cos(x-y)

gives

A^2 + B^2 - 2*A*B*cos(2*Phi) = G^2 + M^2

[YOUSSEF El MOUSSATI]

Yes, but you find the relationship as a function of cos(2*Phi)

[Torsten]

As I already wrote:

If a relation between A, B, G and M contains Phi

A^2 + B^2 - 2*A*B*cos(2*Phi) = G^2 + M^2

, I cannot think of a way to eliminate Phi.

[YOUSSEF El MOUSSATI]

Thank for your , I think that the key to solve this system is the relationship : Sin^2(x)+cos^2(x)=1

[Torsten]

If each of the equations could be solved uniquely for Phi, you could get what you want.

Assume that the first equation would uniquely yield Phi = f(A,B,G) and the second equation would uniquely yield Phi = g(A,B,M), then f(A,B,G) - g(A,B,M) = 0 would be your relation. But unfortunately, the equations cannot be solved uniquely for Phi.

syms A B G M Phi
eqn = A*sin(3*Phi)-B*sin(Phi) == G ;
sol1 = solve(eqn,Phi,'ReturnConditions',1,'MaxDegree',3)

sol1 = struct with fields:
           Phi: [6×1 sym]
    parameters: k
    conditions: [6×1 sym]

sol1.Phi

ans = 

sol1.conditions

ans = 

[YOUSSEF El MOUSSATI]

Ok , thanks for your help , it's very important to me

Answer 2

[John D'Errico]

This is not even remotely a question about MATLAB. As such, it should arguably not even be on Answers. But I have a minute to respond, so I will choose to do so.

Trivial! What is the relation? Admittedly, the relation itself is a slightly complex thing, composed of two equations. The relations are:

A*sin(3*Phi)-B*sin(Phi) = G

A*cos(3*Phi)-B*cos(Phi) = M

which is exactly what you wrote.

It is not a nonlinear system of equations though. Not at all! Phi there is simply a parameter, not one of the parameters involved. That makes your problem fully a LINEAR system of equations. As such, if you want to view it in that form, then we could write:

M = [sin(3*Phi), -sin(Phi); ..

cos(3*Phi), -cos(Phi)]

So M is a matrix function of the parameter Phi. Biven the matrix M, then we could write:

M*[A;B] = [G;M]

There is no simpler relation between those variables. And it is NOT at all nonlinear. Purely linear.

[YOUSSEF El MOUSSATI]

Ok ,thanks for your answer , yes this system is linear but the parameters A,B and C,D are Nonlinear A=f(x^3,x^2,x); B=g(x^3,x^2,x); C=h(x^3,x^2,x); D=l(x^3,x^2,x);