How do I solve Two Simultaneous Equations?

[therealkilkenny]

Hi All,

I have the following equation (X and Y are variables and A and B are constants)

Y=B*Cos(X-A)

and i have two values of Y (0.683,1.199) and two values of X (-10.232,-10.505 in Radians), thus two equations:

0.683=B*Cos(-10.232-A)
1.199=B*Cos(-10.505-A)

How do i get A and B?

Thanks for your help,

[johnsold]

I think a direct algebraic solution is not possible. Thus an iterative numeric solution seems like the best approach. Select an initial guess for one of the parameters, perhaps Bo. Using the Inverse Cosine function, obtain two values for Ao:

Ao1 = x1 - arccos(y1/Bo)

A02 = x2 - arccos(y2/Bo)

Use those vaues to obtain the improved estimate of B, B1. (I am not sure what to recommend here. One of the numerical calculation gurus may be able to help.) Repeat until An and Bn are close enough.

Lynn

[JeanPierre]

Starting with

Y0 = B*cos(X0 - A) and
Y1 = B*cos(X1 - A)

you can get a closed form solution:

B = Y0/cos(X0 - A) = Y1/cos(X1 - A)  (1)

then

Y1*cos(X0 - A) = Y0*cos(X1 - A)

expanding the cosine difference, (please double check...) and dividing by cos(A) you'll get

tan(A) = (Y1*cos(X0) - Y0*cos(X1))/(Y1*sin(X0) - Y0*cos(X1))

Compute A and replace in (1)

LEK

[chilly_charly]

There is a vi in the math sub-palette (with the zeroes functions), named "Nonlinear System Solver" that seems to be the right tool for this sort of job...

[johnsold]

Jean-Pierre,

Thank you. I suspected that a closed form solution might be possible, but I did not spend very much time seeking it out.

I get a sign difference when I derive the result. Cos(a-b) = cos(a)*cos(b) + sin (a)*sin(b).

Then tan(A) = (Y1*cos(X0) - Y0*cos(X1))/(-Y1*sin(X0) + Y0*cos(X1))

Lynn

[JeanPierre]

---

johnsold a écrit:
Jean-Pierre,

Thank you. I suspected that a closed form solution might be possible, but I did not spend very much time seeking it out.

I get a sign difference when I derive the result. Cos(a-b) = cos(a)*cos(b) + sin (a)*sin(b).

Then tan(A) = (Y1*cos(X0) - Y0*cos(X1))/(-Y1*sin(X0) + Y0*cos(X1))

Lynn

---

Yeah exactly the formula in my math table that I looked twice into before deriving....
And an other error it should be:

tan(A) = (Y1*cos(X0) - Y0*cos(X1))/(-Y1*sin(X0) + Y0*sin(X1))

CC: numerical solutions are for the lazy...

[chilly_charly]

JPD,

I am lazy...

[tbob]

OK, I'll join in.  here is what I get:

Start with Y1*cos(X0-A) = Y0*cos(X1-A) and the well known derivation cos(a-b) = cos(a)*cos(b) + sin(a)*sin(b)

then upon expanding:  Y1*[cos(X0)*cos(A) + sin(X0)*sin(A)] = Y0*[cos(X1)*cos(A) + sin(X1)*sin(A)]

then getting rid of [ ]:  Y1*cos(X0)*cos(A) + Y1*sin(X0)*sin(A) = Y0*cos(X1)*cos(A) + Y0*sin(X1)*sin(A)

then divide by cos(A):  Y1*cos(X0) + Y1*sin(X0)*tan(A) = Y0*cos(X1) + Y0*sin(X1)*tan(A)

rearrange terms:  Y1*sin(X0)*tan(A) - Y0*sin(X1)*tan(A) = Y0*cos(X1) - Y1*cos(X0)

tan(A) is common:  tan(A)*[Y1*sin(X0) - Y0*sin(X1)] = Y0*cos(X1) - Y1*cos(X0)

solve for tan(A):  tan(A) = [Y0*cos(X1) - Y1*cos(X0)] / [Y1*sin(X0) - Y0*sin(X1)]

So, Mr. Lynn is wrong, and Monseiur Jean Pierre is still wrong, unless I made a mistake and then I'm wrong!!!

[JeanPierre]

That is the same answer and we are both right (unless proven otherwise) the apparent difference is that the signs of both numerator and divider are reversed (mutiplied by( -1)/(-1))

[johnsold]

Jean-Pierre and tbob,

I tried to copy and paste and modify and did not check carefully enough. Your final answers are both correct and the same within (-1)/(-1), which is close enough except for those doing numerical simulations where 1 +/- 2E-15 might be better.

I guess tbob does not need to address me as Mr. Lynn any longer.

Lynn