I do not understand why I get imaginary numbers as answers to this simultaneous equation I am trying to solve

syms x y
eqn1 = 87*sind(x)+ 65*sind(y) == (-92*sind(200))-(73*sind(0))
eqn2 = 87*cosd(x) + 65*cosd(y) == (92*cosd(200)) - (73*cosd(0))
sol = solve(eqn1,eqn2,x,y)
xSol = sol.x
ySol = sol.y
vpa(xSol) 
vpa(ySol)

Answer 1

[William Rose]

@mintyfrezh,

When you run the code you provided, eqns 1 and 2 are displayed:

87 sin x + 65 sin y = 31.47 (1)

87 cos x + 65 cos y = -159.45 (2)

Eqn 2 tells you right away that there is no real solution, because cos(x) and cos(y) can never be more negative than -1. So the left hand side of eqn 2 can never be more negative than -152, and so can never equal -159.45, if x and y are real. That's why you get a complex solution.

Idea for alternate solution (summary):

1. Apply simple algebra to equations 1 and 2 above.
2. Use the identity cos^2 x = 1-sin^2 x to get an equation that involves sin y only.
3. Do some more algebra to get a quadratic equation for the variable z=sin y.
4. Solve for z using the quadratic formula. The solution is z1=..., z2=.... (I haven't done it.)
5. Solve for y: y = arcsin(z), since z is defined as z=sin y. I predict that the solutions z1 and z2 from step 4 will not be in the range[-1,+1]. If my prediction is right, then there is no real angle y such that sin y=z1 or sin y=z2. So there's no real solution.

[William Rose]

I did the alternate solution which I mentioned above, and it worked. You get a quadratic equation for z=sin y, where the constants A,B,C in the quadratic depend on the constants in the original problem statement. The radicand, B^2-4AC, is negative. Therefore z=sin y is complex, so y is also complex. There is no real solution.

Alternate solution:

a sin x + b sin y = c (1)

a cos x + b cos y = d (2)

where a=87, b=65, c=31.47, and d=-159.45

Divide eqn 1 and eqn 2 by a to get

sin x = e - g sin y (3)

cos x = f - g cos y (4)

where e=c/a=.3617, f=d/a=-1.8328, and g=b/a=.7471. Square (3) and square (4) to get:

(5)

(6)

Apply the identity , to LHS of eqn (6), and use RHS of eqn (5) for :

(7)

Note that we have eliminated sin x from the equation. The only unknown in eqn (7) is (and , but ). So we sunstitute z for sin y and we substitute for cos y, in eqn (7), to get

(8)

Do some algebra on eqn (8) to get

(11)

where

(12A)

(12B)

(12C)

Solve for z:

(13)

When you plug in the known values for e,f,g (see above) to eqns (12A,12B, 12C), you find that

A=7.792, B=-3.295, C=1.791, which means that the term under the square root in eqn (13) is negative:

That means z=sin y is complex. Therefore y=arcsin(z) is also complex. There is no real solution.