Simultaneous Trigonometric Equations - solving for angles

[m101]

Summary:: I have a series of three equations that transform three angles of a system (J1, J2, J3), into three spatial x, y, z coordinates. I want to invert them to find the angles from the coordinates.

Reference: https://www.physicsforums.com/forums/general-math.73/post-thread

I have a series of three equations that transform three angles of a system (J1, J2, J3), into three spatial x, y, z coordinates as follows:

x = A + B cos(J3) + C sin(J2) cos (J1)

y = A + B cos(J3) + C sin (J2) sin (J1)

z = C (cos(J2) - 1) - B sin (J3)

Here A, B and C are numerical constants that I know the value of. What I want is to be able to invert the problem, namely find expressions to calculate the values of J1 to J3 given a set of coordinates (with the angles within the range -180 to 180, where a solution exists). I am struggling a bit with this. Is this possible and if so how can I go about it. Thank you in advance for any help or hints.

 

[fresh_42]

What did you try so far? Do you have the trig function values as a quotient of lengths?

 

[m101]

So far I have tried to rearrange the equations and use trig identities to simplify them.

 

[fresh_42]

And ...? Let us see what you have done. You could e.g. subtract the first two equations and get
$$
\dfrac{x-y}{c}=\sin(J2)\cdot (\cos(J1)-\sin(J1))
$$

At first sight, I would assume that it is not uniquely solvable. My attempt would be to use the exponential complex function to replace the trig functions, but that's just an idea and I don't know whether you know the formulas.

 

[vela]

In vector form, you have
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix} A \\ A \\ -C \end{pmatrix} +
B\begin{pmatrix} \cos j_3 \\ \cos j_3 \\ -\sin j_3 \end{pmatrix} +
C\begin{pmatrix} \sin j_2 \cos j_1 \\ \sin j_2 \sin j_1 \\ \cos j_2 \end{pmatrix}$$ From the last term, $j_1$ and $j_2$ seem to be the azimuthal and polar angles in spherical coordinates. Looking at the problem in terms of vectors might help you find a solution.