How can simultaneous sinusoidal equations be solved?

Question

I have come across a set of simultaneous equations which I can't figure out how to solve. I have 3 equations and only two unknowns, but they are angular quantities and feature in the equations as sinusoidal functions of the angular quantities.

The system of equations is:

$$ \begin{Bmatrix}\cos\psi\sin\theta+\cos\theta\sin\phi\sin\psi \\ \sin\psi\sin\theta-\cos\psi\cos\theta\sin\phi \\ \cos\phi\cos\theta \end{Bmatrix} = \begin{Bmatrix} x \\ y \\ z \end{Bmatrix} $$

Where: x, y, z and psi are known and phi and theta are unknown.

Is it possible to rearrange these equations to solve for theta and phi using the other terms?

Answer 1

The system is not always solvable. The two first equations factor as $$\tag{1} \begin{pmatrix} \cos\psi & \sin\psi \\ \sin\psi & -\cos\psi \end{pmatrix} \begin{pmatrix} \sin\theta \\ \cos\theta \sin\phi \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} $$ where the first factor is just a reflection in a line through the origin, so $$ \sin^2\theta + \cos^2\theta\sin^2\phi = x^2 + y^2 $$ Squaring the third equation gives $$ \cos^2\theta \cos^2\phi = z^2 $$ and adding these equations give $$ x^2+y^2+z^2 = \sin^2\theta + \cos^2\theta(\sin^2\phi + \cos^2\phi) = 1 $$ so your system only has a solution when $(x,y,z)$ is a point on the unit sphere.

If your constants satisfy this, a natural approach would be to divide out the reflection matrix in $(1)$, producing $$ \begin{cases} \sin \theta = x' \\ \cos\theta\sin\phi = y' \\ \cos\theta \cos\phi = z \end{cases} $$ for some $x'$ and $y'$. From here you get $\theta = \arcsin x'$ and and $\phi = \arctan \frac{y'}{z}$, both up to a choice of quadrants that have to be made in a matching way, possibly just by trial and error.

Answer 2

Note that $$ \begin{array}{l} \left( {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} {\cos \psi \sin \theta + \cos \theta \sin \phi \sin \psi } \\ {\sin \psi \sin \theta - \cos \psi \cos \theta \sin \phi } \\ {\cos \phi \cos \theta } \\ \end{array}} \right) = \\ = \left( {\begin{array}{*{20}c} {\sin \theta \cos \psi + \cos \theta \sin \phi \sin \psi } \\ {\sin \theta \sin \psi - \cos \theta \sin \phi \cos \psi } \\ {\cos \theta \cos \phi } \\ \end{array}} \right) = \\ = \sin \theta \left( {\begin{array}{*{20}c} {\cos \psi } \\ {\sin \psi } \\ 0 \\ \end{array}} \right) + \cos \theta \left( {\begin{array}{*{20}c} {\sin \phi \sin \psi } \\ { - \sin \phi \cos \psi } \\ {\cos \phi } \\ \end{array}} \right) = \\ = \sin \theta \left( {\begin{array}{*{20}c} {\cos \psi } \\ {\sin \psi } \\ 0 \\ \end{array}} \right) + \cos \theta \left( {\begin{array}{*{20}c} {\sin \phi \cos \left( {\psi + \pi /2} \right)} \\ {\sin \phi \sin \left( {\psi + \pi /2} \right)} \\ {\cos \phi } \\ \end{array}} \right) = \\ = \sin \theta \;{\bf u} + \cos \theta \;{\bf v} \\ \end{array} $$

where $\bf u$ and $\bf v$ are given in spherical coordinates, and can be seen to be unitary vectors and orthogonal to each other.
Their combination with $sin \theta$ and $cos \theta$ is therefore a circle in their plane.
For the system to have solutions, the vector $\bf p=(x,y,z)$ shall be on the unitary sphere. That given, there will be multiple solutions for the three angles, corresponding to the infinite main circles passing through $\bf p$.