Solving the system $x_0=\cos\theta_1\cos\theta_0$, $x_1=\sin\theta_1\cos\theta_0$, $x_2=\cos\theta_2\sin\theta_0$, $x_3=\sin\theta_2\sin\theta_0$

Question

I have the following system of equations:

$$ \begin{cases} x_0=\cos\theta_1\cos\theta_0 \\[4pt] x_1=\sin\theta_1\cos\theta_0 \\[4pt] x_2=\cos\theta_2\sin\theta_0 \\[4pt] x_3=\sin\theta_2\sin\theta_0 \end{cases} $$

The $x_i$ are known and $\sum_i x_i^2 =1$, while the $\theta_i$ are between -$\pi$ and $\pi$ but not known. How can they be found?

It is easy to get $\theta_1$ and $\theta_2$ by using $\arcsin$ and $\arccos$, but both depend on $\theta_0$, and I do not know how to continue. I have also tried using SymPy and Wolfram Alpha, but they were not successful.

Answer 1

Assuming $x_0x_2\neq 0$, dividing the second one by the first one and the fourth one by the third one we obtain

$$\tan \theta_1=\frac{x_1}{x_0} \implies \theta_1=\arctan\left(\frac{x_1}{x_0}\right)+k_1\pi$$

$$\tan \theta_2=\frac{x_3}{x_2} \implies \theta_2=\arctan\left(\frac{x_3}{x_2}\right)+k_2\pi$$

then we can plug these values in the equations to obtain the right value for $k_1\in\mathbb Z$, $k_2\in\mathbb Z$ and $\theta_0$.

Answer 2

Note $$ x_0+x_1 = \cos(\theta_1 - \theta_0) \\ x_0-x_1 = \cos(\theta_1 + \theta_0) \\ \arccos(x_0+x_1) = \theta_1 - \theta_0 \\ \arccos(x_0-x_1) = \theta_1 + \theta_0 \\ \frac{\arccos(x_0-x_1)+\arccos(x_0+x_1)}{2} = \theta_1 \\ \frac{\arccos(x_0-x_1)-\arccos(x_0+x_1)}{2} = \theta_0 $$ Now we have to adjust for the fact that $\cos$ is not 1-1,
so perhaps other values of the arccos will also be needed.

Answer 3

$x_0^2 = \cos^2 \theta_1 \cos^2 \theta_0$

$x_1^2 = \sin^2 \theta_1 \cos^2 \theta_0$

$x_0^2 + x_1^2 =(\cos^2 \theta_1 + \sin^2 \theta_1)\cos^2 \theta_0$

$x_0^2 + x_1^2=\cos^2 \theta_0$

$\cos \theta_0 = \mp \sqrt{x_0^2 + x_1^2}$

In general,

$\theta_0 = \mp\arccos (\sqrt{x_0^2 + x_1^2}) +2k\pi$

or

$\theta_0 = \mp\arccos (-\sqrt{x_0^2 + x_1^2})+2k\pi=\pi-(\mp\arccos(\sqrt{x_0^2 + x_1^2}))+2k\pi$

$k \in Z$

Answer 4

First choose some $θ_0$ such that $$\cos^2θ_0=x_0^2+x_1^2\quad\text{and}\quad\sin^2θ_0=x_2^2+x_3^2.$$ It is then easy to find the corresponding solutions $(θ_1,θ_2),$ such that your four equations are satisfied.