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I am computationally trying to solve the system of linear equations given below.

$ \frac{1}{1+tan(\theta/2)*tan(\phi/2)*tan(\psi/2)} \begin{bmatrix} tan(\psi/2)-tan(\theta/2)*tan(\phi/2) \\ tan(\phi/2)+tan(\theta/2)*tan(\psi/2) \\ tan(\theta/2)-tan(\phi/2)*tan(\psi/2) \end{bmatrix} = \begin{bmatrix} r_1 \\ r_2 \\ r_3 \end{bmatrix} $

I use the least square approximations to find the solution to this equation on MATLAB. However, no effective solution was found using this method. Considering that the system values for r1, r2, r3 and $\psi$ = 0 are known whereas $\theta$ and $\phi$ are the two unknown values. Is it possible to obtain exact solution for these two unknowns of the system for $r = [0,-0.101233861621737,0.365119069777688]$.

The values I obtain on solving this system is $\theta = 0.6950 $ and $\phi = -0.1785$, Even-though these are good approximate values, I was wondering if it's possible to obtain the exact values using a different approach.

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  • $\begingroup$ I'm confused. If $\psi = 0$ and $r_1 = 0$, the first equation tell us $\tan\frac{\theta}{2}\tan\frac{\phi}{2} = 0$. This is incompatible with your value of $\theta$, $\phi$ (none of them are close to multiples of $2\pi$ which one need to make tan(half-angle) to vanish) $\endgroup$ Sep 21 at 12:07
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    $\begingroup$ Could you tell the origin of the problem, in particular the meaning of angles (Euler angles ? but for which configuration ?). $\endgroup$
    – Jean Marie
    Sep 21 at 15:20
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    $\begingroup$ Replacing the tangents by unknowns, what you have is a system of three cubic equations. By eliminating two unknowns, you can end up with a high degree univariate polynomial. Good luck ! $\endgroup$
    – user958916
    Sep 21 at 15:28
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    $\begingroup$ These formulas look explainable in terms of quaternions. See for example this paper (although not directly connected to Fick's angles) $\endgroup$
    – Jean Marie
    Sep 21 at 15:58
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    $\begingroup$ @JeanMarie Thank you for the beautiful paper suggestion. I actually wanted to work on a reduced version of the quaternion for application in Listing's law. That is why I specifically chose the rotation vectors instead. I naturally set the first component of the rotation vector as zero for the specific application. $\endgroup$
    – Jules
    Sep 22 at 8:26
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Maybe I'm missing something, but if you substitute $\psi = 0$ in the system, and denote $x = \tan(\theta/2), y = \tan(\phi/2)$, you get the system $$ \begin{cases} -xy = r_1 \\ y = r_2\\ x= r_3 \end{cases}. $$

So, either $r_1 = -r_2 r_3$ and the solution is $(x,y)=(r_3, r_2)$ or you have no solution at all.

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Assume $r_1 = 0, r_2, r_3 \ne 0$. Let $t_1 \tan\frac{\psi}{2}$, $t_2 = \tan\frac{\phi}{2}$ and $t_3 = \tan\frac{\theta}{2}$.

With help of a CAS, I get two set of solutions: $$t_2 = \frac{s_2 - \epsilon \sqrt{s_2^2 - 4r_2^2}}{2r_2},\quad t_3 = \frac{s_3 + \epsilon \sqrt{s_3^2 + 4r_3^2}}{2r_3}\tag{*1}$$ where $\epsilon = \pm 1$ and $s_2 = r_2^2+r_3^2+1$, $s_3 = r_2^2+r_3^2-1$.

For $(r_1,r_2,r_3) = (0,−0.101233861621737,0.365119069777688)$, the approximation solution you get corresponds to the $\epsilon = +1$ solution. Numerically, the corresponding angles evaluated to

$$(\psi, \phi, \theta ) \sim (-0.0657294323,-0.1779886273,0.7060269791)$$

The formula in $(*1)$ is relatively simple, it should be possible to derive that by hand. However, I can't figure that out yet. You can plug those expression to a CAS and verify that yourself.

Update

I sort of figure out how to derive the two solutions in $(*1)$.

When $r_1 = 0$, first equation give us $t_1 = t_2t_3$. Substitute this into second and third equation, we have

$$\frac{t_2(1+t_3^2)}{1 + t_2^2t_3^2} = r_2\quad\text{ and }\quad \frac{t_3(1-t_2^2)}{1 + t_2^2t_3^2} = r_3\tag{*2}$$

By brute force, we have $$s_2 = r_2^2 + r_3^2 + 1 = \frac{(1+t_2^2)(1+t_3^2)}{1+t_2^2t_3^2} \implies r_2(1+t_2^2) - s_2t_2 = 0 $$ Similarly, $$s_3 = r_2^2 + r_3^2 - 1 = -\frac{(1-t_2^2)(1-t_3^2)}{ 1 + t_2^2t_3^2} \implies r_3(t_3^2-1) - s_3t_3 = 0$$

Solving these two quadratic equations give us $4$ possible combinations of $(t_2,t_3)$. Throwing them into an CAS, one find only two of them, those appear in $(*1)$, satisfy $(*2)$.

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  • $\begingroup$ This is sort of what I was looking for. Thank you! I would write a code for trying and generalising it for any value of r instead of the given combination. $\endgroup$
    – Jules
    Sep 21 at 15:38

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