If $r$ represents this vector in the rotated device body frame xyz, then $r=\left(\begin{array}{c}0\\ 0\\ -1\end{array}\right)$

The transformation of $r$ due to the rotation about the z axis can be represented by the following rotation matrix.

$A=\left(\begin{array}{ccc}cos\left(\alpha \right)& -sin\left(\alpha \right)& 0\\ sin\left(\alpha \right)& cos\left(\alpha \right)& 0\\ 0& 0& 1\end{array}\right)$

The transformation of $r$ due to the rotation about the x axis can be represented by the following rotation matrix.

$B=\left(\begin{array}{ccc}1& 0& 0\\ 0& cos\left(\beta \right)& -sin\left(\beta \right)\\ 0& sin\left(\beta \right)& cos\left(\beta \right)\end{array}\right)$

The transformation of $r$ due to the rotation about the y axis can be represented by the following rotation matrix.

$C=\left(\begin{array}{ccc}cos\left(\gamma \right)& 0& sin\left(\gamma \right)\\ 0& 1& 0\\ -sin\left(\gamma \right)& 0& cos\left(\gamma \right)\end{array}\right)$

If $R$ resresents the vector $r$ in the earth frame XYZ, then since the inital body frame is aligned with the earth, $R$ is as follows.

$R=ABCr$

$R=\left(\begin{array}{ccc}cos\left(\alpha \right)& -sin\left(\alpha \right)& 0\\ sin\left(\alpha \right)& cos\left(\alpha \right)& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& cos\left(\beta \right)& -sin\left(\beta \right)\\ 0& sin\left(\beta \right)& cos\left(\beta \right)\end{array}\right)\left(\begin{array}{ccc}cos\left(\gamma \right)& 0& sin\left(\gamma \right)\\ 0& 1& 0\\ -sin\left(\gamma \right)& 0& cos\left(\gamma \right)\end{array}\right)\left(\begin{array}{c}0\\ 0\\ -1\end{array}\right)$

$R=\left(\begin{array}{ccc}cos\left(\alpha \right)& -sin\left(\alpha \right)& 0\\ sin\left(\alpha \right)& cos\left(\alpha \right)& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& cos\left(\beta \right)& -sin\left(\beta \right)\\ 0& sin\left(\beta \right)& cos\left(\beta \right)\end{array}\right)\left(\begin{array}{c}-sin\left(\gamma \right)\\ 0\\ -cos\left(\gamma \right)\end{array}\right)$

$R=\left(\begin{array}{ccc}cos\left(\alpha \right)& -sin\left(\alpha \right)& 0\\ sin\left(\alpha \right)& cos\left(\alpha \right)& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}-sin\left(\gamma \right)\\ sin\left(\beta \right)cos\left(\gamma \right)\\ -cos\left(\beta \right)cos\left(\gamma \right)\end{array}\right)$

$R=\left(\begin{array}{c}-cos\left(\alpha \right)sin\left(\gamma \right)-sin\left(\alpha \right)sin\left(\beta \right)cos\left(\gamma \right)\\ -sin\left(\alpha \right)sin\left(\gamma \right)+cos\left(\alpha \right)sin\left(\beta \right)cos\left(\gamma \right)\\ -cos\left(\beta \right)cos\left(\gamma \right)\end{array}\right)$

The compass heading $\theta$ is given by

$\theta ={tan}^{-1}\left(\frac{R_x}{R_y}\right)={tan}^{-1}\left(\frac{-cos\left(\alpha \right)sin\left(\gamma \right)-sin\left(\alpha \right)sin\left(\beta \right)cos\left(\gamma \right)}{-sin\left(\alpha \right)sin\left(\gamma \right)+cos\left(\alpha \right)sin\left(\beta \right)cos\left(\gamma \right)}\right)$

provided that $\beta$ and $\gamma$ are not both zero.

As a consistency check, if we set $\gamma =0$, then

$\theta ={tan}^{-1}\left(\frac{-sin\left(\alpha \right)sin\left(\beta \right)}{cos\left(\alpha \right)sin\left(\beta \right)}\right)=-\alpha$

as expected.