I would like to have more generalized version of the answer of Lord Shark the Unknown for this question .. The concept of algebraic integer is a novelty for me and I'm still uncomfortable with it not knowing what are the exact rules for determining this feature for a given some specific expression (if such rules exist please let me know).
Suppose that we have composition of two 3-d rotations by $\theta$ about $z$ and $x$ axis what gives rotation matrix $R(v,\alpha)$.
$R(v,\alpha)=\begin{bmatrix} \ \ \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{bmatrix}= \begin{bmatrix} \cos\theta & -\cos\theta \sin\theta & \sin\theta \sin\theta\\ \sin\theta& \cos\theta \cos\theta & -\cos\theta \sin\theta\\ 0 & \sin\theta & \cos\theta\end{bmatrix} $
(additionally it was assumed that for some $k$ we have cyclicity $\ \ R^k(z,\theta)=R^k(x,\theta)=I$ what means that $\theta$ is a rational times $\pi$)
Matrix $R(v,\alpha)$ is the matrix of rotation about some axis $v$ by $\alpha$ where $\alpha$ and $\theta$ are linked by the equation
$\text{tr}(R)=2\cos\theta+\cos^2\theta=1+2\cos\alpha$.
Hence $\cos\alpha=\dfrac{1}{2}(2\cos\theta+\cos^2\theta-1)= \dfrac{1}{2}(2\cos\theta-\sin^2\theta)$.
Now according to Lord Shark the Unknown's answer in order to determine the cyclicity of $R(v,\alpha)$ we need to check when $ a=2\cos\theta-\sin^2\theta$ is an algebraic integer.
Question ( consider the case when $\theta$ is rational times $\pi$ )
- is it true that number $ a=2\cos\theta-\sin^2\theta$ is an algebraic integer
iff $ \cos\theta $ is an integer (i.e. has values from the set $\{-1,0,1\})$? - how to prove that the expression $a$ can't be a zero of monic polynomial with integer coefficients beyond this situation ?