How do we prove that $\theta$ is an irrational multiple of $2\pi$ given $\cos(\theta/2)\equiv \cos^2(\pi/8)$?
With $\operatorname{SU}(2)$ rotations,
\begin{align}R_z(\pi/4)R_x(\pi/4)&=[\cos(\pi/8)I-i\sin(\pi/8)Z][\cos(\pi/8)I-i\sin(\pi/8)X]\\ &=\cos^2(\pi/8)I-i[\cos(\pi/8)X+\sin(\pi/8)Y+\cos(\pi/8)Z]\sin(\pi/8)\\ &=\cos(\theta/2)I-i(\hat{n}.\vec{\sigma})\sin(\theta/2)=R_\hat{n}(\theta)\end{align}
where $\vec{n}=(\cos(\pi/8),\sin(\pi/8),\cos(\pi/8))$ and $\hat{n}=\frac{\vec{n}}{||\vec{n}||}$, and $\vec{\sigma}=(X,Y,Z)$ where $X,Y,Z$ are Pauli matrices. Thus $\cos(\theta/2)\equiv\cos^2(\pi/8)$ and $\sin(\theta/2)\equiv\sin(\pi/8)\sqrt{1+\cos^2(\pi/8)}$.
Original Context in my Reference
Ref. to Page 196, 214 of QC and QI by Nelsen and Chuang
Any hint on the possible ways to approach this could be appreciated.
Let $p,q$ be positive coprime integers and suppose by contradiction that $\theta=2\pi p/q$. Since $\cos^2(\pi/8)=(2+\sqrt2)/4$, it has algebraic degree $2$. By considering Galois conjugates of primitive roots of unity, we can show that over the rationals, the algebraic degree of $\cos(p\pi/q)$ is $\phi(q)$ when $q$ is even, and $\phi(q)/2$ when $q$ is odd.
Therefore, we have $\phi(q)=2$ when $q$ is even, giving $q=4,6$ (when $q$ is odd there are no solutions to $\phi(q)=4$). As none of $\cos(\pi p_1/4)$ or $\cos(\pi p_2/6)$ equals $(2+\sqrt2)/4$ for each integer $p_1\in\{1,3\}$ and $p_2\in\{1,5\}$, we obtain the desired contradiction.