I have read in several sources, such as this Math Stack Exchange answer and these slides from a computer animation class (p. 44), that given two quaternions $\mathbf{q}_1, \mathbf{q}_2 \in \mathbb{R}^4$, the angle between the 4D quaternion vectors is equal to half the angle between their corresponding 3D orientations. In other words, letting $\mathbf{R}_1, \mathbf{R}_2 \in SO(3)$ denote the rotation matrices corresponding to $\mathbf{q}_1$ and $\mathbf{q}_2$ respectively, the angle between $\mathbf{q}_1$ and $\mathbf{q}_2$ is half the angle of rotation required to get from $\mathbf{R}_1$ to $\mathbf{R}_2$ (the "geodesic distance"). Note that due to the fact that the maximum angle between any 3D orientations is $\pi$, I believe this fact only holds when the angle between the 4D quaternion vectors is less than or equal to $\pi/2$.

I have done an exhaustive search and can't find any proof of why this is true. So I'm looking for proof.

**my best insight/attempt:**

So far, I have only been able to prove this fact for the case in which $\mathbf{q}_2$ equals the identity quaternion. Start by writing out the quaternions vectors in terms of axis-angles, where $\theta_1$ and $\mathbf{e}_1 \in \mathbb{R}^3$ denote the angle and axis of $\mathbf{q}_1$, and $\theta_2$ and $\mathbf{e}_2 \in \mathbb{R}^3$ denote the angle and axis of $\mathbf{q}_2$. We have $$ \mathbf{q}_1 = \begin{bmatrix} \cos(\theta_1/2) \\ \sin(\theta_1/2) \mathbf{e}_1 \end{bmatrix}, ~~~ \mathbf{q}_2 = \begin{bmatrix} \cos(\theta_2/2) \\ \sin(\theta_2/2) \mathbf{e}_2 \end{bmatrix} . $$ Taking the dot product and using that fact that the quaternions are unit length, we can solve for the angle $\phi$ between the 4D quaternion vectors: $$ \begin{align} \cos \phi &= \frac{\mathbf{q}_1 \cdot \mathbf{q}_2}{\lVert \mathbf{q}_1 \rVert \lVert \mathbf{q}_2 \rVert} \\ &= \cos(\theta_1/2) \cos(\theta_2/2) + \mathbf{e}_1 \cdot \mathbf{e}_2 \sin(\theta_1/2) \sin(\theta_2/2) . \end{align} $$ If $\mathbf{q}_2$ is taken as the identity quaternion, we have $\theta_2 = 0$ and $\mathbf{q}_2 = [1 ~~ 0 ~~ 0 ~~ 0]^T$, and the equation above becomes $$ \begin{align} \cos \phi &= \cos(\theta_1/2) . \end{align} $$ Since we have defined $\phi$ to be non-negative, for $-\pi \leq \theta_1 \leq \pi$ we have $\phi = | \theta_1/2 |$. So this seems to offer some insight, but I'm not sure where to go from here.