A wonderful explanation of why $latex q p q^{-1}$ is rotation of $latex p$, where $latex q$ is most easily interpreted with the complex-number like structure of $latex \cos(\theta/2) + \sin(\theta/2)(a i + b j + c k)$ with $latex a i + bj +ck$ normalized (i.e., $latex a^2+b^2+c^2=1$). Note that $latex q^{-1}$ is then just $latex latex \cos(\theta/2) – \sin(\theta/2)(a i + b j + c k)$ (since $latex (a i + b j + c k)^2=-a^2 – b^2 -c^2 = -1$ just like the imaginary $latex i$ for complex number).
Basically quarternion is 4D stereographic projection back to 3D space. When multiply by a quarternion $latex \cos(\alpha) + \sin(\alpha) {\bf v}$ from the left, points parallel to $latex \bf v$ will be translate along $\bf v$ and points penpendicular to $latex \bf v$ will rotate around $latex \bf v$ (following the right hand rule). Similarly, when multiplying from the right by the same quarternion, points parallel to $latex \bf v$ will be translated in the same direction, but points perpendicular to $latex \bf v$ will be rotated in the opposite direction (following left hand rule). So if we multiple $latex q$ and $latex q^{-1}$ from both left and right hand sides, the translation effect will be exactly cancelled, and the the rotation will be doubled. That is why the angle in $latex q$ should be halved.