Quaternions as rotations: Notation I am reading through this notes on quaternions. I am trying to understand how they work as rotations as long as their norm is always 1.
First, in page 4 above the figure, the quaternion rotation operator is defined as follows
$$q = \cos(\theta) + u\sin(\theta)$$
But later, in equation 4 it is defined with half the angle:
$$q = \cos(\frac{\theta}{2}) + u\sin(\frac{\theta}{2})$$
Is this a mistake or are they telling the same?
 A: It doesn't matter if your quaternion has norm $1$ or not. It always induces a rotation on the subset of pure quaternions.
This is because if $q$ is nonzero, $qxq^{-1}=\frac{|q|}{|q|}qxq^{-1}=\frac{q}{|q|}x\left(\frac{q}{|q|}\right)^{-1}$ and of course $\frac{q}{|q|}$ is a unit quaternion.
So saying "it's a rotation as long as it's a unit quaternion" does not really carry any weight.
But choosing a unit-length representative reduces the number of quaternions being considered all the way down to two (instead of uncountably many.) In fact one can go a step further and insist that the real part is positive to get a single representative for a nontrivial rotation.

Strictly speaking, there is no mistake, but there is an awkwardness being caused by using a $\theta$ in both places.  The question is "how is $\theta$ related to the angle of rotation caused by the unit quaternion?"
One could write $q=\cos(\theta)+u\sin(\theta)$ with $\theta\in [0,\pi/2]$ and rightly say "it rotates by an angle of $2\theta$".
OR one could have written $q=\cos(\theta/2)+u\sin(\theta/2)$ with $\theta\in [0,2\pi]$ to begin with and said "this rotates by an angle of $\theta$.
A: In addition to @rschwieb's explanation, there is always the possibility that two different actions of quaternions (on themselves...) are in play. One is simply by multiplication, the other by conjugation. The latter "doubles" a natural angular parameter...
In summary, one has to figure this out from context... and allow for typos.
