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From a textbook on graphics:

To represent a rotation of $\theta$ degrees about a unit length axis $\mathbf{\hat{k}}$, we use the quaternion

$$ \begin{bmatrix} \cos(\theta / 2) \\ \sin(\theta / 2) \mathbf{\hat{k}} \end{bmatrix} $$

A rotation of $\theta + 4\pi$ degrees about an axis $\mathbf{\hat{k}}$ also gives us the same quaternion...Oddly, a rotation of $\theta + 2 \pi$ degrees about an axis $\mathbf{\hat{k}}$, which in fact is the same rotation, gives us the negated quaternion

$$ \begin{bmatrix} -\cos(\theta / 2) \\ -\sin(\theta / 2) \mathbf{\hat{k}} \end{bmatrix} $$

Question: This quote is from a textbook on graphics that lacks rigor; hence this statement is asserted without proof. So I understand conceptually that $\theta + 2\pi$ represents the same rotation as $\theta$ (a trivial fact). But why do we use this second (negated) matrix to represent this equivalent rotation instead of the original one? What is this textbook getting at?

EDIT: Here is an attempt to get further with the problem:

If $\mathbf{\hat{p}}$ represents a vector in $\mathbb{R}^3$ and $q \in H$ represents the quaternion displayed in the first matrix above, then it seems the claim made above would be true if

$$ q \mathbf{\hat{p}} q^{-1} = (-q) \mathbf{\hat{p}} (-q)^{-1} $$

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Yes, your last equation is true. That is essentially why the unit quaternions are a "double cover" of the rotation group, because each unit quaternion $u$ and its additive inverse $-u$ represent the same rotation.

The justification is trivial algebra:

First of all $(-q)(- q^{-1})=(-1)^2 q q^{-1}=1$, so $(-q)^{-1}=-(q^{-1})$

Using that,

$$(-q)p(-q)^{-1}=(-1)^2qpq^{-1}=qpq^{-1}$$

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