negative quaternion I often see claims that the negative of a quaternion represents the same rotation, just that the axis and angle have both been reversed. However, if I look at the axis-angle representation of the quaternion $q=[\cos(\theta/2),\vec{n}\sin(\theta/2)]$ and then reverse the rotation angle $\theta=-\theta$ and rotation axis $\vec{n}=-\vec{n}$ and plug it in I get 
$$q(-\theta,-\vec{n})=[\cos(-\theta/2),-\vec{n}\sin(-\theta/2)]=[\cos(\theta/2),\vec{n}\sin(\theta/2)]$$ but this is not the same as $-q=[-\cos(\theta/2),-\vec{n}\sin(\theta/2)]$. I must be missing something obvious but cannot figure it out. To me, negating the axis and angle gives $q$ not $-q$, so where am I going wrong? 
 A: Your calculations are correct.  
However, if you replace $\theta$ by $\theta + 2\pi$ (which is equivalent to $\theta$), then you will get $-q$.
In your example, if you replace $n$ by $-n$, and $\theta$ by $-\theta + 2\pi$ (which is equivalent to $-\theta$), then you will also get $-q$. That's where the claims are coming from.
A: In addition to what Ted said that statement may mean the following. The rotation $R$ related to quaternion $q$ sends a vector $\vec{v}$ to the vector $$R(\vec{v})=q\,\vec{v}\,\overline{q}.$$ Because $\overline{-q}=-\overline{q}$, the rotation $R'$ related to $-q$ sends the vector $\vec{v}$ to
$$
\begin{aligned}
R'(\vec{v})&=(-q)\,\vec{v}\,\overline{-q}\\
&=(-1)^2q\,\vec{v}\,\overline{q}\\
&=q\,\vec{v}\,\overline{q}\\
&=R(\vec{v}).
\end{aligned}
$$
The multiplication of quaternions is bilinear, so we can move those $-1$ factors to the front when they cancel each other.
Anyway, the calculation shows that $R$ and $R'$ are the same rotation. 
A: I appreciate the previous answers. But I think for better clarification, it is better to start with the relation of the quaternion and the rotation first, and then try to change the sign of the quaternion and come up with the same rotation. From Using quaternions as rotations, we have:
$q=[\cos(\theta/2),\vec{n}\sin(\theta/2)]=e^{\frac{\theta}{2}\vec{n}}$
Then we can start with the same rotation ($2k\pi+\theta$) as follows:
$e^{\frac{2k\pi+\theta}{2}\vec{n}}=e^{(k\pi+\frac{\theta}{2})\vec{n}}=e^{k\pi\vec{n}}e^{\frac{\theta}{2}\vec{n}}=(\cos(k\pi),\vec{n}\sin(k\pi))e^{\frac{\theta}{2}\vec{n}}=e^{\frac{\theta}{2}\vec{n}}\cos(k\pi)=q\cos(k\pi)$
So, for odd $k \in \{1,3,5,...\}$ we have $e^{\frac{2k\pi+\theta}{2}\vec{n}}=-q$. But for even $k$, they are the same.
