\(
% Arcus cosine.
\def\acos{\cos^{-1}}
% Vector projection.
\def\projection#1#2{{proj_{#1}\left(#2\right)}}
% Vector rejection.
\def\rejection#1#2{{rej_{#1}\left(#2\right)}}
% Norm.
\def\norm#1{{\left\|#1\right\|}}
% Cross product.
\def\cross#1#2{\mathit{cross}\left(#1,#2\right)}
% Dot product.
\def\dot#1#2{{#1 \cdot #2}}
% Magnitude.
\def\mag#1{{\left|#1\right}}
\def\group#1{\left(#1\right)}}
\def\sbgrp#1{\left\{#1\right\}}
\)

## Quaternion to axis angle

A unit quaternion \(\mathbf{q}=\group{\cos\group{\frac{\alpha}{2}},\hat{r}\sin\group{\frac{\alpha}{2}}}\) represents a rotation by an angle \(\alpha\) around an axis \(\hat{r}\)
where

\[\begin{aligned}
&\alpha = 2\acos{q_w}
\end{aligned}\]

\[\begin{aligned}
&\hat{v}_x = \frac{q_x}{\sqrt{1-q_w^2}}
\end{aligned}\]

\[\begin{aligned}
&\hat{v}_y = \frac{q_y}{\sqrt{1-q_w^2}}
\end{aligned}\]

\[\begin{aligned}
&\hat{v}_z = \frac{q_z}{\sqrt{1-q_w^2}}
\end{aligned}\]

### Proof

Recall how an axis angle \(\hat{a},\alpha\) is converted to a unit quaternion \(q\):

\[\begin{aligned}
&q_x = \hat{a}_x \sin\left(\alpha/2\right)
\end{aligned}\]

\[\begin{aligned}
&q_y = \hat{a}_y \sin(\alpha/2)
\end{aligned}\]

\[\begin{aligned}
&q_z = \hat{a}_z \sin(\alpha/2)
\end{aligned}\]

\[\begin{aligned}
&q_w = \cos(\alpha/2)
\end{aligned}\]

To obtain \(\alpha\) from the quaternion solve
\(q_w = \cos(\alpha/2)\) for \(\alpha\):
\[\begin{aligned}
q_w = \cos(\alpha/2)
\end{aligned}\]

\[\begin{aligned}
&\acos{q_w} = \acos{\cos(\alpha/2)}
\end{aligned}\]

\[\begin{aligned}
&\acos{q_w} = \frac{\alpha}{2}
\end{aligned}\]

\[\begin{aligned}
&2\acos{q_w} = \alpha
\end{aligned}\]

\[\begin{aligned}
&\alpha = 2\acos{q_w}
\end{aligned}\]

To obtain \(\hat{v}_{k}\) solve
\(q_k = \hat{v}_k \sin(\alpha/2)\)
for \(\hat{v}_k\):
\[\begin{aligned}
&q_k = \hat{v}_k \sin(\alpha/2)
\end{aligned}\]

\[\begin{aligned}
&\frac{q_k}{\sin(\alpha/2)} = \hat{v}_k
\end{aligned}\]

As
\[\begin{aligned}
&\frac{\alpha}{2} = \frac{2\acos{q_w}}{2} = \acos{q_w}
\end{aligned}\]

\[\begin{aligned}
&\frac{q_k}{\sin(\acos{q_w})} = \hat{v}_k
\end{aligned}\]

As
\[\begin{aligned}
&\sin(\acos{q_w}) = \sqrt{1 - q_w^2}
\end{aligned}\]

\[\begin{aligned}
&\frac{q_k}{\sqrt{1 - q_w^2}} = \hat{v}_k
\end{aligned}\]

As
\[\begin{aligned}
&\sin(\acos{q_w}) = \sqrt{1 - q_w^2}
\end{aligned}\]

\[\begin{aligned}
&\hat{v}_k = \frac{q_k}{\sqrt{1 - q_w^2}}
\end{aligned}\]

The expression \(\hat{v}_k=\frac{q_k}{\sqrt{1-q_w^2}}\) for
\(q_w = \pm1\) is undefined. However, we claim that a
unit quaterion with \(q_w = \pm 1\) represents a rotation
by zero degrees. If that claim holds, we can avoid the
evaluation of the undefined expression by picking an
arbitrary unit vector for \(\hat{v}\).

We convince us that for \(q_w = \pm 1\), \(\alpha\) is some integer multiple of \(2\pi\)
and hence the quaternion represents a rotation by zero degrees:
First, observe the cosine is one at even non-negative integer multiples of \(pi\) e.g.
\(0\), \(2\pi\), \(4\pi\), \(6\pi\), \(8\pi\), ... Second, the cosine is minus one at odd
positive integer multiples of \(pi\) e.g. \(1 \pi\), \(3 \pi\), \(5 \pi\), ... Because of
\(\cos(-x)) = cos(x)\) all this is also true for negative multiples.
Hence \(\alpha/2\) must be some integer multiple of \(0,\pm\pi,
\pm 2\pi, \pm 3\pi,\ldots\) for that \(\pm 1 = \cos\left(\alpha/2\right)\) holds.
Consequently \(\alpha\) must be \(0\), \(\pm 2\pi\), \(\pm 4\pi\), \(\pm 6\pi\), ...
Hence a unit quaternion with \(q_w = 1\) represents a rotation by \(\alpha = 2 n \pi\)
radians. And these rotations are all equivalent to the rotation by zero degrees.