We could "derive" the Rodrigues formula by "vectorizing" the quaternion sandwitch product. First, let us recall the sadwitch product:
$v' = Q \ v \ Q^*$
Where $v$ is a pure quaternion (its real part equal to zero) and $Q$ is a unit quaternion and $Q^*$ is its conjugate:
$Q^{*} = q_0 - i q_1 - j q_2 - k q_3$
Aplying the Euler relation:
$v' = Q \ v \ Q^{-1} = (\cos \frac{\theta}{2} + \sin \frac{\theta}{2} \ b) \ v \ (\cos \frac{\theta}{2} - \sin \frac{\theta}{2} \ b)$
$v' = (Q \ v) \ Q^{-1} = (\cos \frac{\theta}{2} \ v + \sin \frac{\theta}{2} \ b \ v) \ (\cos \frac{\theta}{2} - \sin \frac{\theta}{2} \ b)$
$v' = \cos^2 \frac{\theta}{2} \ v - (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ v \ b + (\sin \frac{\theta}{2} \ \cos \frac{\theta}{2}) \ b \ v - \sin^2 \frac{\theta}{2} \ b \ v \ b$
The product $v \ b$ that apperars in the above expressions is the product of two
pure quaternions and it is defined in vector form as:
$v \ b = (i v_0 + j v_1 + k v_2) \ (i b_0 + j b_1 + k b_2) = -v \cdot b + v \times b$,
where $v \cdot b$ is the dot product and $v \times b$ is the cross product.
Notice that the algebraic sum $-v \ b + b \ v$ is not zero, since the cross product is non-commutative. As a consequence $-v \ b + b \ v = (v \cdot b - v \times b) + (-b \cdot v + b \times v) = 0 + 2 \ b \times v = 0 - 2 \ v \times b$.
Replacing that into our last equation we get:
$v' = \cos^2 \frac{\theta}{2} \ v + 2 \ (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ b \times v - \sin^2 \frac{\theta}{2} \ b \ v \ b$
Interestingly the expression $b \ v \ b$ is nothing more than the reflection of $v$ in the plane with normal $b$. Where $b$ and $v$ are pure quaternions. Reflection can be expressed in vector form as:
$b \ v \ b = 2 \ (-v \cdot b) \ b + v$
where the vector $v$ is translated in direction of the negative normal $b$ twice the distance of $v$ projected in $b$. Finally the equation is:
$v' = \cos^2 \frac{\theta}{2} \ v + 2 \ (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ b \times v - \sin^2 \frac{\theta}{2} \ (2 \ (-v \cdot b) \ b + v)$
$v' = \cos^2 \frac{\theta}{2} \ v - \sin^2 \frac{\theta}{2} \ v + 2 \ (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ b \times v + 2 \ \sin^2 \frac{\theta}{2} \ (v \cdot b) \ b$
Applying the following trigonometric identities:
$\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2}$
$\sin \theta = 2 \cos \frac{\theta}{2} \ \sin \frac{\theta}{2}$
$1 - \cos \theta = 2 \ \sin^2 \frac{\theta}{2}$
We get the Rodrigues formula:
$v' = \cos \theta \ v + \sin \theta \ b \times v + (1 - \cos \theta) \ ( v \cdot b ) \ b $
It was discovered by Olinde Rodriguez three years before Hamilton discover quaternions.
