I was studying the book Linear And Geometric Algebra and I've been stuck at this problem for couple of days.
Assume $\{e_1,e_2,e_3\}$ is an orthonormal basis in $\mathbb{R}^3$, and two quaternions $q_1$ is a rotation of $90^\circ$ around $e_3$ and $q_2$ is a rotation of $90^\circ$ around $e_1$.
$$\mathbf{I} = e_1e_2e_3$$ $$q_1 = e^{-e_3\mathbf{I}45^\circ}$$ $$\quad q_2 = e^{-e_1\mathbf{I}45^\circ}$$
Now calculate the composition $q_2q_1$. The answer in the book is:
$$cos60^\circ-\dfrac{e_1-e_2+e_3}{\sqrt 3}\mathbf{I}\,sin60^\circ$$
But I couldn't figure the steps to get there.
This is what I can do so far,
\begin{align} q_2q_1 &= (cos45^\circ - e_1\mathbf{I}\,sin45^\circ)(cos45^\circ-e_3\mathbf{I}\,sin45^\circ) \\ &= cos45^\circ cos45^\circ-e_1\mathbf{I}\,sin45^\circ cos45^\circ-e_3\mathbf{I}\,cos45^\circ sin45^\circ + e_1\mathbf{I}e_3\mathbf{I}\,sin45^\circ sin45^\circ \\ &= cos45^\circ cos45^\circ-e_1\mathbf{I}\,sin45^\circ cos45^\circ-e_3\mathbf{I}\,cos45^\circ sin45^\circ + e_2\mathbf{I}\,sin45^\circ sin45^\circ \end{align}
How do I proceed?
Thanks in advance.
$$\begin{align}e^{-e_1I\pi/4}e^{-e_3I\pi/4} &= \big(\cos(\pi/4)-e_1I\sin(\pi/4)\big)\big(\cos(\pi/4)-e_3I\sin(\pi/4)\big) \\ &= \frac 12\big(1 -e_3I-e_1I+e_1Ie_3I\big) \\ &= \frac 12\big(1 -e_3I-e_1I+e_2I\big) \\ &= \cos\left(\frac{\pi}{3}\right) - \frac{e_1-e_2+e_3}{2}I\left(\frac{\sqrt{3}}{\sqrt{3}}\right) \\ &= \cos\left(\frac{\pi}{3}\right) - \frac{e_1-e_2+e_3}{\sqrt{3}}I\left(\frac{\sqrt{3}}{2}\right) \\ &= \cos\left(\frac{\pi}{3}\right) - \frac{e_1-e_2+e_3}{\sqrt{3}}I\sin\left(\frac{\pi}{3}\right)\end{align}$$