I am doing a project about quaternions and their rotation. I am trying to get the proof of quaternion rotation identity by Wikipedia: https://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation to make sense. But I don't understand how they get from line $3$ and down. (Line $3$ which starts with: $$v(\cos^2(\frac{a}{2})-\sin^2(\frac{a}{2})\dots)$$
I also found a similar try on this website: https://erkaman.github.io/posts/quaternion_rotation.html, but don't understand the simplification there either.
Many thanks!
Best Answer
$$ \begin{aligned} {\vec {v'}}&= {\vec {v}}\cos ^{2}{\frac {\alpha }{2}}+({\vec {u}}{\vec {v}}-{\vec {v}}{\vec {u}})\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}-{\vec {u}}{\vec {v}}{\vec {u}}\sin ^{2}{\frac {\alpha }{2}}\\[6pt] &=\color{blue}{{\vec {v}}\cos ^{2}{\frac {\alpha }{2}}}+2({\vec {u}}\times {\vec {v}})\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}-(\color{blue}{{\vec {v}}({\vec {u}}\cdot {\vec {u}})}-\color{red}{2{\vec {u}}({\vec {u}}\cdot {\vec {v}})})\color{purple}{\sin ^{2}{\frac {\alpha }{2}}}\\[6pt] &=\color{blue}{{\vec {v}}(\cos ^{2}{\frac {\alpha }{2}}-\sin ^{2}{\frac {\alpha }{2}})}+({\vec {u}}\times {\vec {v}})(2\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}})+\color{red}{{\vec {u}}({\vec {u}}\cdot {\vec {v}})(2\sin ^{2}{\frac {\alpha }{2}})}\\[6pt] &={\vec {v}}\cos \alpha +({\vec {u}}\times {\vec {v}})\sin \alpha +{\vec {u}}({\vec {u}}\cdot {\vec {v}})(1-\cos \alpha )\\[6pt]&=({\vec {v}}-{\vec {u}}({\vec {u}}\cdot {\vec {v}}))\cos \alpha +({\vec {u}}\times {\vec {v}})\sin \alpha +{\vec {u}}({\vec {u}}\cdot {\vec {v}})\\[6pt]&={\vec {v}}_{\bot }\cos \alpha +({\vec {u}}\times {\vec {v}})\sin \alpha +{\vec {v}}_{\|} \end{aligned} $$
Since $\vec u$ is a unit vector, $\vec u\cdot\vec u=1$. Is it clear now?