I'm studiyng some notes of differential geometry and stuck at this point:
Being $\alpha$ the smallest angle between $a$ and $b$, we have $\cos \alpha=\langle a, b\rangle $, so $|a-b|^2=4\sin^2\bigg(\dfrac{\alpha}{2}\bigg)$.
First, I know that $|a||b|\cos \alpha=\langle a, b\rangle $, maybe the vectors are units, but I could not imagine from what I get the relation with sin.
Many Thanks.
The question assumes $a,\,b$ are unit vectors so $|a-b|^2=2(1-\cos\alpha)$, which is famously equal to what you want. If you want a geometric interpretation, draw an isosceles triangle to show the base $|a-b|$ is $2\sin\frac{\alpha}{2}$.