Excuse me.
Let in $\mathbb{R}^2$ there is a circle $\{(x,y)\in\mathbb{R}^2~|~x^2+y^2=r^2\}$.
Then, let the position of the point $A$ in this space is $\mathbf{a}=r(\mathbf{i}\cos\alpha+\mathbf{j}\sin\alpha)$
and the position of the point $B$ is $\mathbf{b}=r(\mathbf{i}\cos\beta-\mathbf{j}\sin\beta)$.
Of course,
$\mathbf{a}\cdot\mathbf{i}=r\cos\alpha$ and $\mathbf{b}\cdot\mathbf{i}=r\cos\beta$
so $\mathbf{a}\cdot\mathbf{b}=r^2\cos(\alpha+\beta)$.
If an angle which be formed by the vector $\mathbf{a}-\mathbf{b}$ and $r\mathbf{i}-\mathbf{b}$ is $\gamma$, then here is the expression
$(\mathbf{a}-\mathbf{b})\cdot(r\mathbf{i}-\mathbf{b})=|\mathbf{a}-\mathbf{b}|\,|r\mathbf{i}-\mathbf{b}|\cos\gamma$ alias
$$ \cos\gamma=\frac{1+\cos\alpha-\cos\beta-\cos(\alpha+\beta)}{2\sqrt{1-\cos(\alpha+\beta)}\sqrt{1-\cos\beta}}. $$
In sight, the last equation is stranged enough.
But, if we plot the graph $\gamma$ vs $\beta$, we get that
$\gamma$ has value $\alpha/2$ for $0<\beta<2\pi-\alpha$,
and has value $\pi-\alpha/2$ for $2\pi-\alpha<\beta<2\pi$.
How can it? Until now, I am still confused.
Thank you very much.
For $\vec{OA}=\bf{a}$, $\vec{OB}=\bf{b}$, we have, from the representation $\textbf{a}=\hat{i}\cos(\alpha)+\hat{j}\sin(\alpha)$, $\vec{OA}$ makes an angle of $\alpha$ with the positive $x-$axis. Similarly $\vec{OB}$ makes an angle of $-\beta$ with the positive $x-$axis.
The following picture represents the condition represents the situation when $0<\beta < 2\pi-\alpha$.
$\vec{DB}=r\textbf{i}-\textbf{b}$
$\vec{BA}=\textbf{a}-\textbf{b}$
so the angle between them is $\gamma=\angle ABD$ which is the angle the minor circle segment $AD$ subtends at the circumference, which is half of the angle the segment subtends at the center, so it must be $\dfrac{\angle AOD}2=\dfrac{\alpha}2$.
So, $0 < \beta < 2\pi-\alpha$ is the situation where point $B$ lies on the major segment $AD$ in this case, and
$2\pi-\alpha<\beta<2\pi$ is the case where $B$ lies on the minor segment $AD$.
(Referring to the location of $B$ as on the minor or major segment $AD$ makes sense only for particular values of $\alpha$, to be more rigorous, $0<\beta<2\pi-\alpha$ means if you traverse counterclockwise along the circumference from $D$ you will encounter $A$ before $B$, and if $2\pi-\alpha<\beta<2\pi$ you will encounter $B$ before $A$.)
For the case $2\pi-\alpha < \beta<2\pi$, consider the following figure, take any point $C$ on the major segment $AD$, then the required angle $$\gamma=\angle ABD = \pi - \angle ACD = \pi - \dfrac{AOD}2=\pi-\dfrac{\alpha}2$$ where $ABCD$ is a cyclic quadrilateral, so it's opposite angles sum up to $\pi$.