Prove that $\sin \left(\alpha - \beta\right) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$
Consider an unit circle with center $O \left(0,0\right)$.
Consider two points $P$ and $Q$ on this circle.
Then, $\left|\overrightarrow{OP}\right| = 1$ $\;$ and $\;$ $\left|\overrightarrow{OQ}\right| = 1$
Let $OP$ and $OQ$ make angles $\alpha$ and $\beta$ respectively with the $X$ axis.
Then the coordinates of points $P$ and $Q$ are $\left(\cos \alpha, \sin \alpha\right)$ and $\left(\cos \beta, \sin \beta\right)$ respectively.
Let $\hat{i}$ and $\hat{j}$ be unit vectors along the $X$ and $Y$ axes respectively.
Now, $\overrightarrow{OP} = \overrightarrow{OM} + \overrightarrow{MP} = \hat{i} \cos \alpha + \hat{j} \sin \alpha$
and $\overrightarrow{OQ} = \overrightarrow{ON} + \overrightarrow{NQ} = \hat{i} \cos \beta + \hat{j} \sin \beta$
$\begin{aligned}
\therefore \; \overrightarrow{OQ} \times \overrightarrow{OP} & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\cos \beta & \sin \beta & 0 \\
\cos \alpha & \sin \alpha & 0
\end{vmatrix} \\\\
& = \hat{i} \left(0\right) - \hat{j} \left(0\right) + \hat{k} \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right) \\\\
& = \hat{k} \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right) \;\;\; \cdots \; (1)
\end{aligned}$
$\begin{aligned}
\text{By definition, } \overrightarrow{OQ} \cdot \overrightarrow{OP} & = \left|\overrightarrow{OQ}\right| \left|\overrightarrow{OP}\right| \sin \left(\alpha - \beta\right) \hat{k} \\\\
& = 1 \times 1 \times \sin \left(\alpha - \beta\right) \hat{k} \\\\
& = \sin \left(\alpha - \beta\right) \hat{k} \;\;\; \cdots \; (2)
\end{aligned}$
$\therefore$ $\;$ From equations $(1)$ and $(2)$ we have
$\sin \left(\alpha - \beta\right) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$
Hence proved.