Prove by vector method $\;$ $\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$
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 $\overrightarrow{OQ} = \overrightarrow{ON} + \overrightarrow{NQ} = \hat{i} \cos \beta - \hat{j} \sin \beta$
$\begin{aligned}
\therefore \; \overrightarrow{OP} \cdot \overrightarrow{OQ} & = \left(\hat{i} \cos \alpha + \hat{j} \sin \alpha\right) \cdot \left(\hat{i} \cos \beta - \hat{j} \sin \alpha\right) \\\\
& = \cos \alpha \cos \beta - \sin \alpha \sin \beta \;\;\; \cdots \; (1)
\end{aligned}$
$\begin{aligned}
\text{By definition, } \overrightarrow{OP} \cdot \overrightarrow{OQ} & = \left|\overrightarrow{OP}\right| \cdot \left|\overrightarrow{OQ}\right| \cos \left(\alpha + \beta\right) \\\\
& = 1 \times 1 \times \cos \left(\alpha + \beta\right) \\\\
& = \cos \left(\alpha + \beta\right) \;\;\; \cdots \; (2)
\end{aligned}$
$\therefore$ $\;$ From equations $(1)$ and $(2)$ we have
$\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$
Hence proved.