Using vectors,prove that $\cos (A-B)=\cos A \cos B+\sin A \sin B$.

(N/A) Let $\widehat{OP}$ and $\widehat{OQ}$ be unit vectors making angles $A$ and $B$,respectively,with the positive direction of the $x$-axis. Then $\angle QOP = A-B$ (as shown in the figure).
We know that $\widehat{OP} = \cos A \hat{i} + \sin A \hat{j}$ and $\widehat{OQ} = \cos B \hat{i} + \sin B \hat{j}$.
By the definition of the dot product,$\widehat{OP} \cdot \widehat{OQ} = |\widehat{OP}| |\widehat{OQ}| \cos(A-B)$.
Since $\widehat{OP}$ and $\widehat{OQ}$ are unit vectors,$|\widehat{OP}| = 1$ and $|\widehat{OQ}| = 1$.
Therefore,$\widehat{OP} \cdot \widehat{OQ} = \cos(A-B) \quad \dots(1)$.
In terms of components,$\widehat{OP} \cdot \widehat{OQ} = (\cos A \hat{i} + \sin A \hat{j}) \cdot (\cos B \hat{i} + \sin B \hat{j})$.
Using the property $\hat{i} \cdot \hat{i} = 1, \hat{j} \cdot \hat{j} = 1, \hat{i} \cdot \hat{j} = 0$,we get:
$\widehat{OP} \cdot \widehat{OQ} = \cos A \cos B + \sin A \sin B \quad \dots(2)$.
From equations $(1)$ and $(2)$,we get:
$\cos(A-B) = \cos A \cos B + \sin A \sin B$.