Question

Prove by vector method : $\cos (A-B)=\cos A.\cos B+\sin A.\sin B$.

Answer 1

Let $i$ and $j$ are unit vector of $x$ and $y$ axis respectively. If $OP$ and $OQ$ makes angle $A$ and $B$ with x-axes respectively
Let $\overline{OC}$ and $\overline{OD}$ are unit vector with respect to $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ respectively.
Hence coordinate of $C(\cos A,\sin A)$ and $D(\cos A,\sin A)$ and
$\overrightarrow{OC}=i\cos A+j\sin A$
$\overrightarrow{OC}=i\cos B+j\sin B$
Here $(A-B)$ is the angle between $\overrightarrow{OC}$ and $\overrightarrow{OD}$ we have
$\cos (A-B)=\frac{\overrightarrow{OC}.\overrightarrow{OD}}{|\overrightarrow{OC}||\overrightarrow{OD}|}$
$\therefore \cos (A-B)=\left[\right(\cos A)\hat{i}+(\sin A\left)j\right]$
$\left[\right(\cos B)\hat{i}+(\sin B\left)j\right]$
$=\cos A\cos B+\sin A\sin B$.