Question
Using vector method prove that in $\triangle ABC$, $\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta$
✓
Answer
Let $\hat{i}$ and $\hat{j}$ be the unit vectors along X -axis and Y -axis respectively.
$
\begin{array}{ll}
\text { Let } & OP=OQ=1 \text { unit } \\
\therefore & \overrightarrow{OP}=(\cos \alpha) \hat{i}+\cos (90-\alpha) \hat{j} \\
\Rightarrow & \overrightarrow{OP}=\hat{i} \cos \alpha+\hat{j} \sin \alpha
\end{array}
$
$
\begin{array}{ll}
\text { Let } & OP=OQ=1 \text { unit } \\
\therefore & \overrightarrow{OP}=(\cos \alpha) \hat{i}+\cos (90-\alpha) \hat{j} \\
\Rightarrow & \overrightarrow{OP}=\hat{i} \cos \alpha+\hat{j} \sin \alpha
\end{array}
$
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