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Using vector method prove that : $c)\;\cos (A+B)=\cos A\cos B-\sin A\sin B$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Toolbox:
• $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|cos\theta$
• $\overrightarrow a.\overrightarrow b=a_1b_1+a_2b_2+a_3b_3$.
Ans:
Let $P(x_1,y_1)$ be a point in first quadrant which is at a distance of 1 unit from origin O
and let angle made by OP with X axis be A.
Similarly let $Q(x_2,-y_2)$ be a point in $4^{th}$ quadrant which is at a distance of 1 unit from origin O
and let angle made by OQ with X axis be B.
$\Rightarrow |\overrightarrow {OP}|=|\overrightarrow {OQ}|=1$ and $<\:POQ=A+B$
$\Rightarrow \overrightarrow {OP}=cosA\hat i+sinA\hat j$ and $\overrightarrow {OQ}=cosB\hat i-sinB\hat j$.
We know that $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|cos\theta$ where $\theta$ is the angle between the vectors.
$\Rightarrow\overrightarrow {OP}.\overrightarrow {OQ}=|\overrightarrow {OP}||\overrightarrow {OQ}|cos(A+B)$
$\Rightarrow\:|\overrightarrow {OP}.\overrightarrow {OQ}|=cos(A+B)$...............(i)
We also know that $\overrightarrow a.\overrightarrow b=(a_1\hat i+a_2\hat j+a_3\hat k).(b_1\hat i+b_2\hat j+b_3\hat k)$
$\Rightarrow\:\overrightarrow {OP}.\overrightarrow {OQ} (cosA\hat i+sinA\hat j).(cosB\hat i-sinB\hat j)$
$=cosA\:cosB+sinA\:(-sinB)$
$=cosAcosB-sinAsinB$
$\Rightarrow \:|\overrightarrow {OP}.\overrightarrow {OQ}|=cosAcosB-sinAsinB$................(ii)
From (i) and (ii) we get the result
$cos(A+B)=coscosB-sinAsinB$.
Hence proved.

edited Apr 22, 2013