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

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

Toolbox:
• $|\overrightarrow a\times\overrightarrow b|=|\overrightarrow a||\overrightarrow b|sin\theta$ where $\theta$ is the angle between the vectors.
• $\overrightarrow a\times\overrightarrow b=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right|$ where $\overrightarrow a=a_1\hat i+a_2\hat j+a_3\hat k$ and $\overrightarrow b=b_1\hat i+b_2\hat j+b_3\hat k$
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 $1^{st}$ 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 $|\overrightarrow a\times\overrightarrow b|=|\overrightarrow a||\overrightarrow b|sin\theta$ where $\theta$ is the angle between the vectors.
$\Rightarrow |\overrightarrow {OP}\times\overrightarrow {OQ}|=|\overrightarrow {OP}||\overrightarrow {OQ}|sin(A-B)$
$\Rightarrow\:|\overrightarrow {OP}\times\overrightarrow {OQ}|=sin(A+B)$...............(i)
We also know that $\overrightarrow a\times\overrightarrow b=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right|$ where $\overrightarrow a=a_1\hat i+a_2\hat j+a_3\hat k$ and $\overrightarrow b=b_1\hat i+b_2\hat j+b_3\hat k$
$\Rightarrow\:\overrightarrow {OP}\times\overrightarrow {OQ} =\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\cosA&sinA&0\\cosB&sinB&0\end{array}\right|$
$=\hat i(0)-\hat j(0)+\hat k(cosAsinB-sinAcosB)$
$\Rightarrow \:|\overrightarrow {OP}\times\overrightarrow {OQ}|=sinAcosB-cosAsinB$................(ii)
From (i) and (ii) we get the result
$sin(A-B)=sinAcosB-cosAsinB$.
Hence proved.

edited Apr 22, 2013