By vector method, provethat sin(a-b)= sina.cosb - cosa.sinb where a,b belong to (0,pi)

Consider a unit circle, x² + y² = 1 


Let O be the centre of this circle. Let P and Q be points on the circle.

Then, $$\begin{gathered} \overrightarrow{OP}=\left(\cos A, \sin A\right)=\cos A\hat{i}+\sin A\hat{j} \\ \overrightarrow{OQ}=\left(\cos B, \sin B\right)=\cos B\hat{i}+\sin B\hat{j} \end{gathered}$$

Now, $$\begin{gathered} \overrightarrow{OP}\times \overrightarrow{OQ}=\left|\overrightarrow{OP}\right| \left|\overrightarrow{OQ}\right| \sin \angle POQ \hat{n}, where \widehat{n } is the unit vector in the direction of \overrightarrow{OP}\times \overrightarrow{OQ} \\ \Rightarrow \overrightarrow{OP}\times \overrightarrow{OQ}=\left|\overrightarrow{OP}\right| \left|\overrightarrow{OQ}\right| \sin \left(A-B\right)\hat{n} \\ \Rightarrow \left|\overrightarrow{OP}\times \overrightarrow{OQ}\right|=\left|\overrightarrow{OP}\right| \left|\overrightarrow{OQ}\right| \sin \left(A-B\right) -------\left(1\right) \end{gathered}$$

Also, $$\begin{gathered} \overrightarrow{OP}\times \overrightarrow{OQ}=\left(\cos A\hat{i}+\sin A\hat{j}\right)\times \left(\cos B\hat{i}+\sin B\hat{j}\right) \\ \Rightarrow \overrightarrow{OP}\times \overrightarrow{OQ}=\cos A\sin B\hat{k}-\sin A\cos B\hat{k} \\ \Rightarrow \left|\overrightarrow{OP}\times \overrightarrow{OQ}\right|=\sqrt{{\left(\cos A\sin B-\sin A\cos B\right)}^2} \\ \Rightarrow \left|\overrightarrow{OP}\times \overrightarrow{OQ}\right|=\sin A\cos B-\cos A\sin B --------\left(2\right) \end{gathered}$$

From (1) & (2), we have:-

sin(A-B)=sinAcosB-cosAsinB

​Hence Proved.