# Prove that sin $(A - B)$= sin $A$ cos $B$ - cos $A$ sin $B$.

Toolbox:
• For two vectors $\overrightarrow a \: and \: \overrightarrow b$, the vector product $\overrightarrow a$ x $\overrightarrow b=|\overrightarrow a||\overrightarrow b| \sin \theta \hat n$ with $\hat n \perp$ to $\overrightarrow a \: and \: \overrightarrow b\: and \: \overrightarrow a, \overrightarrow b, \hat n$ forming a right handed system.
• If $\overrightarrow a = a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k, \: \overrightarrow b = b_1\overrightarrow i+b_2\overrightarrow j+b_3\overrightarrow k$ then $\overrightarrow a$ x $\overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Step 1
In the $XOY$ plane consider the radii $OL, OL'$ of the unit circle where $<{XOL} = A , \: <{XOL'} = B.$ Let $\overrightarrow i, \overrightarrow j, \overrightarrow k$ be the unit vectors along the coordinate axes.
Draw $LM, L'M' \perp OX.$
Step 2
$\overrightarrow {OL} = \overrightarrow {OM}+\overrightarrow {ML}$
$= OM\overrightarrow i + ML \overrightarrow j$
$\overrightarrow {OL'} = OM'\overrightarrow i+M'L'\overrightarrow j$
Now in $\Delta OLM \large\frac{OM}{OL} = OM = \cos A$
$\large\frac{ML}{OL} = ML = \sin A$ ( since $OL = 1$)
Similarly, $OM' = \cos B, M'L' = \sin B$
Step 3
By definition $\overrightarrow {OL'} \times \overrightarrow {OL} = (OL')(OL) \sin (A-B) \overrightarrow k = \sin (A-B) \overrightarrow k$ (i)
But $\overrightarrow {OL} = \cos A\overrightarrow i+ \sin A \overrightarrow j$
$\overrightarrow {OL'} = \cos B \overrightarrow i + \sin B \overrightarrow j$
$\therefore \overrightarrow {OL} \times \overrightarrow {OL'} = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ \cos A & \sin A & 0 \\ \cos B & \sin B & 0 \end{vmatrix} = - ( \cos A \sin B - \sin A \cos B) \overrightarrow k$
or $\overrightarrow {OL'} \times \overrightarrow {OL} = ( \sin A \cos B-\cos A \sin B) \overrightarrow k$ (ii)
From (i) and (ii) $\sin(A-B) = \sin A \cos B - \cos A \sin B$ Hence proved

edited Jun 23, 2013