Question
If $A = B = 45^\circ ,$show that$:\sin (A - B) = \sin A \cos B - \cos A \sin B$
✓
Answer
Given that $A = B = 45^\circ $
$\text{LHS} = \sin (A – B)$
$= \sin ( 45^\circ – 45^\circ )$
$= \sin 0^\circ $
$= 0$
$\text{RHS} = \sin A \cos B – \cos A \sin B$
$= \sin 45^\circ \cos 45^\circ – \cos 45^\circ \sin 45^\circ $
$=\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}$
$=0$
$\text { LHS = RHS }$
$\text{LHS} = \sin (A – B)$
$= \sin ( 45^\circ – 45^\circ )$
$= \sin 0^\circ $
$= 0$
$\text{RHS} = \sin A \cos B – \cos A \sin B$
$= \sin 45^\circ \cos 45^\circ – \cos 45^\circ \sin 45^\circ $
$=\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}$
$=0$
$\text { LHS = RHS }$
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