Question
Evaluate the following:
If A = 60° and B = 30°, verify that:
$\sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}$

Answer

$\text{A}=60^\circ$ and $\text{B}=30^\circ$
$\sin(\text{A}-\text{B})=\sin30^\circ=\frac12$
$\sin\text{A}\cos\text{B}=\cos\text{A}\sin\text{B}=\sin60^\circ\sin30^\circ$
$=\Big(\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}-\frac12\times\frac12\Big)=\Big(\frac{3}{4}-\frac{1}{4}\Big)$
$=\frac{2}{4}=\frac12$
$\therefore\ \sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}$

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