Question
If A = B = 60°, verify that.
$\sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}$
$\sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}$
✓
Answer
$\sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}$
$\Rightarrow\sin(60^\circ-60^\circ)=\sin60^\circ\cos60^\circ-\cos60^\circ\sin^\circ60^\circ$
$\Rightarrow\sin0^\circ=\frac{\sqrt{3}}{2}\times\frac{1}{2}-\frac{1}{2}\times\frac{\sqrt{3}}{2}$
$\Rightarrow0=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$\Rightarrow0=0$
$\Rightarrow\text{L.H.S} = \text{R.H.S}$
$\Rightarrow\sin(60^\circ-60^\circ)=\sin60^\circ\cos60^\circ-\cos60^\circ\sin^\circ60^\circ$
$\Rightarrow\sin0^\circ=\frac{\sqrt{3}}{2}\times\frac{1}{2}-\frac{1}{2}\times\frac{\sqrt{3}}{2}$
$\Rightarrow0=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$\Rightarrow0=0$
$\Rightarrow\text{L.H.S} = \text{R.H.S}$
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