Question : If $\sin(60^{\circ}-x)=\cos(y+60^{\circ})$, then the value of $\sin(x-y)$ is:
Option 1: $\frac{1}{\sqrt2}$
Option 2: $\frac{1}{2}$
Option 3: $\frac{\sqrt3}{2}$
Option 4: $1$
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Correct Answer: $\frac{1}{2}$
Solution : Given: $\sin(60^{\circ}-x)=\cos(y+60^{\circ})$ ⇒ $\cos(90^{\circ}-60^{\circ}+x)=\cos(y+60^{\circ})$ ⇒ $30^{\circ}+x=y+60^{\circ}$ ⇒ $x-y=30^{\circ}$ Now, $\sin(x-y)$ = $\sin(30^{\circ})$ = $\frac{1}{2}$ Hence, the correct answer is $\frac{1}{2}$.
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