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If sin$\left(\frac{2A+B}{2}\right)$ = cos$\left(\frac{2A-B}{2}\right)$ = $\frac{\sqrt{3}}{2}$, 0° < $\frac{2A+B}{2}$ < 90° and 0° < $\frac{2A+B}{2}$ < 90° then find the value of sin[3(A - B)].