In triangle ABC, find the range of sinAsinC. In $\triangle$ABC, $\angle$B=$\frac{\pi}{3}$. Find the range of sinAsinC.
I used C=120-A to  simplify the equation in sinC and then applied $-1<sinC<1$ but answer did not matched.

From given, $A+C = \frac{2\pi}3$.
$$\sin A\sin C = \frac{\cos(A-C)-\cos(A+C)}{2} = \frac{\cos(\frac{2\pi}3-2C)-\cos\frac{2\pi}3}{2}$$
Find the maximum and the minimum of the first $\cos$ term.

For maximum, choose $C = \frac\pi3$, which gives $\cos(\frac{2\pi}3-2C) = 1$.
For minimum, choose $C = 0$ or $\frac{2\pi}{3}$, which gives $\cos(\frac{2\pi}3-2C) = -\frac12$.
$$0 =\frac{-1/2+1/2}{2} < \frac{\cos(\frac{2\pi}3-2C)-\cos\frac{2\pi}3}{2} \le \frac{1+1/2}2 = \frac34$$
