(1) If: $\sin \theta=\frac{2}{3}, 0^{\circ}<\theta<90^{\circ}$, then $\sin 2 \theta$ equals: (A) $\frac{4 \sqrt{5}}{3}$ (B) $\pm \frac{\sqrt{5}}{3}$ (C) $\frac{4 \sqrt{5}}{9}$ (D) $\frac{-4 \sqrt{5}}{9}$

Step 1 ：Given that \(\sin \theta = \frac{2}{3}\), we need to find \(\sin 2\theta\) using the double angle formula: \(\sin 2\theta = 2\sin \theta \cos \theta\)

Step 2 ：We can find \(\cos \theta\) using the Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\). Solving for \(\cos \theta\), we get \(\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \frac{\sqrt{5}}{3}\)

Step 3 ：Now, we can find \(\sin 2\theta\) using the double angle formula: \(\sin 2\theta = 2\sin \theta \cos \theta = 2\left(\frac{2}{3}\right)\left(\frac{\sqrt{5}}{3}\right) = \boxed{\frac{4 \sqrt{5}}{9}}\)