If \( \sin \theta=\frac{2}{3}, 0<\theta<\frac{\pi}{2} \), find the exact value of each of the following. \( \begin{array}{lll}\text { (a) } \sin (2 \theta) & \text { (b) } \cos (2 \theta) & \text { (c) } \sin \frac{\theta}{2}\end{array} \) (d) \( \cos \frac{\theta}{2} \)\( ~ \begin{array}{l}\text { (a) } \sin (2 \theta)=\frac{4 \sqrt{5}}{9} \\ \text { (Type an exact answer, using radicals as needed.) } \\ \text { (b) } \cos (2 \theta)=\frac{1}{9} \\ \text { (Type an exact answer, using radicals as needed.) } \\ \text { (c) } \sin \frac{\theta}{2}=\sqrt{\frac{3-\sqrt{5}}{6}} \\ \text { (Type an exact answer, using radicals as needed.) } \\ \text { (d) } \cos \frac{\theta}{2}=\square \\ \text { (Type an exact answer, using radicals as needed.) }\end{array} \)

To find the values requested, we can start by using some trigonometric identities. First, since \( \sin \theta = \frac{2}{3} \), we can calculate \( \cos \theta \) using the Pythagorean identity: \[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}. \] (a) To find \( \sin(2\theta) \): \[ \sin(2\theta) = 2 \sin \theta \cos \theta = 2 \left(\frac{2}{3}\right)\left(\frac{\sqrt{5}}{3}\right) = \frac{4\sqrt{5}}{9}. \] (b) Now for \( \cos(2\theta) \): \[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta = \left(\frac{\sqrt{5}}{3}\right)^2 - \left(\frac{2}{3}\right)^2 = \frac{5}{9} - \frac{4}{9} = \frac{1}{9}. \] (c) To find \( \sin\left(\frac{\theta}{2}\right) \), we can use: \[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}} = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{2}} = \sqrt{\frac{\frac{3 - \sqrt{5}}{3}}{2}} = \sqrt{\frac{3 - \sqrt{5}}{6}}. \] (d) Finally, we can find \( \cos\left(\frac{\theta}{2}\right) \): \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}} = \sqrt{\frac{1 + \frac{\sqrt{5}}{3}}{2}} = \sqrt{\frac{\frac{3 + \sqrt{5}}{3}}{2}} = \sqrt{\frac{3 + \sqrt{5}}{6}}. \] Putting it all together, we'll have the exact values: \[ \begin{array}{l} \text{(a) } \sin (2\theta) = \frac{4\sqrt{5}}{9} \\ \text{(b) } \cos (2\theta) = \frac{1}{9} \\ \text{(c) } \sin \frac{\theta}{2} = \sqrt{\frac{3 - \sqrt{5}}{6}} \\ \text{(d) } \cos \frac{\theta}{2} = \sqrt{\frac{3 + \sqrt{5}}{6}}. \end{array} \]