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) \( \cos \frac{\theta}{2} \) \( \begin{array}{l}\text { (a) } \sin (2 \theta)=\frac{4 \sqrt{5}}{9} \\ \text { (Type an exact answer, using radicals as needed.) } \\ \text { (b) } \cos (2 \theta)=\frac{1}{9} \\ \text { (Type an exact answer, using radicals as needed.) } \\ \text { (c) } \sin \frac{\theta}{2}=\square\end{array} \)

To find the exact values, we start by using the given \( \sin \theta = \frac{2}{3} \) and applying trigonometric identities. (a) For \( \sin (2\theta) \), we use the double angle formula: \[ \sin (2\theta) = 2 \sin \theta \cos \theta. \] First, we need to find \( \cos \theta \). Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we have: \[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9}. \] Thus, \( \cos \theta = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \). Now, substituting back: \[ \sin (2\theta) = 2 \times \frac{2}{3} \times \frac{\sqrt{5}}{3} = \frac{4\sqrt{5}}{9}. \] (b) For \( \cos (2\theta) \), we use the identity: \[ \cos (2\theta) = \cos^2 \theta - \sin^2 \theta. \] Using \( \cos^2 \theta = \frac{5}{9} \) and \( \sin^2 \theta = \frac{4}{9} \): \[ \cos (2\theta) = \frac{5}{9} - \frac{4}{9} = \frac{1}{9}. \] (c) Now, for \( \sin \frac{\theta}{2} \), we use the half-angle formula: \[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}}. \] We already found \( \cos \theta = \frac{\sqrt{5}}{3} \). So, \[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{2}} = \sqrt{\frac{\frac{3 - \sqrt{5}}{3}}{2}} = \sqrt{\frac{3 - \sqrt{5}}{6}}. \] (d) Lastly, for \( \cos \frac{\theta}{2} \), we use: \[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}}. \] Substituting \( \cos \theta = \frac{\sqrt{5}}{3} \): \[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \frac{\sqrt{5}}{3}}{2}} = \sqrt{\frac{\frac{3 + \sqrt{5}}{3}}{2}} = \sqrt{\frac{3 + \sqrt{5}}{6}}. \] So, the final answers are: (a) \( \sin (2\theta) = \frac{4\sqrt{5}}{9} \) (b) \( \cos (2\theta) = \frac{1}{9} \) (c) \( \sin \frac{\theta}{2} = \sqrt{\frac{3 - \sqrt{5}}{6}} \) (d) \( \cos \frac{\theta}{2} = \sqrt{\frac{3 + \sqrt{5}}{6}} \)