Final Answer:Given: The trigonometric function
[tex]sin\theta=\frac{3\sqrt{2}}{5}\text{ and }\frac{\pi}{2}<\theta<\pi[/tex]Required: To determine the exact value of
[tex]cos2\theta\text{ and sin\lparen}\frac{\theta}{2})[/tex]Explanation: Using the trigonometric identity.
[tex]cos2\theta=1-2sin^2\theta[/tex]We get,
[tex]\begin{gathered} cos2\theta=1-2\times(\frac{3\sqrt{2}}{5})^2 \\ =1-\frac{36}{5} \\ =-\frac{31}{36} \end{gathered}[/tex]Since theta lies in the second quadrant cos will be negative.
Now,
[tex]\begin{gathered} cos\theta=\sqrt{1-sin^2\theta} \\ =\sqrt{1-\frac{18}{25}} \end{gathered}[/tex]which gives
[tex]cos\theta=-\frac{\sqrt{7}}{5}[/tex]And,
[tex]cos\theta=1-2sin^2\frac{\theta}{2}[/tex]or
[tex]sin\frac{\theta}{2}=\sqrt{\frac{1-cos\theta}{2}}[/tex]Putting the values we get
[tex]sin\frac{\theta}{2}=\sqrt{\frac{1+\frac{\sqrt{7}}{5}}{2}}[/tex][tex]sin\frac{\theta}{2}=\sqrt{\frac{5+\sqrt{7}}{10}}[/tex]Final Answer:
[tex]\begin{gathered} cos\theta=-\frac{\sqrt{7}}{5} \\ s\imaginaryI n\frac{\theta}{2}=\sqrt{\frac{5+\sqrt{7}}{10}} \end{gathered}[/tex]