Final Answer:Given: The trigonometric function
![sin\theta=(3√(2))/(5)\text{ and }(\pi)/(2)<\theta<\pi](https://img.qammunity.org/2023/formulas/mathematics/college/d4990ac60yztcqzm3ee8f3nnx1faaegcer.png)
Required: To determine the exact value of
![cos2\theta\text{ and sin\lparen}(\theta)/(2))](https://img.qammunity.org/2023/formulas/mathematics/college/zxzarmiwot7yssyuxzz27ywaauyscqt1gw.png)
Explanation: Using the trigonometric identity.
![cos2\theta=1-2sin^2\theta](https://img.qammunity.org/2023/formulas/mathematics/college/hxivt0vthzqv0pi0bblykr739lvniwretw.png)
We get,
![\begin{gathered} cos2\theta=1-2*((3√(2))/(5))^2 \\ =1-(36)/(5) \\ =-(31)/(36) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/q1p1bu1a89enumd4w2qvz4lm1g2ejxchu7.png)
Since theta lies in the second quadrant cos will be negative.
Now,
![\begin{gathered} cos\theta=√(1-sin^2\theta) \\ =\sqrt{1-(18)/(25)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rhnmgjcphe11h0bm0fosethygbw9smcjof.png)
which gives
![cos\theta=-(√(7))/(5)](https://img.qammunity.org/2023/formulas/mathematics/college/joahxl4gki9a056l3uraoihebc9w1b81jb.png)
And,
![cos\theta=1-2sin^2(\theta)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/t4ftg0jlkdbyf1zmnbp1b0tlzitm6q8l8e.png)
or
![sin(\theta)/(2)=\sqrt{(1-cos\theta)/(2)}](https://img.qammunity.org/2023/formulas/mathematics/college/3qe4cxtaal8l7dw9dutfuiscwqgk2bdmr0.png)
Putting the values we get
![sin(\theta)/(2)=\sqrt{(1+(√(7))/(5))/(2)}](https://img.qammunity.org/2023/formulas/mathematics/college/88ygkg9hordz4m6xea0ymwnt4rf3644rxo.png)
![sin(\theta)/(2)=\sqrt{(5+√(7))/(10)}](https://img.qammunity.org/2023/formulas/mathematics/college/odrxd56999r50n16g1v0n654t75be91ie7.png)
Final Answer:
![\begin{gathered} cos\theta=-(√(7))/(5) \\ s\imaginaryI n(\theta)/(2)=\sqrt{(5+√(7))/(10)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ng41y7dj5inz97629aozplissxo3kkcmvc.png)