Since
[tex]\frac{\pi}{2}<\theta<\pi[/tex]θ is in the second quadrant; in such quadrant, the cosine function is negative whereas the sine function is positive.
a) In general,
[tex]\cos (a+b)=\cos a\cos b-\sin a\sin b[/tex]Thus, in our case,
[tex]\Rightarrow\cos (2\theta)=\cos (\theta+\theta)=\cos \theta\cdot\cos \theta-\sin \theta\cdot\sin \theta=\cos ^2\theta-\sin ^2\theta[/tex]Furthermore,
[tex]\begin{gathered} \sin ^2\theta+\cos ^2\theta=1 \\ \Rightarrow\cos ^2\theta=1-\sin ^2\theta \end{gathered}[/tex]In our case,
[tex]\begin{gathered} \Rightarrow\cos ^2\theta=1-(\frac{3\sqrt[]{2}}{5})^2=1-\frac{18}{25}=\frac{7}{25} \\ \Rightarrow\cos ^2\theta=\frac{7}{25} \\ \Rightarrow\cos 2\theta=\frac{7}{25}-\frac{18}{25}=-\frac{11}{25} \end{gathered}[/tex]Thus, cos(2θ)=-11/25
b) On the other hand,
[tex]\cos 2x=\cos ^2x-\sin ^2x=(1-\sin ^2x)-\sin ^2x=1-2\sin ^2x[/tex]Set
[tex]\begin{gathered} \theta=2x \\ \Rightarrow x=\frac{\theta}{2} \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} \Rightarrow\cos \theta=1-2\sin ^2(\frac{\theta}{2}) \\ \Rightarrow\sin ^2(\frac{\theta}{2})=\frac{1-\cos \theta}{2} \end{gathered}[/tex]We found in part a) that
[tex]\begin{gathered} \cos ^2\theta=\frac{7}{25} \\ \text{and} \\ \theta\to\text{ in second quadrant} \\ \Rightarrow\cos \theta=-\frac{\sqrt[]{7}}{5} \end{gathered}[/tex]On the other hand, since theta is in the second quadrant, theta/2 is in the first quadrant; thus, sin(theta/2) has to be positive as it is in the first quadrant.
Thus,
[tex]\begin{gathered} \frac{\theta}{2}\to\text{ first quadrant} \\ \Rightarrow\sin (\frac{\theta}{2})\ge0 \\ \Rightarrow\sin ^2(\frac{\theta}{2})=\frac{1-(\frac{-\sqrt[]{7}}{5})}{2}=\frac{1+\frac{\sqrt[]{7}}{5}}{2} \end{gathered}[/tex][tex]\begin{gathered} \Rightarrow\sin (\frac{\theta}{2})=+\sqrt{\frac{1+\frac{\sqrt[]{7}}{5}}{2}} \\ \Rightarrow\sin (\frac{\theta}{2})=\sqrt[]{\frac{1+\frac{\sqrt[]{7}}{5}}{2}}=\sqrt[]{\frac{5+\sqrt[]{7}}{10}} \end{gathered}[/tex]Hence, sin(theta/2)=sqrt((5+sqrt7)/10)
