Show: $\cos \left( \frac{ 3\pi }{ 8 } \right) = \frac{1}{\sqrt{ 4 + 2 \sqrt{2} }}$ I'm having trouble showing that:
$$\cos\left(\frac{3\pi}{8}\right)=\frac{1}{\sqrt{4+2\sqrt2}}$$
The previous parts of the question required me to find the modulus and argument of $z+i$ where $z=\operatorname{cis{\theta}}$. Hence, I found the modulus to be $2\cos{\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}$ units and that the argument would be $\operatorname{arg}(z+i)=\frac{\pi}{4}+\frac{\theta}{2}$.
Now, the next step that I took was that I replaced every theta with $\frac{3\pi}{8}$ in the polar form of the complex number $z+i$. So now it would look like this:
$$z+i=\left[2\cos{\left(\frac{\pi}{8}\right)}\right]\operatorname{cis}{\left(\frac{3\pi}{8}\right)}$$
Then, I expanded the $\operatorname{cis}{\left(\frac{3\pi}{8}\right)}$ part to become $\cos{\left(\frac{3\pi}{8}\right)}+i\sin{\left({\frac{3\pi}{8}}\right)}$. So now I've got the $\cos\left({\frac{3\pi}{8}}\right)$ part but I don't really know what to do next. I've tried to split the angle up so that there would be two angles so I can use an identity, however, it would end up with a difficult fraction instead. So if the rest of the answer or a hint would be given to finish the question, that would be great!!
Thanks!!

$$
\begin{aligned}
\text{By}\cos ^{2} x &=\frac{1+\cos 2 x}{2}, \textrm{ we have } \\
\cos \left(\frac{3 \pi}{8}\right) &=\sqrt{\frac{1+\cos \frac{3 \pi}{4}}{2}} \\
&=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}} \\
&=\frac{\sqrt{2-\sqrt{2}}}{2}
\end{aligned}
$$
