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!!

As $\frac{3\pi}{8}$ and $\frac{\pi}{8}$ are complementary angles, we get
$$\begin{align}
\cos\frac{3\pi}{8}&=\sin\frac{\pi}{8}\\
&=\sin\frac{\pi/4}{2}\\
&=\sqrt{\frac{1-\cos(\pi/4)}{2}}\\
&=\sqrt{\frac{1-(1/\sqrt{2})}{2}}\\
&=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}\\
&=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}\cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}}\\
&=\sqrt{\frac{1}{4+2\sqrt{2}}}\\
&=\frac{1}{\sqrt{4+2\sqrt{2}}}
\end{align}$$
