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!!
 A: Hint: by the double angle formula:
$$-\,\frac{1}{\sqrt{2}}=\cos\left(\frac{3\pi}{4}\right)=2\,\cos^2\left(\frac{3\pi}{8}\right)-1 \;\;\implies\;\; \cos\left(\frac{3\pi}{8}\right) = \sqrt{\frac{1}{2}\left({1-\frac{1}{\sqrt{2}}}\right)}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$
The above is equivalent to the posted form since $\sqrt{2-\sqrt{2}} \cdot \sqrt{4+2\sqrt2} = 2\,$.
A: As $\frac{3\pi}{8}$ and $\frac{\pi}{8}$ are complementary angles, we get
$$\begin{align}
\cos\frac{3\pi}{8}&=\sin\frac{\pi}{8}\\
&=\sin\frac{\pi/4}{2}\\
&=\sqrt{\frac{1-\cos(\pi/4)}{2}}\\
&=\sqrt{\frac{1-(1/\sqrt{2})}{2}}\\
&=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}\\
&=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}\cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}}\\
&=\sqrt{\frac{1}{4+2\sqrt{2}}}\\
&=\frac{1}{\sqrt{4+2\sqrt{2}}}
\end{align}$$
A: Problem statement
$$
\cos \left( \frac{3\pi}{8} \right) = \cos \left( \frac{\pi}{4} - \frac{\pi}{8} \right)
$$
Basic formulas
Use the $\color{blue}{angle \ addition}$ formula
$$
\cos \left( \alpha + \beta \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta,
$$
and the $\color{blue}{half\ angle}$ formula
$$
  \cos 2 \theta = \sqrt{\frac{1\color{red}{+}\cos \theta}{2}}, \qquad
  \sin 2 \theta = \sqrt{\frac{1\color{red}{-}\cos \theta}{2}}
$$
And unit circle
$$
\cos \left( \frac{\pi}{4} \right) =
\sin \left( \frac{\pi}{4} \right) =
\frac{1}{\sqrt{2}}
$$
Solution
$$
\begin{align}
\cos \left( \frac{3\pi}{8} \right) 
  &= \cos \left( \frac{\pi}{4} - \frac{\pi}{8} \right) \\[3pt]
%%
  &= \cos \left( \frac{\pi}{4} \right)\cos \left(\frac{\pi}{8} \right) 
 - 
    \sin \left( \frac{\pi}{4} \right)\sin \left(\frac{\pi}{8} \right)\\[3pt]
%%
  &= \cos \left( \frac{\pi}{4} \right) \sqrt{\frac{1\color{red}{+}\cos \frac{\pi}{4}}{2}}
   -
     \sin \left( \frac{\pi}{4} \right)\sqrt{\frac{1\color{red}{-}\cos \frac{\pi}{4}}{2}} \\[4pt]
%%
  &= \frac{1}{\sqrt{2}} \frac{\sqrt{\sqrt{2}+2}}{2}
   -
   \frac{1}{\sqrt{2}} \frac{\sqrt{\sqrt{2}-2}}{2} \\[5pt]
%%
  &= \frac{\sqrt{2 - \sqrt{2}}} {2} \\[3pt]
%%
 &= \frac{1}{\sqrt{4+2 \sqrt{2}}}
%%
\end{align}
$$
A: $$
\begin{aligned}
\text{By}\cos ^{2} x &=\frac{1+\cos 2 x}{2}, \textrm{ we have } \\
\cos \left(\frac{3 \pi}{8}\right) &=\sqrt{\frac{1+\cos \frac{3 \pi}{4}}{2}} \\
&=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}} \\
&=\frac{\sqrt{2-\sqrt{2}}}{2}
\end{aligned}
$$
