# 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!!

• Typo in the denominator of your title. – dantopa Mar 28 '17 at 5:01
• Title suggestion: Show $\cos \left( \frac{ 3\pi }{ 8 } \right) = \frac{1}{\sqrt{ 4 + 2 \sqrt{2} }}$ – dantopa Mar 28 '17 at 5:05
• Thanks, I'll edit it right away! – Maths Matador Mar 28 '17 at 5:15

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}

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 $\sqrt{2-\sqrt{2}} \cdot \sqrt{4+2\sqrt2} = 2\,$.

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}