How to evaluate $\cos{\frac{\pi}{8}}$?

Question

I have to evaluate $\cos{\frac{\pi}{8}}$ and I'm supposed to do so evaluating first $\cos^2{\frac{\pi}{8}}$ (since it's an exercise to practice half-angle formulas). Solving this second formula I get to

$\cos^2{\frac{\pi}{8}} = \frac{1}{2} + \frac{1}{2\sqrt[]{2}}$

where I'm stuck. I'm not sure it is an useful evaluation and, worst of all, I don't think that could help me solving $\cos{\frac{\pi}{8}}$. I don't know how to get to $\frac{\sqrt[]{2 + \sqrt[]{2}}}{2}$, which is given as the proper answer. Could anyone explain it to me?

Thank you very much in advance.

Answer 1

You're on the right track: note that $$ \cos^2\Big(\frac{\pi}{8}\Big)=\frac{1}{2}+\frac{1}{2\sqrt{2}}=\frac{2+\sqrt{2}}{4} $$ and now take a square root to get $\cos(\frac{\pi}{8})$ (which is positive).

Answer 2

$$\cos2\theta = 2\cos^2\theta -1$$

Put $\theta = \frac{\pi}{4}$,

$$\frac{1}{\sqrt{2}} = 2\cos^2\frac{\pi}{8} - 1$$

$$\cos^2\frac{\pi}{8} = \frac{\sqrt{2}+1}{2\sqrt{2}} = \frac{2 + \sqrt{2}}{4}$$

$$\implies \cos\frac{\pi}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$$

Answer 3

Since $\cos^2\frac\pi8=\frac12+\frac1{2\sqrt2}=\frac{2+\sqrt2}4$, $\cos\frac\pi8=\frac{\sqrt{2+\sqrt2}}2$.