Simplifying $\scriptsize\sqrt{2+\sqrt{2}} + \sqrt{2+\sqrt{2+\sqrt{2}}} + \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} + \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}$

Question

The question is in the title: is there a simpler form or result for

$$\sqrt{2+\sqrt{2}} + \sqrt{2+\sqrt{2+\sqrt{2}}} + \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} + \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}\quad?$$

Answer 1

$ \cos 2\theta =2\cos ^2\theta -1\Rightarrow \cos\theta=\sqrt{\dfrac{\cos 2\theta+1}{2}} $

Therefore $$\cos \dfrac{\pi}{8}=\sqrt{\dfrac{\frac{1}{\sqrt{2}}+1}{2}}=\sqrt{\dfrac{1+\sqrt{2}}{2\sqrt{2}}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}.$$

Hence $$\cos \dfrac{\pi}{16}=\sqrt{\dfrac{\dfrac{\sqrt{2+\sqrt{2}}}{2}+1}{2}}=\dfrac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}.$$

Then $$\cos \dfrac{\pi}{32}=\sqrt{\dfrac{\dfrac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}+1}{2}}=\dfrac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}.$$

Since $\sin^2 \dfrac{\pi}{32}+\cos^2 \dfrac{\pi}{32}=1$ we have $$\sin \dfrac{\pi}{32}=\dfrac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}.$$

So $$\sqrt{2+\sqrt{2}} + \sqrt{2+\sqrt{2+\sqrt{2}}} + \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} + \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}\\ =2\left(\cos \dfrac{\pi}{8}+\cos \dfrac{\pi}{16}+\cos \dfrac{\pi}{32}+\sin \dfrac{\pi}{32}\right)\dots$$