Some results!

Recall that, $\cos^2 \phi -\sin ^2 \phi= \cos 2\phi$ .Dividing angle by $2$ we have $\cos^2 \frac{\phi}{2} -\sin ^2 \frac{\phi}{2}= \cos \phi$ . In additon, we call it as half angle formula . Thus we have that $\cos^2 \frac{\phi}{2} -\sin ^2 \frac{\phi}{2}= \cos \phi\implies \cos\phi = \pm \sqrt{\dfrac{1 +\cos \phi}{2}}$ Now set $\phi = \frac{\pi}{4}$ which results $\cos \frac{\pi}{8} = \left(\dfrac{1+\sqrt 2}{2\sqrt 2}\right)^{2^{-1}}= \dfrac{\sqrt{1+\sqrt 2}}{\sqrt {2\sqrt 2}}$ Plug $\phi = \frac{\pi}{8}$ which shows that $\cos \frac{\pi}{16}= \left(\dfrac{1+\cos \frac{\phi}{8}}{2}\right)= \left(\dfrac{\sqrt{\sqrt {1+\sqrt 2}+\sqrt {2\sqrt 2}}}{\sqrt{2\sqrt {2\sqrt 2}}}\right)$ Further set $\phi = \frac{\pi}{16}, \frac{\pi}{32},\frac{\pi}{64}\cdots \frac{\pi}{2^{n}}$ where $n\in\mathbb N$ . $\begin{aligned}\cos \frac{\pi}{32}& = \dfrac{\sqrt{\sqrt{\sqrt{1+\sqrt 2}+\sqrt{2\sqrt{2}}}+\sqrt{2\sqrt{2{\sqrt{2}}}}} }{\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}} }} \\ \cos\frac{\pi}{64} & = \dfrac{\sqrt{\sqrt{\sqrt{\sqrt{1+\sqrt 2}+\sqrt{2\sqrt{2}}}+\sqrt{2\sqrt{2{\sqrt{2}}}}} + \sqrt{2\sqrt{2\sqrt{2\sqrt2}}}}}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}} }}\\ \cos \frac{\pi}{128} & = \left(\dfrac{\sqrt{\sqrt{\sqrt{\sqrt{1+\sqrt 2}+\sqrt{2\sqrt{2}}}+\sqrt{2\sqrt{2{\sqrt{2}}}}} + \sqrt{2\sqrt{2\sqrt{2\sqrt2}}}}+\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}} }}\right)^{2^{-1}} \end{aligned}$ Thus $\cos \frac{\pi}{128} =\cos \frac{\pi}{2^7}$ . Thus $n=7$ .