Nested sine and cosine half angle formulas [closed]

By evaluating each radical we have, $$\sqrt{\dfrac{1+\frac{\sqrt3}2}2}=\sqrt{\dfrac{1+\cos\frac{\pi}6}{2}}=\cos\frac{\pi}{12}$$ $$\sqrt{\dfrac{1-\cos\frac{\pi}{12}}{2}}=\sin\frac{\pi}{24}$$

$$\sqrt{\dfrac{1-\sin\frac{\pi}{24}}{2}}=\sqrt{\dfrac{\sin^2\frac{\pi}{48}+\cos^2\frac{\pi}{48}-2\sin\frac{\pi}{48}\cos\frac{\pi}{48}}{2}}=\dfrac{|\sin\frac{\pi}{48}-\cos\frac{\pi}{48}|}{\sqrt2}$$$$=\frac1{\sqrt2}.\cos\frac{\pi}{48}-\frac1{\sqrt2}\sin\frac{\pi}{48}=\cos(\frac{\pi}4+\frac{\pi}{48})=\cos\frac{13\pi}{48}$$ And finally, $$\sqrt{\dfrac{1+\cos{\frac{13\pi}{48}}}{2}}=\cos\frac{13\pi}{96}$$ Hence, $\arccos(\cos\frac{13\pi}{96})=\frac{13\pi}{96}$.