Interesting solutions for cyclic infinite nested square roots of 2 I have derived cosine values for following cyclic infinite nested square roots of 2 ( Hereafter simply referred as $cin\sqrt2$)
$cin\sqrt2[1-]$ represents $\sqrt{2-\sqrt{2-...}}$
$cin\sqrt2[1-1+]$ represents $\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+...}}}}$
$cin\sqrt2[1-1+]$ = $2\cos\frac{2\pi}{5}$
$cin\sqrt2[2-2+]$ = $2\cos\frac{4\pi}{15}$
$cin\sqrt2[3-3+]$ = $2\cos\frac{24\pi}{65}$
$cin\sqrt2[4-4+]$ = $2\cos\frac{16\pi}{51}$
$cin\sqrt2[5-5+]$ = $2\cos\frac{352\pi}{1025}$
$cin\sqrt2[6-6+]$ = $2\cos\frac{64\pi}{195}$
$cin\sqrt2[7-7+]$ = $2\cos\frac{5504\pi}{16385}$
$cin\sqrt2[8-8+]$ = $2\cos\frac{256\pi}{771}$
$cin\sqrt2[9-9+]$ = $2\cos\frac{87552\pi}{262145}$
$cin\sqrt2[10-10+]$ = $2\cos\frac{1024\pi}{3075}$
.
.
.
We need pattern for $cin\sqrt2[n-n+]$ ($n\in N$)
At present I have figured out the pattern as follows
If n is even $$cin\sqrt2[n-n+] = 2\cos(\frac{2^n\pi}{3\times(2^n+1)})$$
If n is odd $$cin\sqrt2[n-n+] = 2\cos(\frac{\frac{2^n+1}{3}\times 2^n\times\pi}{2^{2n}+1})$$
In both the cases, it is quite obvious that
$$\lim_{n\to\infty} cin\sqrt2[n-n+]=2\cos\frac{\pi}{3}=1$$

My question is how to combine both the equations into a single one?


We can multiply the fraction in the even-$n$ case by $2^n-1$ on top and bottom, yielding
$$\operatorname{cin}\sqrt2[n-n+]=2\cos\frac{2^n\pi(2^n-(-1)^n)}{3(2^{2n}-(-1)^n)}$$
