Interesting infinite nested square roots of 2 for $2\cos1°$ and $2\sin1°$

Question

It is interesting to note that any angle between 45° to 90° satisfying $1\over4$ < $p \over q$ <$1\over2$ where $ p \over q$ is of form $p = 2^n $ and $q$ is an odd number satisfying $2^{n+1} <q <2^{n+2}$ can be represented as cyclic infinite nested square roots of 2 ( Hereafter referred as $cin\sqrt2$)

Interestingly 64° falls between 45° and 90° and can be represented in radians as $16\pi \over 45$

Expansion of $2\cos\frac{16\pi}{45}$ happens as follows

$2\cos\frac{16\pi}{45} = \sqrt{2+2\cos\frac{32\pi}{45}} =\sqrt{2-2\cos\frac{13\pi}{45}} $ $=\sqrt{2- \sqrt{2+2\cos\frac{26\pi}{45}}} = \sqrt{2- \sqrt{2-2\cos\frac{19\pi}{45}}}$ $=\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{38\pi}{45}}}} = \sqrt{2-\sqrt{2-\sqrt{2-2\cos\frac{7\pi}{45}}}}$ $=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{14\pi}{45}}}}}$ $=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+2\cos\frac{28\pi}{45}}}}}}$ $=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2\cos\frac{17\pi}{45}}}}}}$ $=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+2\cos\frac{34\pi}{45}}}}}}}$ $=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-2\cos\frac{11\pi}{45}}}}}}}$ $=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{22\pi}{45}}}}}}}}$ $=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+2\cos\frac{44\pi}{45}}}}}}}}}$ $=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2\cos\frac{\pi}{45}}}}}}}}}$....1

For the sake of simplicity last nested radical can be represented as $n\sqrt2[3-1+2-1+1-]$(nested square roots of 2 having $3-1+2-1+1-$)

$2\cos\frac{\pi}{45}$ is represented as $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2\cos\frac{16\pi}{45}}}}}$ or simply as $n\sqrt2(4+2\cos\frac{16\pi}{45})$...2

Combining 1 & 2 will be single cycle of nested radical for $2\cos\frac{16\pi}{45}$ simply represented as $n\sqrt2[3-1+2-1+1-4+]$

Now we can represent $2\cos\frac{16\pi}{45}$ as cyclic infinite nested square roots of 2 as $cin\sqrt2(3-1+2-1+1-4+)$

As $64^\circ$ is $(2^6)^\circ$ taking half angle for 6 times will give noncyclic nested square roots of 2 as $n\sqrt2(6+)$

Therefore $2\cos1°$ can be represented as $n\sqrt2(6+)cin\sqrt2[3-1+2-1+1-4+]$ and

$2\sin1°$ can be represented as $n\sqrt2(1-5+)cin\sqrt2[3-1+2-1+1-4+]$

This opens the world of nested square roots of 2 for calculating $2\cos1°$ or $2\sin1°$ without any need by taking root of cubic equation which involves imaginary component. And no need to know the value of$\pi$ to evaluate trig values as in Taylor series expansion.

Calculating single cycle itself provides result with accuracy of 7 digits after decimal point. (Six '2's in noncyclic part ten '2's in first cycle in infinite radical.)

With available scientific calculators and by programming, I have confirmed the results (for a long Post like this I feel it is difficult to incorporate those things)

My question is, is it possible to simplify above procedures by some other means?

Answer 1

(Revising and extending some comments.)

We can generate a version of these representations thusly:

• Let $\theta_0$ be the angle whose double-cosine we seek.
• Let $\theta_{n+1}:=\min(2\theta_n,180^\circ-2\theta_n)$ and $\sigma_n := \operatorname{sgn}( 90^\circ-2\theta_n)$
• Generate $\theta$- and $\sigma$-values, stopping either when $\sigma_N=0$ (the "terminating" case) or when $\theta_N=\theta_{m}$ for some $m<N$ (the "non-terminating" case).
• Then either $\sigma(\theta_0) := [\sigma_0, \sigma_1, \ldots, \sigma_N=0|]$ (terminating) or $\sigma(\theta_0) := [\sigma_0, \sigma_1, \sigma_2, \ldots, \sigma_{m} \;|\; \sigma_{m+1}, \sigma_{m+2}, \ldots, \sigma_N ]$ (non-terminating) corresponds to a nested-radical representation of $2\cos \theta_0$: Each $\sigma_n$ gives the sign at the corresponding nesting level, as in $$2\cos \theta_0 = \sqrt{ 2 + \sigma_0 \sqrt{2 + \sigma_1 \sqrt{ 2+\sigma_2 \sqrt{\cdots} }}}$$ and the "$|$" separates a finite initial sequence from an infinitely-repeating cycle. (Caveat: The index arithmetic for my placement of the "$|$" may be off by one.)

For instance, $$\sigma(45^\circ) = [0|] \qquad \sigma(22.5^\circ) = [1\bar{0}|] \qquad \sigma(64^\circ) = [|\bar{3}1\bar{2}1\bar{1}4] \qquad \sigma(1^\circ) = [2|4\bar{3}1\bar{2}1\bar{1}]$$ where, for the sake of compactness, I'm collapsing strings of $k$ "$-1$"s to "$\bar{k}$" and strings of $k$ "$+1$"s to simply "$k$". Also, I'm writing a terminating "$0$" as either "$0$" or "$\bar{0}$", whichever contrasts with the "sign" of the preceding string; we can think of this as indicating a collapsed string of no "$+1$"s or no "$-1$"s. (BTW: Please pardon the potential confusion of the over-bar notation with repeating-decimal notation.)

---

The acute integer-degree angles separate into a few families of "doubled" angles (adjusted by the "min" step in the algorithm, if necessary) with related repeating cycles. For now, I've rendered the representations as Mathematica generated them, but one could make the cycles identical by augmenting the initial sequences with portions of them and manipulating appropriately; eg, $$\begin{align} \sigma(8^\circ) &= [| 3\bar{3}1\bar{2}1\bar{1}1] = [3\bar{3}1\bar{2}1\bar{1}1| 3\bar{3}1\bar{2}1\bar{1}1] = [3\bar{3}1\bar{2}1\bar{1}| 4\bar{3}1\bar{2}1\bar{1}] \\ \sigma(16^\circ) &= [| 2\bar{3}1\bar{2}1\bar{1}2] = [2\bar{3}1\bar{2}1\bar{1}2| 2\bar{3}1\bar{2}1\bar{1}2] = [2\bar{3}1\bar{2}1\bar{1}|4\bar{3}1\bar{2}1\bar{1}] \end{align}$$
so that the last versions of $\sigma(8^\circ)$ and $\sigma(16^\circ)$ have the same repeating cycle as $\sigma(1^\circ)$, $\sigma(2^\circ)$, and $\sigma(4^\circ)$. It's not necessarily clear what the "canonical cycle" should be, but I might suggest that such a cycle would begin and end with different "signs" (ie, "$p\cdots\bar{q}$" or "$\bar{p}\cdots q$", but not "$p\cdots q$" or "$\bar{p}\cdots\bar{q}$").

Each table begins with an orbit of doubled-and-adjusted angles. It happens that these are all the angles whose corresponding $\sigma$s can be written with empty initial sequences.

Note that the formula $2\cos\theta=\sqrt{2+2\cos2\theta}$ implies that (some form of) $\sigma(\min(2\theta,180^\circ-2\theta))$ is obtained from (some form of) $\sigma(\theta)$ by reducing the first digit; eg, $[2\bar{3}\cdots]\to[1\bar{3}\cdots]\to [\bar{3}\cdots]$.

$$\begin{array}{rrcrrcrrcr} \theta & \sigma(\theta) && \theta & \sigma(\theta) && \theta & \sigma(\theta) \\ \hline 4^\circ & [| 4\bar{3}1\bar{2}1\bar{1}] &\to& \phantom{1}8^\circ & [| 3\bar{3}1\bar{2}1\bar{1}1] &\to& 16^\circ & [| 2\bar{3}1\bar{2}1\bar{1}2] \\ 32^\circ & [| 1\bar{3}1\bar{2}1\bar{1}3] &\to& 64^\circ & [| \bar{3}1\bar{2}1\bar{1}4] &\to& 52^\circ & [| \bar{2}1\bar{2}1\bar{1}4\bar{1}] \\ 76^\circ & [| \bar{1}1\bar{2}1\bar{1}4\bar{2}] &\to& 28^\circ & [| 1\bar{2}1\bar{1}4\bar{3}] &\to& 56^\circ & [| \bar{2}1\bar{1}4\bar{3}1] \\ 68^\circ & [| \bar{1}1\bar{1}4\bar{3}1\bar{1}] &\to& 44^\circ & [| 1\bar{1}4\bar{3}1\bar{2}] &\to& 88^\circ & [| \bar{1}4\bar{3}1\bar{2}1] &\to& (4^\circ) \\ \hline 1^\circ & [2| 4\bar{3}1\bar{2}1\bar{1}] &\to& 2^\circ & [1| 4\bar{3}1\bar{2}1\bar{1}] &\to& (\phantom{1}4^\circ) \\ \hline 7^\circ & [2| 1\bar{2}1\bar{1}4\bar{3}] &\to& 14^\circ & [1| 1\bar{2}1\bar{1}4\bar{3}] &\to& (28^\circ) \\ \hline 11^\circ & [2| 1\bar{1}4\bar{3}1\bar{2}] &\to& 22^\circ & [1| 1\bar{1}4\bar{3}1\bar{2}] &\to& (44^\circ) \\ \hline 13^\circ & [2| \bar{2}1\bar{2}1\bar{1}4\bar{1}] &\to& 26^\circ & [1| \bar{2}1\bar{2}1\bar{1}4\bar{1}] &\to& (52^\circ) \\ \hline 17^\circ & [2| \bar{1}1\bar{1}4\bar{3}1\bar{1}] &\to& 34^\circ & [1| \bar{1}1\bar{1}4\bar{3}1\bar{1}] &\to& (68^\circ) \\ \hline 19^\circ & [2| \bar{1}1\bar{2}1\bar{1}4\bar{2}] &\to& 38^\circ & [1| \bar{1}1\bar{2}1\bar{1}4\bar{2}] &\to& (76^\circ) \\ \hline 23^\circ & [1\bar{1}| \bar{1}4\bar{3}1\bar{2}1] &\to& 46^\circ & [\bar{1}| \bar{1}4\bar{3}1\bar{2}1] &\to& (88^\circ) \\ \hline 29^\circ & [1\bar{1}| \bar{3}1\bar{2}1\bar{1}4] &\to& 58^\circ & [\bar{1}| \bar{3}1\bar{2}1\bar{1}4] &\to& (64^\circ) \\ \hline 31^\circ & [1\bar{1}| \bar{2}1\bar{1}4\bar{3}1] &\to& 62^\circ & [\bar{1}| \bar{2}1\bar{1}4\bar{3}1] &\to& (56^\circ) \\ \hline 37^\circ & [1\bar{1}| 1\bar{3}1\bar{2}1\bar{1}3] &\to& 74^\circ & [\bar{1}| 1\bar{3}1\bar{2}1\bar{1}3] &\to& (32^\circ) \\ \hline 41^\circ & [1\bar{1}| 2\bar{3}1\bar{2}1\bar{1}2] &\to& 82^\circ & [\bar{1}| 2\bar{3}1\bar{2}1\bar{1}2] &\to& (16^\circ) \\ \hline 43^\circ & [1\bar{1}| 3\bar{3}1\bar{2}1\bar{1}1] &\to& 86^\circ & [\bar{1}| 3\bar{3}1\bar{2}1\bar{1}1] &\to& (\phantom{1}8^\circ) \\ \hline 47^\circ & [\bar{2}| 3\bar{3}1\bar{2}1\bar{1}1] &\to& (86^\circ) \\ \hline 49^\circ & [\bar{2}| 2\bar{3}1\bar{2}1\bar{1}2] &\to& (82^\circ) \\ \hline 53^\circ & [\bar{2}| 1\bar{3}1\bar{2}1\bar{1}3] &\to& (74^\circ) \\ \hline 59^\circ & [\bar{2}| \bar{2}1\bar{1}4\bar{3}1] &\to& (62^\circ) \\ \hline 61^\circ & [\bar{2}| \bar{3}1\bar{2}1\bar{1}4] &\to& (58^\circ) \\ \hline 67^\circ & [\bar{2}| \bar{1}4\bar{3}1\bar{2}1] &\to& (46^\circ) \\ \hline 71^\circ & [\bar{1}1| \bar{1}1\bar{2}1\bar{1}4\bar{2}] &\to& (38^\circ) \\ \hline 73^\circ & [\bar{1}1| \bar{1}1\bar{1}4\bar{3}1\bar{1}] &\to& (34^\circ) \\ \hline 77^\circ & [\bar{1}1| \bar{2}1\bar{2}1\bar{1}4\bar{1}] &\to& (26^\circ) \\ \hline 79^\circ & [\bar{1}1| 1\bar{1}4\bar{3}1\bar{2}] &\to& (22^\circ) \\ \hline 83^\circ & [\bar{1}1| 1\bar{2}1\bar{1}4\bar{3}] &\to& (14^\circ) \\ \hline 89^\circ & [\bar{1}1| 4\bar{3}1\bar{2}1\bar{1}] &\to& (\phantom{1}2^\circ) \\ \hline \end{array}$$

$$\begin{array}{rrcrrcrrcrrcr} \theta & \sigma(\theta) && \theta & \sigma(\theta) && \theta & \sigma(\theta) && \theta & \sigma(\theta) \\ \hline 12^\circ & [| 2\bar{2}] &\to& 24^\circ & [| 1\bar{2}1] &\to& 48^\circ & [| \bar{2}2] &\to& 84^\circ & [| \bar{1}2\bar{1}] &\to& (12^\circ) \\ \hline 3^\circ & [2| 2\bar{2}] &\to& 6^\circ & [1| 2\bar{2}] &\to& (12^\circ) \\ \hline 21^\circ & [2| \bar{1}2\bar{1}] &\to& 42^\circ & [1| \bar{1}2\bar{1}] &\to& (84^\circ) \\ \hline 33^\circ & [1\bar{1}| \bar{2}2] &\to& 66^\circ & [\bar{1}| \bar{2}2] &\to& (48^\circ) \\ \hline 39^\circ & [1\bar{1}| 1\bar{2}1] &\to& 78^\circ & [\bar{1}| 1\bar{2}1] &\to& (24^\circ) \\ \hline 51^\circ & [\bar{2}| 1\bar{2}1] &\to& (78^\circ) \\ \hline 55^\circ & [\bar{2}| 1\bar{1}1] &\to& (70^\circ) \\ \hline 57^\circ & [\bar{2}| \bar{2}2] &\to& (66^\circ) \\ \hline 69^\circ & [\bar{1}1| \bar{1}2\bar{1}] &\to& (42^\circ) \\ \hline 87^\circ & [\bar{1}1| 2\bar{2}] &\to& (\phantom{1}6^\circ) \\ \hline \end{array}$$

$$\begin{array}{rrcrrcrrcr} \theta & \sigma(\theta) && \theta & \sigma(\theta) && \theta & \sigma(\theta) \\ \hline 20^\circ & [| 2\bar{1}] &\to& 40^\circ & [| 1\bar{1}1] &\to& 80^\circ & [| \bar{1}2] &\to& (20^\circ) \\ \hline 5^\circ & [2| 2\bar{1}] &\to& 10^\circ & [1| 2\bar{1}] &\to& (20^\circ) \\ \hline 25^\circ & [1\bar{1}| \bar{1}2] &\to& 50^\circ & [\bar{1}| \bar{1}2] &\to& (80^\circ) \\ \hline 35^\circ & [1\bar{1}| 1\bar{1}1] &\to& 70^\circ & [\bar{1}| 1\bar{1}1] &\to& (40^\circ) \\ \hline 65^\circ & [\bar{2}| \bar{1}2] &\to& (50^\circ) \\ \hline 85^\circ & [\bar{1}1| 2\bar{1}] &\to& (10^\circ) \\ \hline \end{array}$$

$$\begin{array}{rrcrrcr} \theta & \sigma(\theta) && \theta & \sigma(\theta) \\ \hline 36^\circ & [| 1\bar{1}] &\to& 72^\circ & [| \bar{1}1] &\to& (36^\circ) \\ \hline 9^\circ & [2| 1\bar{1}] &\to& 18^\circ & [1| 1\bar{1}] &\to& (36^\circ) \\ \hline 27^\circ & [1\bar{1}| \bar{1}1] &\to& 54^\circ & [\bar{1}| \bar{1}1] &\to& (72^\circ) \\ \hline 63^\circ & [\bar{2}| \bar{1}1] &\to& (54^\circ) \\ \hline 81^\circ & [\bar{1}1| 1\bar{1}] &\to& (18^\circ) \\ \hline \end{array}$$

$$\begin{array}{rrcrrcr} \theta & \sigma(\theta) && \theta & \sigma(\theta) \\ \hline 60^\circ & [| \bar{1}] &\to& (60^\circ) \\ \hline 15^\circ & [2| \bar{1}] &\to& 30^\circ & [1| \bar{1}] &\to& (60^\circ) \\ \hline 75^\circ & [\bar{1}1| \bar{1}] &\to& (30^\circ) \\ \hline \end{array}$$

$$\begin{array}{rr} \theta & \sigma(\theta) \\ \hline 45^\circ & [0|] \\ \hline \end{array}$$

Answer 2

I have found and proved the following theorem on July, 2021.

[Theorem 1] For any rational number $\frac{n}{m}$, both $\cos \left(\frac{n}{m}\pi\right)$ and $\sin \left(\frac{n}{m}\pi\right)$ can be represented as cyclic infinite nested square roots of $2$, of which the cyclic period is less than $\frac{m-1}{2}$.

[Example 1]

$ 2 \cos \left(\frac{\pi }{3}\right)=\sqrt{2-2 \cos \left(\frac{\pi }{3}\right)} $

$ 2 \cos \left(\frac{\pi }{5}\right)=\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{5}\right)}} $

$ 2 \cos \left(\frac{\pi }{7}\right)=\sqrt{2+\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{7}\right)}}} $

$ 2 \cos \left(\frac{\pi }{9}\right)=\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{9}\right)}}} $

$ 2 \cos \left(\frac{\pi }{11}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{11}\right)}}}}} $

$ 2 \cos \left(\frac{\pi }{13}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{13}\right)}}}}}} $

$ 2 \cos \left(\frac{\pi }{15}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{15}\right)}}}} $

$ 2 \cos \left(\frac{\pi }{17}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{17}\right)}}}} $

$ 2 \cos \left(\frac{\pi }{19}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{19}\right)}}}}}}}}} $

$ 2 \cos \left(\frac{2 \pi }{19}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+2 \cos \left(\frac{2 \pi }{19}\right)}}}}}}}}} $

$ 2 \cos \left(\frac{3 \pi }{19}\right)=\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{3 \pi }{19}\right)}}}}}}}}} $

$ 2 \cos \left(\frac{4 \pi }{19}\right)=\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+2 \cos \left(\frac{4 \pi }{19}\right)}}}}}}}}} $

$ 2 \cos \left(\frac{5 \pi }{19}\right)=\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-2 \cos \left(\frac{5 \pi }{19}\right)}}}}}}}}} $

$ 2 \cos \left(\frac{6 \pi }{19}\right)=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+2 \cos \left(\frac{6 \pi }{19}\right)}}}}}}}}} $

$ 2 \cos \left(\frac{7 \pi }{19}\right)=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-2 \cos \left(\frac{7 \pi }{19}\right)}}}}}}}}} $

$ 2 \cos \left(\frac{8 \pi }{19}\right)=\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2 \cos \left(\frac{8 \pi }{19}\right)}}}}}}}}} $

$ 2 \cos \left(\frac{9 \pi }{19}\right)=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{9 \pi }{19}\right)}}}}}}}}} $

$ 2 \cos \left(\frac{\pi }{21}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{21}\right)}}}}}} $

$ 2 \cos \left(\frac{\pi }{23}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{23}\right)}}}}}}}}}}} $

$ 2 \cos \left(\frac{\pi }{25}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{25}\right)}}}}}}}}}} $

$ 2 \cos \left(\frac{\pi }{27}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{27}\right)}}}}}}}}} $

$ 2 \cos \left(\frac{\pi }{29}\right)=\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+\sqrt{2-2 \cos \left(\frac{\pi }{29}\right)}}}}}}}}}}}}}} $

$ 2 \cos \left(\frac{\pi }{31}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{31}\right)}}}}} $

$ 2 \cos \left(\frac{\pi }{33}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{33}\right)}}}}} $

$ 2 \cos \left(\frac{\pi }{35}\right)=\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 \cos \left(\frac{\pi }{35}\right)}}}}}}}}}}}} $

[Example 2]

$ 2 \cos \left(\frac{355 \pi }{113}\right)=-\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+\sqrt{2-2 \cos \left(\frac{355 \pi }{113}\right)}}}}}}}}}}}}}} $

$ 2 \cos \left(\frac{17 \pi }{65537}\right)=\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+\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{17 \pi }{65537}\right)}}}}}}}}}}}}}}}} $

$2\cos(\frac{\pi}{1729})=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ \sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\ \sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\ \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2\cos(\frac{\pi}{1729})}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

$2\cos(\frac{\pi}{641})=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ \sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\ \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\ \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2\cos(\frac{\pi}{641})}}}}}}}}}}}}}}}}}}}\ }}}}}}}}}}}}}}$

[Example 3]

$ 2 \cos \left(\frac{520 \pi }{1314}\right)=\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+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+2 \cos \left(\frac{520 \pi }{1314}\right)}}}}}}}}}}}}}}}}}} $

$ 2 \sin \left(\frac{520 \pi }{1314}\right)=\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+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+2 \cos \left(\frac{520 \pi }{1314}\right)}}}}}}}}}}}}}}}}}} $

$ 2 \cos \left(\frac{1314 \pi }{520}\right)=-\sqrt{2-\sqrt{2+2 \cos \left(\frac{7 \pi }{65}\right)}} $

$ 2 \sin \left(\frac{1314 \pi }{520}\right)=\sqrt{2+\sqrt{2+2 \cos \left(\frac{7 \pi }{65}\right)}} $

$ 2 \cos \left(\frac{7 \pi }{65}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-2 \cos \left(\frac{7 \pi }{65}\right)}}}}}} $

(Note: In Chinese, 520 means "I Love you". 1314 means "Forever".)

I found the above Theorem 1 and Examples by using my following Theorem 2.

[Theorem 2] Let $p_i$ be odd primes, $k_i$ be positive integers. Then

(1) for any integer $h>1$

$\\2^{2 \prod _{i=1}^h \frac{p_i-1}{2} p_i^{k_i-1}}-1\equiv 0 \left(mod \prod _{i=1}^h p_i^{k_i}\right)$

(2) for $p_1\equiv \pm 1 (mod\ 8)$

$\\2^{\frac{p_1-1}{2} p_1^{k_1-1}}-1\equiv 0 \left(mod\ p_1^{k_1}\right)$

(3) for $p_1\equiv \pm 3 (mod\ 8)$

$\\2^{\frac{p_1-1}{2} p_1^{k_1-1}}+1\equiv 0 \left(mod\ p_1^{k_1}\right)$

On the other hand, we have

[Theorem 3]

Any cyclic infinite nested square roots of 2 can be represented as $\cos \left(q\pi\right)$ and $\sin \left(q\pi\right)$, where q is rational number.

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Answer 3

Other similar results: https://math.stackexchange.com/a/4248080/954936

Examples:

$ 4 \cos \left(\frac{4 \pi }{9}\right)+1=\sqrt{11-2 \sqrt{11+2 \sqrt{11-2 \left(4 \cos \left(\frac{4 \pi }{9}\right)+1\right)}}} $

$ 4 \cos \left(\frac{2 \pi }{9}\right)+1=\sqrt{11+2 \sqrt{11-2 \sqrt{11-2 \left(4 \cos \left(\frac{2 \pi }{9}\right)+1\right)}}} $

$ 4 \cos \left(\frac{\pi }{9}\right)-1=\sqrt{11-2 \sqrt{11-2 \sqrt{11+2 \left(4 \cos \left(\frac{\pi }{9}\right)-1\right)}}} $

$ 4 \cos \left(\frac{\pi }{4}\right)+1=\sqrt{11+2 \sqrt{11-2 \left(4 \cos \left(\frac{\pi }{4}\right)+1\right)}} $

$ 4 \cos \left(\frac{\pi }{4}\right)-1=\sqrt{11-2 \sqrt{11+2 \left(4 \cos \left(\frac{\pi }{4}\right)-1\right)}} $

Answer 4

For nested square roots of n>=3, I have found some new result

$ \sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\left(2 \cos \left(\frac{\pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{8 \pi }{15}\right)\right)}}}}=2 \cos \left(\frac{\pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{8 \pi }{15}\right) $

$ \sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\left(2 \cos \left(\frac{4 \pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{7 \pi }{15}\right)\right)}}}}=2 \cos \left(\frac{4 \pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{7 \pi }{15}\right) $

$ \sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\left(2 \cos \left(\frac{2 \pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{11 \pi }{15}\right)\right)}}}}=2 \cos \left(\frac{2 \pi }{15}\right)+2 \cos \left(\frac{5 \pi }{15}\right)+2 \cos \left(\frac{11 \pi }{15}\right) $

$ \sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\left(2 \cos \left(\frac{\pi }{15}\right)+2 \cos \left(\frac{6 \pi }{15}\right)+2 \cos \left(\frac{7 \pi }{15}\right)\right)}}}}=2 \cos \left(\frac{\pi }{15}\right)+2 \cos \left(\frac{6 \pi }{15}\right)+2 \cos \left(\frac{7 \pi }{15}\right) $

please see math.stackexchange.com/a/4233146/954936