Solving cyclic infinite nested square roots of 2 as cosine functions

Solution 1:

Solving cyclic infinite nested square roots of 2

Half angle cosine formula offers easy solution to nested square roots of 2 as follows

$$2\cos\frac{\theta}{2} = \sqrt{2+2\cos\theta }$$ and $$2\sin\frac{\theta}{2} = \sqrt{2-2\cos\theta }$$

Substitution of $x$ with $2\cos\theta$ in "infinite nested square roots of 2"

Simplest example is $\sqrt{2-\sqrt{2-...}}$ $--->$ in this infinite nested square roots of 2, $x = \sqrt{2-x}$ can be expressed as $2\cos\theta = \sqrt{2-2cos\theta}$ which can be simplified as $2\cos\theta = 2\sin\frac{\theta}{2} = 2\cos(\frac{\pi}{2}-\frac{\theta}{2})$ Now $\theta$ may be solved as $\frac{3\theta}{2}=\frac{π}{2}$ and $\theta$ = $\pi\over3$ which is 60° Now the solution is obvious $2\cos60° = 1$

Checking for other simple nested square roots of 2 having alternate $'+'$ and $'-'$ signs as follows $$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-...}}}}$$ this can be solved as follows by substituting $2\cos\theta$ for $x$ $2\cos\theta = \sqrt{2+\sqrt{2-2cos\theta}}$ $==>$ $2\cos\theta = \sqrt{2+2\sin\frac{\theta}{2}}$ $2\cos\theta = \sqrt{2+2\cos(\frac{π}{2}-\frac{\theta}{2}})$$==>$ $2\cos\theta = 2\cos(\frac{π}{4}-\frac{\theta}{4})$ Now solving for$\theta$ as follows $\theta = (\frac{π}{4}-\frac{\theta}{4})$ ==> $5\theta \over 4$ = $\frac{\pi}{4}$ and the result is $\theta = \frac{π}{5}$ which is $2\cos36° = \phi = \frac{\sqrt5 +1}{2}$ i.e. golden ratio which is also well known answer

Therefore substitution of $x$ with $2\cos\theta$ opens up a new way to solve infinite nested square roots of 2 which can be cyclic in nature

Exploring further we can solve any combination of cyclic nested square roots of 2 as follows

If we remember some basic rules we can generalise the solving of these cyclic nested square roots of 2. For the sake of simplicity further they are represented as $cin\sqrt2$ (cyclic infinite nested square roots of 2)

$cin\sqrt2[1+1-]$ represents $\sqrt{2+\sqrt{2-...}}$ and $cin\sqrt2[2-2+]$ represents $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+...}}}}$

By this method if we solve

$cin\sqrt2[1-2+]$ will be $2\cos\frac{2^2}{2^3+1}$ which is 2cos80°

$cin\sqrt2[1-3+]$ will be $2\cos\frac{2^3}{2^4+1}\pi$

$cin\sqrt2[1-4+]$ will be $2\cos\frac{2^4}{2^5+1}\pi$

$cin\sqrt2[1-5+]$ will be $2\cos\frac{2^5}{2^6+1}\pi$

Clearly it shows pattern as follows $$cin\sqrt2[1-n+]$$ will be represented in cosine terms as $2\cos(\frac{2^{n}\cdot\pi}{2^{n+1}+1})$

Next level of solution $cin\sqrt2[2-1+]$ will be solved as follows

$$2\cos\theta = \sqrt{2-\sqrt{2-\sqrt{2+2\cos\theta}}}$$ Subsequent steps as follows $2\cos\theta = \sqrt{2-\sqrt{2-2\cos\frac{\theta}{2}}}$ $2\cos\theta = \sqrt{2-2\sin\frac{\theta}{4}}$ $==>$ $2\cos\theta = \sqrt{2-2\cos(\frac{\pi}{2}- \frac{\theta}{4}})$ $==>$ $2\sin(\frac{\pi}{4}- \frac{\theta}{8})$ $==>$ $2\cos(\frac{\pi}{2}- \frac{\pi}{4}+ \frac{\theta}{8})$

Solving further for $\theta$ will result in $\theta = \frac{2\pi}{7}$

$\therefore$ $2\cos\frac{2\pi}{7}$ can be expanded as $cin\sqrt2[2-1+]$

Solving $cin\sqrt2[2-2+]$ as follows

$2\cos\theta = \sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+2\cos\theta}}}}$ $==>$ $\sqrt{2-\sqrt{2-2\cos\frac{\theta}{4}}}$ $==>$ $\sqrt{2-2\sin\frac{\theta}{8}}$ $==>$ $\sqrt{2-2\cos(\frac{\pi}{2}-\frac{\theta}{8})}$ $==>$ $2\sin(\frac{\pi}{4}-\frac{\theta}{16})$ $==>$ $2\cos(\frac{\pi}{2}-\frac{\pi}{4}+\frac{\theta}{16})$

Further solving for $\theta$ will lead to $\theta = \frac{4\pi}{15}$ which is 48°

Solving $cin\sqrt2[2-3+]$ as follows

$2\cos\theta = \sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2\cos\theta}}}}}$ $==>$ $\sqrt{2-\sqrt{2-2\cos\frac{\theta}{8}}}$ $==>$ $\sqrt{2-2\sin\frac{\theta}{16}}$ $==>$ $\sqrt{2-2\cos(\frac{\pi}{2}-\frac{\theta}{16})}$ $==>$ $2\sin(\frac{\pi}{4}-\frac{\theta}{32})$ $==>$ $2\cos(\frac{\pi}{2}-\frac{\pi}{4}+\frac{\theta}{32})$

Further solving for $\theta$ will lead to $\theta = \frac{8\pi}{31}$ which is $\theta = \frac{2^3}{2^5-1}\pi$

$cin\sqrt2[2-4+]$ will be $2\cos\frac{2^4}{2^6-1}\pi$

$cin\sqrt2[2-5+]$ will be $2\cos\frac{2^5}{2^7-1}\pi$

$cin\sqrt2[2-6+]$ will be $2\cos\frac{2^6}{2^8-1}\pi$

We can observe the pattern and it is possible to generalise as follows

$$cin\sqrt2[2-n+] = 2\cos(\frac{2^n}{2^{n+2}-1})\pi$$

Significance

1. No need for solving cubic equation to get the value of $\cos10°$ - straight we can solve from $2\cos80°$(as $cin\sqrt2[1-2+]$ - first few digits can be calculated with 2 to 3 cycles of cyclic nested square roots of 2 (and with the help of half angle cosine formula we can easily calculate $2\cos10°$)
2. Of course as an alternative method to Taylor series expansion this method can be utilised to calculate the cosine values. (Taylor series expansion needs the value of $\pi$ to calculate the value of Trigonometric functions. Here we don't need)
3. Most interestingly Fermat numbers also included in representing the angles like $2\cos\frac{2}{5}\pi$, $2\cos\frac{8}{17}\pi$, $2\cos\frac{128}{257}\pi$, $2\cos\frac{32768}{65537}\pi$... can be represented as cyclic infinite nested square roots of 2 as follows $cin\sqrt2[1-1+]$, $cin\sqrt2[1-3+]$,$cin\sqrt2[1-7+]$,$cin\sqrt2[1-15+]$... Respectively ( Deriving the values with python program will be effortless for the angles like $2\cos\frac{\pi}{65537}$ which otherwise look very much complicated

(The code below required very small fraction of a second to calculate $2\cos\frac{32768}{65537}\pi$)

Python code for solving $2\cos\frac{32768}{65537}\pi$ 4. Conventionally we don't have exact representation of angles like $\cos\frac{\pi}{7}$, $\cos\frac{\pi}{11}$. But intuitively or with some effort these can be derived and represented as cyclic infinite nested square roots of 2.

From above discussion, it is observable that any cyclic nested square roots of 2 can be solved to $2\cos\theta$ where $90° > \theta > 45°$ when the first sign is always negative $"-"$. By this method literally any positive integer angle can be represented as either finite nested radicals(like $2\cos45°$ as $\sqrt2$, $2\cos30°$ as $\sqrt3$ or $2\cos22.5°$ as finite nested radical $\sqrt{2+\sqrt2}$ or cyclic infinite nested square roots of 2 without requiring any imaginary numbers or Taylor series expansion as we saw in deriving $2\cos80°$ or $2\cos\frac{4π}{9}$ as $cin\sqrt2[1-2+]$

Disclaimer: This is a part of my research on an attempt to solve interesting cyclic infinite nested square roots of 2 This is my partial answer for my question

I hope many of the people interested in nested radicals can reimagine the solving of various permutations and combinations of cyclic infinite nested square roots of 2 in terms of cosine angles