Let us consider interesting factors about cosine angles
$2\cos(\frac{\pi}{2^3}) = \sqrt{2+\sqrt2}$
$2\cos(\frac{\pi}{2^4}) = \sqrt{2+\sqrt{2+\sqrt2}}$
To generalize
$2\cos(\frac{\pi}{2^n}) = \sqrt{2+\sqrt{2+\sqrt{2+...\text{(n-1) times}}}}$
Let us imagine the angles $p \over q$ radians where $1 \over 4$ < $p \over q$ < $1 \over 2$ which satisfy 2^n as denominator and numerator as odd number p satisfying $2^{(n-2)} < p < 2^{(n-1)}$
Let us see the simplest example $2\cos(\frac{3\pi}{2^3}) = \sqrt{2-\sqrt2}$ where $3 = 2^2-1$
$2\cos(\frac{5\pi}{2^4}) = \sqrt{2-\sqrt{2-\sqrt2}}$ where $5 = 2^3-(2^2-1)$ ---> let us represent as $n\sqrt2(2-)$
$2\cos(\frac{7\pi}{2^4}) = \sqrt{2-\sqrt{2+\sqrt2}}$ where $7 = 2^3-1$ ---> let us represent simply as $n\sqrt2(1-1+)$
We have $2^n$ number of odd numbers between $2^{n+1}$ to $2^{n+2}$
Let us explore little further
We have $4$ odd numbers from $8$ as $9, 11, 13, 15$ to $16$ where we can represent cosine angles as nested square roots of 2 as follows
$2\cos(\frac{9\pi}{2^4})$ = $n\sqrt2(1-1-1+)$ or $n\sqrt2(2-1+)$ where $9 = 2^4-(2^3-1)$
$2\cos(\frac{11\pi}{2^4})$ = $n\sqrt2(1-1-1-)$ or $n\sqrt2(3-)$ where $11 = 2^4-(2^3-(2^2-1))$
$2\cos(\frac{13\pi}{2^4})$ = $n\sqrt2(1-1+1-)$ where $13 = 2^4-(2^2-1)$
$2\cos(\frac{15\pi}{2^4})$ = $n\sqrt2(1-1+1+)$ or $n\sqrt2(1-2+)$ where $15 = 2^4-1)$
Here are the observations
- All odd numbers can be represented as sum (addition and subtraction) of $2^n$ in descending order.
- The total number of signs ($+$ and $-$) is $a$ where $2^a < p < 2^{(a+1)}$ and total number of $2$s inside the nested radical is $(a+1)$
- The missing power of 2 is $+$s and others are $-$s
What if the odd number is 57 or 149 or big numbers?
The wonder is we can make the odd numbers as sums of 2^n in decreasing order
For example if odd number is 57 then the cosine angle is $57 \over 2^7$ radians which can be derived as finite nested square roots of 2 as follows
$57 = 2^6-7$ and $7 = 2^3-1$ and $\therefore 57 = 2^6-(2^3-1)$
Now $2\cos(\frac{57\pi}{2^7}) = \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt2}}}}}$ or simply as $n\sqrt2(1-2+1-1+)$
$149 = 2^8-107$ and $107 = 2^7-21$ and $21 = 2^5-11$ and $11 = 2^4-5$ and $5 = 2^3-3$ and $3 = 2^2-1$
$\therefore 149 = 2^8-(2^7-(2^5-(2^4-(2^3-(2^2-1)))))$
Now $2\cos(\frac{149\pi}{2^9}) = \sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt2}}}}}}}$ or simply as $n\sqrt2(2-1+4-)$
One more easier method that I tried as follows
E.g
$2\cos(\frac{3\pi}{2^3})$ = $\sqrt{2+2\cos(\frac{3\pi}{2^2}})$ = $\sqrt{2-2\cos(\frac{\pi}{2^2}})$ = $\sqrt{2-\sqrt2}$
$2\cos(\frac{5\pi}{2^4}) = \sqrt{2+2\cos(\frac{5\pi}{2^3}}) = \sqrt{2-2\cos(\frac{3\pi}{2^3}}) = \sqrt{2-\sqrt{2+2\cos(\frac{3\pi}{2^2}}}) = \sqrt{2-\sqrt{2-2\cos(\frac{\pi}{2^2}}}) = \sqrt{2-\sqrt{2-\sqrt2}}$
$\therefore$ for any odd number we can derive the finite nested square roots of 2 easily
As it is too lengthy I'll restrict with 1 big number example
For $2\cos(\frac{57\pi}{2^7}) = \sqrt{2+2\cos(\frac{57\pi}{2^6}}) = \sqrt{2-2\cos(\frac{7\pi}{2^6}}) = \sqrt{2-\sqrt{2+2\cos(\frac{7\pi}{2^5}}}) = \sqrt{2-\sqrt{2+\sqrt{2+2\cos(\frac{7\pi}{2^4}}}}) = \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2\cos(\frac{7\pi}{2^3}}}}}) = \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-2\cos(\frac{\pi}{2^3}}}}}) = \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+2\cos(\frac{\pi}{2^2}}}}}}) = \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt2}}}}}$
Conclusion : For a given odd number $'p'$ in numerator of angle with corresponding denominator $'q'$ as $2^n$ in the angle as radians, satisfying $1 \over 4$ < $p \over q$ < $1 \over 2$, the cosine angle can be represented as finite nested square roots of 2
These steps I have derived myself and verified the results. Is there any more simpler method to derive the finite nested square roots of 2?
