I recently asked a question about how to find value of $\sin(\pi/18)$ and I understand that there is no expression for $\sin(\pi/18)$ that uses the ordinary arithmetical operations. but I know that we have exact value by nested radicals. I want to know how we get this value ?
2 Answers
$\begingroup$Theorem 1: For $a_1=\sqrt a$, $a_2=\sqrt{a-\sqrt{a}}, a_3=\sqrt{a-\sqrt{a+\sqrt a}},\ldots,$ We have $a_n$ as$$\lim_{n\rightarrow\infty}a_n=\frac {A-1}6+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt3}\right)\tag1$$where $A=\sqrt{4a-7}$.
Proof: I present to you the proof for the theorem.
Start with $$x^2=y+a,\qquad y^2=z+a\qquad z^2=x+a\tag{2.1}$$
Then we can factor the resulting $8$ degree polynomial into factors $$x^3+\frac 12x^2\left\{1+\sqrt{4a-7}\right\}-\frac 12x\left\{2a+1-\sqrt{4a-7}\right\}+\frac 12\left\{a-2-a\sqrt{4a-7}\right\}\tag{2.2}$$ and $$x^3+\frac 12x^2\left\{1-\sqrt{4a-7}\right\}-\frac 12x\left\{2a+1+\sqrt{4a-7}\right\}+\frac 12\left\{a-2+a\sqrt{4a-7}\right\}\tag{2.3}$$. For brevity, we can let $A=\sqrt{4a-7}$ and use the identity$$\sin^3\theta-\frac 34\sin\theta+\frac 14\sin3\theta=0$$
We first consider $(2.2)$. Substituting $x=s-\frac {A+1}{6}$ and recalling that $A^2=4a-7$, we find that$$s^3+\left(\frac {A-4a}3\right)s+\frac {12a-14-A-8Aa}{27}=0\tag{2.4}$$
And similarly, setting $x=s-\frac {1-A}6$ in $(2.3)$, we get$$s^3+\left(\frac {-A-4a}3\right)s+\frac {12a-14+A+8Aa}{27}=0\tag{2.5}$$ Which we notice is the same as $(2.4)$, but with $A$ replaced with $-A$. Next, setting $s=\frac 23t\sqrt{4a-A}$ and $s=\frac 23t\sqrt{4a+A}$ in $(2.4)$ and $(2.5)$ respectively, we deduce$$\begin{align*} & t^3-\frac 34t+\frac {12a-14-A-8Aa}{8(4a-A)^{3/2}}=0\\ & t^3-\frac 34t+\frac {12a-14+A+8Aa}{8(4a+A)^{3/2}}=0\end{align*}\tag{2.6}$$ Setting $t=\sin\theta$ and using the identity, we get the solutions to $(2.6)$. Simplifying and more simplifying, we get Theorem 1.
Simply plugging in $a=2$ gives you $2\sin\frac {\pi}{18}$ and dividing both sides by two, you get your first equation.$$\frac 12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\ldots}}}}=\sin\frac {\pi}{18}\tag3$$
$\endgroup$ 8 $\begingroup$Deriving $\sin\frac{\pi}{18}$ will be simpler when we implement Half angle cosine formula$$2\sin\frac{\theta}{2} = \sqrt{2-2\cos\theta}$$$$2\cos\frac{\theta}{2} = \sqrt{2+2\cos\theta}$$
and cosine angle identity
$$2\cos(\pi-\theta) = -2\cos\theta$$
We can derive as follows$2\sin(\frac{\pi}{18})$ = $\sqrt{2-2\cos(\frac{\pi}{9})}$ = $\sqrt{2-\sqrt{2+2\cos(\frac{2\pi}{9}})}$ = $\sqrt{2-\sqrt{2+\sqrt{2+2\cos(\frac{4\pi}{9}}})}$ = $\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+2\cos(\frac{8\pi}{9}}}})}$ = $\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-2\cos(\frac{\pi}{9}}}})}$
Now you can observe $2\cos(\frac{\pi}{9})$ repeats in cycle infinitely
$$\therefore 2\sin\frac{\pi}{18} = \sqrt{2-\sqrt{2+\sqrt{2+...}}}$$ $[-++]$ is the repeating pattern in infinite nested radical. Refer here. Welcome to the world of infinite nested square roots of 2 as solution to cosine angles!
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