Value of $\sin(\pi/18)$ by nested radicals 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 ?
http://mathworld.wolfram.com/TrigonometryAnglesPi18.html

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!
