Using half angle identities

2\sin\dfrac{\theta }{2} = \sqrt{2-\cos\theta }

2\cos\dfrac{\theta }{2} = \sqrt{2+\cos\theta }

and

Supplemental Angle identity for cosine function

\cos(\pi-\theta ) = - \cos\pi

Then,

2\sin \dfrac{\pi }{18} = \sqrt{2-\cos\dfrac{\pi}{9} }

=\sqrt{2-\sqrt{2+\cos\dfrac{2\pi}{9} } }

=\sqrt{2-\sqrt{2+\sqrt{2+\cos\dfrac{4\pi}{9} } } }

=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\cos\dfrac{8\pi}{9} } } } }

=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\cos\dfrac{\pi}{9} } } } }

Therefore, the value of \sin\dfrac{\pi }{18} will be the value of the cyclic nested radical. The signs before the nested radicals will repeat in period of 3 in the pattern [−++].

\sin \dfrac{\pi }{18} =\dfrac{1}{2} \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\dots} } } }