Sines are sinister

$z^{18}=1$ has roots $z = 1,e^\dfrac{\pi i}{9},e^\dfrac{2\pi i}{9},e^\dfrac{3\pi i}{9},. . . . . . e^\dfrac{16\pi i}{9},e^\dfrac{17\pi i}{9},$ then $z^{18}-1=(z-1)(z-e^\dfrac{\pi i}{9})(z-e^\dfrac{2\pi i}{9})(z-e^\dfrac{3\pi i}{9}),. . . . . . (z-e^\dfrac{16\pi i}{9})(z-e^\dfrac{17\pi i}{9})$ Now dividing L.H.S. and R.H.S. by (z-1), and taking limits at z = 1, $18=(1-e^\dfrac{\pi i}{9})(1-e^\dfrac{2\pi i}{9})(1-e^\dfrac{3\pi i}{9}),. . . . . . (1-e^\dfrac{16\pi i}{9})(1-e^\dfrac{17\pi i}{9}).$ Multiplying by its conjugate gives, $18^2=2^{17}(1-cos\dfrac{\pi}{9})(1-cos\dfrac{2\pi}{9})(1-cos\dfrac{3\pi}{9}),. . . . . . (1-cos\dfrac{16\pi}{9})(1-cos\dfrac{17\pi}{9}).$ Now putting $1-cos2\theta=2sin^2\theta$ everywhere thus halving angles $18^2 = 2^{17}(2sin^2\dfrac{\pi}{18})(2sin^2\dfrac{2\pi}{18})(2sin^2\dfrac{3\pi}{18}),. . . . . . (2sin^2\dfrac{16\pi}{18})(2sin^2\dfrac{17\pi}{18}).$ Taking square-root $18= 2^{17}(sin\dfrac{\pi}{18})(sin\dfrac{2\pi}{18})(sin\dfrac{3\pi}{18}),. . . . . . (sin\dfrac{16\pi}{18})(sin\dfrac{17\pi}{18}).$ Now use the fact that first and last factors are equal as $sin\theta = sin(\pi-\theta)$ and as $sin\dfrac{9\pi}{18}=1$ , taking square-root once again we get the desired answer as $\sqrt(\dfrac{9}{2^{16}})= \dfrac{3}{256}.$