Question

prove that

sin (5π/18) - cos (4π/9) = √3 sin(π/9)

Answer 1

you can see your answer

Answer 2

Given,

$L.H.S \to \sin \frac{5\pi}{18} \ - \ \cos \frac{4\pi}{9} \\ \\ = \cos (\frac{\pi}{2} - \frac{5\pi}{18}) - \cos \frac{4\pi}{9} \\ \\ = \cos (\frac{9\pi- 5\pi}{18}) - \cos \frac{4\pi}{9} \\ \\ = \cos (\frac{4\pi}{18}) - \cos \frac{4\pi}{9} \\ \\ = \cos (\frac{2\pi}{9}) - \cos \frac{4\pi}{9} \\ \\ = 2 \ \sin \frac{6\pi}{18}. \sin \frac{2\pi}{18} \\ \\ = 2 \ \sin \frac{\pi}{3}. \sin \frac{\pi}{9} \\ \\ =2 \frac{\sqrt{3}}{2}. \sin \frac{\pi}{9} \\ \\ = \sqrt{3} \ \sin \frac{\pi}{9} \\ \\ = R.H.S. \ \ \textbf{Proved}$