Lemma: For all real numbers $x $, we have $\sin(\frac{\pi}{3} - x) \cdot \sin(\frac{\pi}{3} + x) \cdot \sin(x) = \frac{\sin(3x)}{4} $. Apply the Lemma, let $x = \frac{\pi}{9} $, we have: $\sin(\frac{\pi}{9}) \cdot \sin(\frac{2\pi}{9}) \cdot \sin(\frac{4\pi}{9}) = \frac{\sin(\frac{\pi}{3})}{4} = \frac{\sqrt{3}}{8} $. [RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT] |