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avip · 29-12-2010, 05:23 PM

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} $.

29-12-2010, 05:23 PM	#2
avip +Thành Viên+ Tham gia ngày: Sep 2010 Bài gởi: 392 Thanks: 135 Thanked 247 Times in 159 Posts	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} $.