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man111 29-12-2010 04:56 PM

Trigonometric product
 
Calculate value of $sin(\frac{\pi}{9})\times sin(\frac{2\pi}{9})\times sin(\frac{4\pi}{9})= $

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} $.

Persian 29-12-2010 05:45 PM

Trích:

Nguyên văn bởi man111 (Post 76107)
Calculate value of $sin(\frac{\pi}{9})\times sin(\frac{2\pi}{9})\times sin(\frac{4\pi}{9})= $


$Put:A = sin\frac{\pi }{9}.sin\frac{{2\pi }}{9}.sin\frac{{4\pi }}{9}\\
4A = 2[c{\rm{os}}\frac{{ - \pi }}{3} - c{\rm{os}}\frac{{5\pi }}{9}]\sin \frac{{2\pi }}{9} = \sin \frac{{2\pi }}{9} - 2\sin \frac{{2\pi }}{9}c{\rm{os}}\frac{{5\pi }}{9}\\ $
$= \sin \frac{{2\pi }}{9} - \sin \frac{{7\pi }}{9} + \frac{{\sqrt 3 }}{2} = \sin \frac{{2\pi }}{9} - \sin \frac{{2\pi }}{9} + \frac{{\sqrt 3 }}{2} = \frac{{\sqrt 3 }}{2}\\ $
$\Rightarrow A = \frac{{\sqrt 3 }}{8} $


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