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Nguyên văn bởi man111  Calculate value of $ tan^2\frac{\pi}{9}+tan^2\frac{2\pi}{9}+tan^2\frac{ 4\pi}{9}= $ |
$\frac{\pi}{9}, \frac{2\pi}{9},\frac{4\pi}{9} $are roots of equation tan9x=0
$tan9x=0\Leftrightarrow 9x=k\pi\Leftrightarrow 6x=-3x+k\pi\Leftrightarrow tan6x=-tan3x\Leftrightarrow \frac{2tan3x}{1-tan^{2}3x}=-tan3x $
$\Rightarrow tan3x=0 or tan^{2}3x=3 $
Easy to see that $\frac{\pi}{9}, \frac{2\pi}{9},\frac{4\pi}{9} $aren't roots of equation tan3x=0, so they are roots of equation $tan^{2}3x=3 $ (1)
We have $tan3x=\frac{3tanx-tan^3x}{1-3tan^2x} $
So $(1) \Leftrightarrow tan^6x-33tan^4x+27tan^2x-3\Rightarrow t^3-33t^2+27t-3=0(tan^2x=t)(2) $
$Because \frac{\pi}{9}, \frac{2\pi}{9},\frac{4\pi}{9} $are roots of equation $tan^{2}3x=3,tan^2\frac{\pi}{9},tan^2\frac{2\pi}{9} ,tan^2\frac{4\pi}{9} $are roots of equation(2)
=> $ tan^2\frac{\pi}{9}+tan^2\frac{2\pi}{9}+tan^2\frac{ 4\pi}{9}=33 $(Viet)
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]