Trigonometric sum. Calculate value of $ tan^2\frac{\pi}{9}+tan^2\frac{2\pi}{9}+tan^2\frac{ 4\pi}{9}= $ |
Trích:
$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) |
Múi giờ GMT. Hiện tại là 06:10 PM. |
Powered by: vBulletin Copyright ©2000-2024, Jelsoft Enterprises Ltd.