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青青子衿
Posted 2014-1-4 09:46
$\tan x-\cos x=\frac{\sin x-\cos^2x}{\cos x}=\frac{\sin^2x+\sin x-1}{\cos x}.$
$x∈(0，\fracπ2)∪(\fracπ2,π),0<\sin x<1,
f(x)=x^2+x-1在(-\frac12,+∞)上为增函数,且有f(\frac{-1+√5}{2})=0$
∴当$\sin x=\frac{-1+√5}{2}$,令$arcsin(\frac{-1+√5}{2})=α，$
即$x=α$或者$π-α$时$\sin^2x+\sin x-1=0,\tan x=\cos x$
$0<\sin x<\frac{-1+√5}{2}$时,即$0<x<α$或$π-α<x<π$,$\sin^2x+\sin x-1<0,$
Ⅰ)$0<x<α$时,$\cos x>0,\tan x-\cos x<0，\tan x<\cos x$
Ⅱ)$π-α<x<π$时,$\cos x<0,\tan x-\cos x>0，\tan x>\cos x,$
$\frac{-1+√5}{2}<\sin x<1,α<x<\fracπ2,π/2< x<π-α ,\sin^2x+\sin x-1>0$
Ⅲ)$α<x<\fracπ2,\cos x>0,\tan x-\cos x>0,\tan x>\cos x.$
Ⅳ)$\fracπ2< x<π-α,\cos x<0,\tan x-\cos x<0，\tan x<\cos x$
由Ⅰ,Ⅱ,Ⅲ,Ⅳ有
$0<x<α或\fracπ2< x<π-α$时,$\tan x<\cos x$
$α<x<\fracπ2或π-α<x<π$时,$\tan x>\cos x$ |
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kuing
Posted 2014-1-3 00:16
