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高考数学必做百题第32题(理科2017版)

  2016-10-02 17:50:34  

032. 已知

$f\left( x \right)=\left( 1+\dfrac{1}{\tan x} \right){{\sin }^{2}}x-2\sin \left( x+\dfrac{\pi }{4} \right)\sin \left( x-\dfrac{\pi }{4} \right)$。

(1)若$\tan \alpha =2$,求$f\left( \alpha  \right)$的值;

(2)若$x\in \left[ \dfrac{\pi }{12},\dfrac{\pi }{2} \right]$,求$f\left( x \right)$的单调区间。

解:(1)

$f\left( x \right)=\left( 1+\dfrac{1}{\tan x} \right){{\sin }^{2}}x-2\sin \left( x+\dfrac{\pi }{4} \right)\sin \left( x-\dfrac{\pi }{4} \right)$

$=\left( {{\sin }^{2}}x+\sin x\cos x \right)-2\sin \left( x+\dfrac{\pi }{4} \right)\cos \left( x+\dfrac{\pi }{4} \right)$

$=\dfrac{1-\cos 2x}{2}+\dfrac{1}{2}\sin 2x+\sin \left( 2x+\dfrac{\pi }{2} \right)$

$=\dfrac{1}{2}+\dfrac{1}{2}\left( \sin 2x-\cos 2x \right)+\cos 2x$

$=\dfrac{1}{2}\left( \sin 2x+\cos 2x \right)+\dfrac{1}{2}$。

∵$\tan \alpha =2$,

∴$\sin 2\alpha =\dfrac{2\sin \alpha \cos \alpha }{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }=\dfrac{2\tan \alpha }{{{\tan }^{2}}\alpha +1}=\dfrac{4}{5}$,

$\cos 2\alpha =\dfrac{{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha }{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }=\dfrac{1-{{\tan }^{2}}\alpha }{{{\tan }^{2}}\alpha +1}=-\dfrac{3}{5}$,

∴$f\left( \alpha  \right)=\dfrac{1}{2}\left( \sin 2\alpha +\cos 2\alpha  \right)+\dfrac{1}{2}=\dfrac{3}{5}$。

(2)∵$f\left( x \right)=\dfrac{1}{2}\left( \sin 2x+\cos 2x \right)+\dfrac{1}{2}$

$=\dfrac{1}{2}\sin \left( 2x+\dfrac{\pi }{4} \right)+\dfrac{1}{2}$。

∵$x\in \left[ \dfrac{\pi }{12},\dfrac{\pi }{2} \right]$,∴$2x+\dfrac{\pi }{4}\in \left[ \dfrac{5\pi }{12},\dfrac{5\pi }{4} \right]$,

①当$2x+\dfrac{\pi }{4}\in \left[ \dfrac{5\pi }{12},\dfrac{\pi }{2} \right]$,即$x\in \left[ \dfrac{\pi }{12},\dfrac{\pi }{8} \right]$时,$f\left( x \right)$单调递增;

②当$2x+\dfrac{\pi }{4}\in \left[ \dfrac{\pi }{2},\dfrac{5\pi }{4} \right]$,即$x\in \left[ \dfrac{\pi }{8},\dfrac{\pi }{2} \right]$时,$f\left( x \right)$单调递减。

∴$f\left( x \right)$的单调递减区间是$\left[ \dfrac{\pi }{8},\dfrac{\pi }{2} \right]$,单调递增区间是$\left[ \dfrac{\pi }{12},\dfrac{\pi }{8} \right]$。



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