解设f(x)=tan(sinx)- sin(tan x),则 f(x)=sec(sin x)cos x-cos(tan x )secx 2 cosx-cos(tan x)cos"(sin x) cos-(sIn x)cOS-x 当0<x< arctan-时,0<tanx<,0<simx<x 由余弦函数在(0,-)上的凸性有 anx+sinx cos(tan x)cos(sin x)scos(tan x)+ cos(sin x)]scos 3 &o(x)=tan x+2sin x-3x, (x)=secx+2cos x-3= tanx-4sin-->0 于是tnx+2smx>3x所以 cos tanx+2smx<cosx即 I cos( tan x)cos2(smx)<cos3x 于是当x∈(0 arctan,)时,f(x)>0.又f()=0,所以f(x)>0 当x∈ arctan)时, sin(arctan)<smx<1.由于 tan (arctan -) sin(arctan-) 1+tan(arctan =) /4+n24 故<sinx<1.于是1< tan(sin x)<tanl 当x∈ arctan,二)时,f(x)>0 综上可得,当x∈(0,)时,tan(sinx)>sin(tanx)) , tan(sin ) sin(tan ). 2 π , 0, ) , ( ) 0. 2 π , 2 π [arctan sin 1. 1 tan(sin ) tan1. 4 π , 4 π 4 π π 4 π 1 2 π ) 2 π 1 tan (arctan ) 2 π tan(arctan ) 2 π sin(arctan ) sin 1. 2 π ) , sin(arctan 2 π , 2 π [arctan ) ( ) 0, ( 0) 0, ( ) 0. 2 π ( 0, arctan cos , cos(tan ) cos (sin ) cos . 3 tan 2sin tan 2sin 3 , cos 0. 2 ( ) tan 2sin 3 ( ) sec 2cos 3 tan 4sin . 3 tan 2sin [cos(tan ) 2cos(sin )] cos 3 1 cos(tan ) cos (sin ) 2 π 0 . 2 π , 0 sin 2 π 0 tan 2 π 0 arctan . cos (sin ) cos cos cos(tan ) cos (sin ) ( ) sec (sin ) cos cos(tan ) sec 解 设 ( ) tan(sin ) sin(tan ), 则 2 2 2 2 3 2 2 2 3 2 2 2 3 2 2 2 x x x x f x x x x x x f x f f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x f x x x x x f x x x Œ > \ Œ > < < < < > + = + = + = Œ < < Œ ¢ > = > < < + + > = + - ¢ = + - = - > + £ + £ < < < < < < - ¢ = - = = - 综上可得 当 ( 时 当 时 故 于是 当 时 由于 于是当 时, 又 所以 于是 所以 即 设 , 由余弦函数在( , )上的凸性有 当 时, j j