设0
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设0设0设0a-b=2-xsinx-cos^2x=1-xsinx+sin^2x+cos^2x-cos^2x=1-xsinx+sinx^2=1-sinx(x-sinx)首先x>sinx(0b理论上上面的
设0
设0
设0
a-b=2-xsinx-cos^2x
=1-xsinx+sin^2x+cos^2x-cos^2x
=1-xsinx+sinx^2
=1-sinx(x-sinx)
首先x>sinx(0b
理论上上面的证明也适用与0
a-b=2-xsinx-cos^2x
=1-xsinx+sin^2x+cos^2x-cos^2x
=1-xsinx+sinx^2
=1+x^2/4-xsinx+sin^2x-x^2/4
=1-x^2/4+(x/2-sinx)^2
1-x/2对于0
0<1-π/4=1-x/2<1
(1-x/2)(1+x/2)>0
(x/2-sinx)^2>=0
所以，a-b>0
即a>b