[2cos（x/2）^2-sinx-1]/[√2sin（π/4+X）]
=[cosx+1-sinx-1]/[√2sin（π/4+X）]
=[cosx-sinx]/[√2sin（π/4+X）]
=√2[√2/2cosx-√2/2sinx]/[√2sin（π/4+X）]
=√2[cosπ/4cosx-sinπ/4sinx]/[√2sin（π/4+X）]
=√2[cos(π/4+x)]/[√2sin（π/4+X）]
=cos(π/4+x)/sin（π/4+X)
=cot(π/4+x)
化简（2cos（x/2）^2-sinx-1）/（根号2sin（π/4+X））
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