根号1-2sin/2cosx/2+根号1+2sinx/2cosx/2 (0<x<90)
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解：原式
=√{[sin(x/2)]^2+[cos(x/2)]^2-2sin(x/2)cos(x/2)}+√{[sin(x/2)]^2+[cos(x/2)]^2+2sin(x/2)cos(x/2)}
=√{[sin(x/2)-cos(x/2)]^2}+√{[sin(x/2)+cos(x/2)]^2}
∵0<x<90°
∴0<x/2<45°
∴sin(x/2)-cos(x/2)<0，sin(x/2)+cos(x/2)>0
∴原式=[cos(x/2)-sin(x/2)]+[sin(x/2)+cos(x/2)]=2cos(x/2)
=√{[sin(x/2)]^2+[cos(x/2)]^2-2sin(x/2)cos(x/2)}+√{[sin(x/2)]^2+[cos(x/2)]^2+2sin(x/2)cos(x/2)}
=√{[sin(x/2)-cos(x/2)]^2}+√{[sin(x/2)+cos(x/2)]^2}
∵0<x<90°
∴0<x/2<45°
∴sin(x/2)-cos(x/2)<0，sin(x/2)+cos(x/2)>0
∴原式=[cos(x/2)-sin(x/2)]+[sin(x/2)+cos(x/2)]=2cos(x/2)
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