√(1-sinx)=√[(sin^2(x/2)-2sin(x/2)cos(x/2)+cos^2(x/2)].
∴√(1-sinx)=√[(sin(x/2)-cos(x/2)]^2=|six(x/2)-cos(x/2)|.
当0≤x/2≤π/4时,即0≤x≤π/2,sin(x/2)≤cos(x/2),
∴原式=|sin(x/2)-cos(x/2)|=-[sin(x/2)-cos(x/2]
=cos(x/2)-sin(x/2).
当π/4＜x/2≤π/2时,即π/2＜x≤π时,xin(x/2)>cos(x/2).
∴原式=sin(x/2)-cos(x/2).
当π