解：√(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)|.
當0≤x/2≤π/4時，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).
當π/4＜x/2≤π/2時，即π/2＜x≤π時，xin(x/2)>cos(x/2).
∴原式=sin(x/2)-cos(x/2).
當π<x/2≤5π/4時，即2π<x≤5π/2時，sin(x/2)<cos(x/2).
∴原式=cos(x/2)-sin(x/2).
..
以下按此週期性變化着。
√(1+cosx)=√[2cos^2(x/2)]=√2|cos(x/2)|.
當cos(x/2)≥0時，原式=√2cos(x/2).
當cos(x/2)<0時，原式=-√2cos(x/2)
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