化简:根号(1-sinx)+根号(1+sinx),x属于(0,π)

x属于(0,π) ,则x/2属于(0,π/2),则sin(x/2)＞0,cos(x/2)＞0
根号(1-sinx)+根号(1+sinx),
=根号[sin²(x/2)+cos²(x/2)-2sin(x/2)cos(x/2)]+根号[sin²(x/2)+cos²(x/2)+2sin(x/2)cos(x/2)]
=根号[sin(x/2)-cos(x/2)]²+根号[sin(x/2)+cos(x/2)]²
=|sin(x/2)-cos(x/2)|+sin(x/2)+cos(x/2)
当x属于(0,π/2)时,x/2属于(0,π/4),则sin(x/2)＜cos(x/2),
所以|sin(x/2)-cos(x/2)|+sin(x/2)+cos(x/2)=cos(x/2)-sin(x/2)+sin(x/2)+cos(x/2)=2cos(x/2)
当x属于[π/2,π)时,x/2属于(π/4,π/2),则sin(x/2)＞cos(x/2),
所以|sin(x/2)-cos(x/2)|+sin(x/2)+cos(x/2)=sin(x/2)-cos(x/2)+sin(x/2)+cos(x/2)=2sin(x/2)
所以
当x属于(0,π/2)时,原式=2cos(x/2)
当x属于[π/2,π)时,原式=2sin(x/2)