化简：（1+sinx+cosx)(sin(x/2-)-cos(x/2）)/根号（2+2cosx),条件（1＜x＜π 求详解

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1+cosx=2cos²(x/2)
(1+sinx+cosx)(sin(x/2)-cos(x/2）)/√(2+2cosx)
=[2cos²(x/2)+2sin(x/2)cos(x/2)][sin(x/2)-cos(x/2)]/√[2(1+cosx)]
=2cos(x/2) [cos(x/2)+sin(x/2)][sin(x/2)-cos(x/2)]/√[4cos²(x/2)]
=2cos(x/2) [sin²(x/2)-cos²(x/2)]/[2|cos(x/2)|] ( ∵ 1＜x＜π ∴1/斗毕2<x/2<散型π/2)
=cos(x/冲销猜2)(-cosx)/cos(x/2) ( ∵cos(x/2)>0 )
=-cosx

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