若0＜α＜π，则(1+sinα+cosα)(sinα2−cosα2)(2+2cosα)=．

Question

若0＜α＜π，则((1+sinα+cosα)(sin(α)/(2)−cos(α)/(2)))/(√((2+2cosα)))=______．

Answer 1

∵0＜α＜π，
∴0＜(α)/(2)＜(π)/(2)，
∴cos(α)/(2)＞0，
∴原式=((2sin(α)/(2)cos(α)/(2)+2cos2(α)/(2))(sin(α)/(2)−cos(α)/(2)))/(√(2×2cos2(α)/(2)))
=(2cos(α)/(2)(sin(α)/(2)+cos(α)/(2))(sin(α)/(2)−cos(α)/(2)))/(2cos(α)/(2))
=sin2(α)/(2)-cos2(α)/(2)
=-cosα．
故答案为：-cosα．