y=sinx/2-cosx
=sin(x/2)-[1-sin²(x/2)]/2
=sin(x/2)-1/2+sin²(x/2)/2
=1/2[sin²(x/2)+2sin(x/2) +1]-1
=1/2[sin(x/2) +1]²-1
由-1≤sin(x/2) ≤1得
sin(x/2)=-1时,取得最小值-1
sin(x/2)=1时,取得最大值1
所以值域为[-1,1]
y=sinx/2-cosx
=sin(x/2)-[1-sin²(x/2)]/2
=sin(x/2)-1/2+sin²(x/2)/2
=1/2[sin²(x/2)+2sin(x/2) +1]-1
=1/2[sin(x/2) +1]²-1
由-1≤sin(x/2) ≤1得
sin(x/2)=-1时,取得最小值-1
sin(x/2)=1时,取得最大值1
所以值域为[-1,1]