求y=1+sinX+cosX+1/2sin2X的值域,

1/2sin2X+sinX+cosX+1=sinxcosx+sinX+cosX+1=1/2(2+2sinxcosx)+sinX+cosX=1/2(1+1+2sinxcosx)+sinX+cosX=1/2(sinx^2+2sinxcosx+cos^2+1)+sinX+cosX=1/2((sinx+cosx)^2+1)+sinX+cosX=
1/2(sinx+cosx)^2+(sinX+cosX)+1/2
令t=sinX+cosX
y=1/2(t^2+2t+1)=1/2(t+1)^2 由于t=sinX+cosX=根号2（sin(x+45度)）所以t的值域为负根号2到根号2之间 当t=-1y最小为0 t=根号2时y最大为3/2+根号2 所以值域为0—3/2+根号2

y=sinx+cosx+1/2(2sinxcosx)+1
令t=sinx+cosx= √2sin(x+π/4)∈[-√2,√2]
∵(sinx+cosx)^2=sin2x+sinx^2+cos^2=sin2x+1
∴1/2sin2x=1/2(t^2-1)
则y=t+1/2(t^2-1)+1=1/2t^2+t+1/2,y取得最小值0时t=-1
又∵t∈[-√2,√2]
∴y∈[0,3/2+√2]

y=1+sinX+cosX+1/2sin2X=1/2+sinX+cosX+1/2(sinX+cosX)²=1/2(sinX+cosX+1)²
=1/2[√2sin(x+π/4)+1]²,∵-1≤sin(x+π/4)≤1,当sin(x+π/4)=1时，y=1+sinX+cosX+1/2sin2X有最大值
(3+2√2)/2，当sin(x+π/4)=-√2/2，y=1+sinX+cosX+1/2sin2X有最小值0，∴值域[0,(3+2√2)/2]