y=sinx+cosx+sinxcosx
(sinx+cosx)^2=1+2sinxcosx
sinxcosx=[(sinx+cosx)^1-1]/2
设t=sinx+cosx=√2sin(x+π/4)∈[-√2,√2]
sinxcosx=(t^2-1)/2
y=t+(t^2-1)/2=1/2 t^2+t-1/2
=1/2(t^2+2t+1)-1
=1/2(t+1)^2-1
∵t∈[-√2,√2]
∴ t=-1时,ymin=-1
∴ t=√2时,ymax=√2+1/2
∴函数值域为[-1,√2+1/2]