y=(sinx+1)(cosx+1)
=sinxcosx+sinx+cosx+1
=(sinx+cosx)²/2+sinx+cosx+1/2
令t=sinx+cosx∈[-√2,√2]
则y=t²/2+t+1/2
=(t+1)²/2
所以最大值是t=√2时ymax=(√2+1)²/2
最小值是t=-1时ymin=0
y=(sinx+1)(cosx+1)
=sinxcosx+sinx+cosx+1
=(sinx+cosx)²/2+sinx+cosx+1/2
令t=sinx+cosx∈[-√2,√2]
则y=t²/2+t+1/2
=(t+1)²/2
所以最大值是t=√2时ymax=(√2+1)²/2
最小值是t=-1时ymin=0