求函数y=sinx-cosx+sinxcosx,0≤sinx-cosx≤√2

设t = sinx - cosx
0≤t≤√2
t² = (sinx - cosx)²
=1 - 2sinxcosx
sinxcosx = (1 - t²) / 2
y = t + (1 - t²) / 2
= -(t + 1)² / 2 + 1
0≤t≤√2
1≤t + 1≤√2 + 1
1≤(t + 1)²≤(√2 + 1)²
-1 / 2≥-(t + 1)² / 2≥-3 / 2 - √2
1 / 2≥y≥-√2 - 1 / 2
-(1 + √2) / 2≤y≤1 / 2
∴值域[(-(1 + √2) / 2,1 / 2]