换元,可设t=sinx+cosx=(√2)sin[x+(π/4)]∴-√2≤t≤√2.且t²-1=2sinxcosx.∴2f(x)=(t²-1)/(t+1)=t-1.即f(x)=(t-1)/2.∴-(1+√2)/2≤f(x)≤(√2-1)/2.值域为[-(1+√2)/2,(√2-1)/2]...