y=sinxcosx-cos2x
=1/2*sin2x-cos2x
=√[(1/2)^2+1^2]sin(x-z)
=[(√5)/2]sin(x-z)
其中tanz=1/(1/2)=2
-1<=sin(x-z)<=1
所以值域[-(√5)/2,(√5)/2]