f(x)=a·b+1
=(2sinx,2sinx)·(cosx,-sinx)+1
=2sinxcosx-2sinx^2+1
=sin(2x)-(1-cos(2x))+1
=sin(2x)+cos(2x)
=√2sin(2x+π/4)
1
f(x)=1/2,即：
√2sin(2x+π/4)=1/2
即：sin(2x+π/4)=√2/4
-sin(4x)=cos(4x+π/2)
=1-2sin(sin(2x+π/4))^2
=1-2(√2/4)^2
=3/4
即：sin(4x)=-3/4
2
x∈(0,π)
2x+π/4∈(π/4,9π/4)
sin(2x+π/4)∈[-1,1]
故：f(x)∈[-√2,√2]