1.ab=sinwxcoswx+√3cos²wx=1/2sin2wx+√3/2cos2wx+√3/2
f(x)=ab-√3/2=1/2sin2wx+√3/2cos2wx=sin(2wx+π/3)
T=π,所以2π/2w=π,所以w=1,所以f(x)=sin(2x+π/3),
所以单调递增区间为[kπ-5π/12,kπ+π/12]
2.cosA=(b^2+c^2-a^2)/2bc=√3/2A=π/6
f(A)=f(π/6)=sin(2*π/6+π/3)=sin(2π/3)=√3/2