已知函数f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)
问题描述：
已知函数f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)
(1)求函数的单调递增区间.(2)三角形ABC的三个内角A.B.C成等差数列,若A为锐角,f(A)=0,BC=2,求AC的长
答
f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)=-cosx*cosx-sinx*(-cosx)=-(1+cos2x)/2+(1/2)sin2x=(1/2)sin2x-(1/2)cos2x-1/2=(√2/2)*[sin2x*cos(π/4)-cos2x*sin(π/4)]-1/2=(√2/2)sin(2x-π/4)-1/2（1）增区间2kπ-π/...