a=(2cosx,sinx),b=(sin(x+3分之π）,cosx-根号下3sinx),f(x)=a×b.求函数f（x）的最小正周期,求f(x)得值

想成整理后得f(x)=sin2x+√3cos2x=2sin(2x+3/π),周期为π
单调增区间,2x+3/π∈[-π/2+2kπ,π/2+2kπ],x∈[-5π/12+kπ,π/12+kπ]
单调减区间,2x+3/π∈[π/2+2kπ,3π/2+2kπ],x∈[π/12+kπ,7π/12+kπ]