已知函数f(x)=(√3cosx-sinx)sin2x/2cosx+1/2.(1)求f（π/3）的值；（2）求函数f（x

cosαsinβ=[sin(α+β)-sin(α-β)]/2

f(x)=(√3cosx-sinx)sin2x/2cosx+1/2
=(√3cosx-sinx)sinx+1/2
=2cos(x+π/4)sinx+1/2
=sin(x+π/4+x)-sin(x+π/4-x)+1/2
=sin(2x+π/4)-(√2-1)/2
后面的你会的.
再问： 我不会化简和单调区间
再答： f(x)=(√3cosx-sinx)sin2x/2cosx+1/2 =(√3cosx-sinx)sinx+1/2 =2cos(x+π/6sinx+1/2 =sin(x+π/6x)-sin(x+π/6-x)+1/2 =sin(2x+π/6) 则（1）f(π/3)=sin(2π/3+π/6)=1/2 (2)T=2π/2=π 单调递减区间为： [kπ+π/6，kπ+7π/6] ,k∈Z