1f(x)=cos²(x/2)-sin(x/2)*cos(x/2)-1/2 =1/2*[2cos²(x/2)-1]-1/2*sinx =1/2*cosx-1/2*sinx =√2/2*(√2/2*cosx-√2/2*sinx) =√2/2*cos(x+π/4)最小正周期为：T=2π/1=2π∵-1≤cos(x+π/4...追问一个问题如果第二个问求的是sin2α的值该如何计算