x∈[π/4,π/3],f(x)=1/2sin(x-5π/12)cos(5π/12-x)+[(根号3)/2][sin(x+4)]^2

I suppose the last terms is [sin(x+π/4)]^2 f(x)=1/2sin(x-5π/12)cos(5π/12-x)+[√3/2][sin(x+π/4)]^2 = (-1/4)(2sin(5π/12-x)cos(5π/12-x) +[√3/2][sin(x+π/4)]^2 =(-1/4)(sin2(5π/12-x))+[√3/2][sin(x...