f(x)=(2cos²x-1)sin2x+1/2cos4x=cos2xsin2x+1/2cos4x=1/2sin4x+1/2cos4x=√2/2*(sin4x*√2/2+√2/2cos4x)=√2/2sin(4x+π/4)于是最小正周期T=2π/4=π/2因为-1≤sin(4x+π/4）≤1所以最大值为：√2/2...