1、f(x)=√2sin2x+√2cos2x
=2(cosπ/4sin2x+sinπ/4cos2x)
=2sin(2x+π/4)
所以:f(3π/8)=2sin(2X3π/8+π/4)=2sinπ=0
2、因:f(x)=2sin(2x+π/4)
易知,最大值为2,最小正周期为:T=2π/2=π
1、f(x)=√2sin2x+√2cos2x
=2(cosπ/4sin2x+sinπ/4cos2x)
=2sin(2x+π/4)
所以:f(3π/8)=2sin(2X3π/8+π/4)=2sinπ=0
2、因:f(x)=2sin(2x+π/4)
易知,最大值为2,最小正周期为:T=2π/2=π