Question

f(x)=2sinx(cosx+sinx)求f(x)最小正周期和最大值

Answer 1

解:f(x)=2(sinx)^2+sin2x=1-cos2x+sin2x=1+√2sin(2x-π/4)
T=2π/2=π
f(x)max=1+√2

Answer 2

f(x)=2^（3/2）sinxsin(x+π/4)=-2^0.5[cos(2x+π/4)-cosπ/4]
T=2π/2=π
fmax=-2^0.5(-1-2^-0.5)=√2+1