F(X)=(sinx+cosx)'2+cos2x.求F(X)最小周期和值域.快阿

F(X)=(sinx+cosx)^2+cos2x=1+sin2x+cos2x=(√2)sin(2x+π/4)+1
所以,最小正周期是2π/2=π;
由于-1≤sin(2x+π/4)≤1,所以1-(√2)≤F(x)≤1+(√2),
所以,值域是[1-(√2),1+(√2)]