若f(x)=(cosx)^2-(sinx)^2+√3sin2x+1 求f(x)的最大值,最小值及周期
f(x)=cos²x-sin²x+√3sin(2x)+1
=cos(2x)+√3sin(2x)+1
=2[sin(2x)(√3/2)+cos(2x)(1/2)]+1
=2sin(2x+π/6) +1
當(dāng)sin(2x+π/6)=1時,f(x)有最大值[f(x)]max=3
當(dāng)sin(2x+π/6)=-1時,f(x)有最小值[f(x)]min=-1
最小正周期T=2π/2=π
周期為kπ (k∈Z).