f(x)=cosx+sin^2x最小正周期

题目

f(x)=cosx+sin^2x最小正周期

答案

f(x)=cosx+1-cos²x
=-(cos²x-cosx+1/4)+1/4+1
=-(cosx-1/2)²+5/4
所以T=2π/1=2π