f(x)=sin(派-ax)cosax+cos^2(ax),a>0最小正周期是派,求a_数学解答

f(x)=sin(派-ax)cosax+cos^2(ax),a>0最小正周期是派,求a

人气：444 ℃　时间：2020-05-23 23:30:25

解答

f(x)=sin(派-ax)cosax+cos^2(ax)=sinaxcosax+(cos2ax+1)/2=1/2sin2ax+(cos2ax+1)/2
所以2π/2a=π,a=1