1个回答
展开全部
因为
f(x)=cos(xπ/4-π/3)-cos(xπ/4)
=[cos(xπ/4)cos(π/3)+sin(xπ/4)sin(π/3)]-cos(xπ/4)
=[1/2cos(xπ/4)+√3/2sin(xπ/4)]-cos(xπ/4)
=-1/2cos(xπ/4)+√3/2sin(xπ/4)
=cos(xπ/4+2π/3)
所以f(x)的最小正周期
T=2π/(π/4)=8
f(x)=cos(xπ/4-π/3)-cos(xπ/4)
=[cos(xπ/4)cos(π/3)+sin(xπ/4)sin(π/3)]-cos(xπ/4)
=[1/2cos(xπ/4)+√3/2sin(xπ/4)]-cos(xπ/4)
=-1/2cos(xπ/4)+√3/2sin(xπ/4)
=cos(xπ/4+2π/3)
所以f(x)的最小正周期
T=2π/(π/4)=8
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
