已知函数f(x)=2sinwxcoswx+2cos平方wx(w>0),则函数的最小正周期为π 5
1个回答
展开全部
f(x)=2sinwxcoswx+2cos²wx
= sin2wx+cos2wx+1
= √2{sin2wxcosπ/4+cos2wxsinπ/4} + 1
= √2sin(2wx+π/4) + 1
w>0，且函数的最小正周期为π
2π/(2w)=πw=1
f(x) = √2sin(2x+π/4) + 1
x∈[0，π/2]
2x∈[0，π]
2x+π/4∈[π/4，5π/4]
2x+π/4 = π/2时有最大值 = √2*1 + 1 = √2+1
2x+π/4 = 5π/4时有最小值 = √2*(-√2/2) + 1 = -1+1 = 0
值域【0，√2+1】
= sin2wx+cos2wx+1
= √2{sin2wxcosπ/4+cos2wxsinπ/4} + 1
= √2sin(2wx+π/4) + 1
w>0，且函数的最小正周期为π
2π/(2w)=πw=1
f(x) = √2sin(2x+π/4) + 1
x∈[0，π/2]
2x∈[0，π]
2x+π/4∈[π/4，5π/4]
2x+π/4 = π/2时有最大值 = √2*1 + 1 = √2+1
2x+π/4 = 5π/4时有最小值 = √2*(-√2/2) + 1 = -1+1 = 0
值域【0，√2+1】
本回答由网友推荐
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
