一道数学题，急需，在线等
展开全部
f(x)=cos2xcosπ/3+sin2xsinπ/3+2sin(x-π/4)cos[π/2-(x+π/4)]
=cos2x*1/2+sin2x*√3/2+2sin(x-π/4)cos(x-π/4)
=cos2x*1/2+sin2x*√3/2+sin(2x-π/2)
=cos2x*1/2+sin2x*√3/2-cos2x
=sin2x*√3/2-cos2x*1/2
=sin2xcosπ/6-cos2xsinπ/6
=sin(2x-π/6)
T=2π/2=π
-π/12<=x<=π/2
-π/3<=2x-π/6<=5π/6
所以最大=sinπ/2=1
最小=sin(-π/3)=-√3/2
值域[-√3/2,1]
=cos2x*1/2+sin2x*√3/2+2sin(x-π/4)cos(x-π/4)
=cos2x*1/2+sin2x*√3/2+sin(2x-π/2)
=cos2x*1/2+sin2x*√3/2-cos2x
=sin2x*√3/2-cos2x*1/2
=sin2xcosπ/6-cos2xsinπ/6
=sin(2x-π/6)
T=2π/2=π
-π/12<=x<=π/2
-π/3<=2x-π/6<=5π/6
所以最大=sinπ/2=1
最小=sin(-π/3)=-√3/2
值域[-√3/2,1]
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
