已知函数f(x)=4sin²x+2sin2x-2,x∈R
(1)求f(x)的最小正周期及f(x)取得最大值时的集合;(2)求证:函数f(x)的图像关于直线x=-π/8对称....
(1)求f(x)的最小正周期及f(x)取得最大值时的集合;
(2)求证:函数f(x)的图像关于直线x=-π/8对称. 展开
(2)求证:函数f(x)的图像关于直线x=-π/8对称. 展开
1个回答
2011-01-13 · 知道合伙人教育行家
关注
展开全部
f(x)=4sin²x+2sin2x-2
=2(1-cos2x)+2sin2x-2
=2(sin2x-cos2x)
=2根号2(sin2xcosπ/2-cos2xsinπ/2)
=2根号2sin(2x-π/4)
最小正周春陆期:2π/2=π
(2x+π/4)=2kπ-π/2,k∈Z 时,sin(2x-π/4)=1,f(x)有最大值2根号2
此时x=kπ-3π/8,k∈Z
f(-π/8-x)=2根号2sin[2(-π/8-x)-π/4]=2根号2sin[-2x-π/2]=2根号2sin[π-(-2x-π/2)]=2根号2sin[2x+3π/2]=2根号2sin[(2x+3π/2)-2π]=2根号谈帆2sin[2x-π/2]
f(-π/8+x)=2根号2sin[2(-π/8+x)-π/扒侍顷4]=2根号2sin[2x-π/2]
f(-π/8-x) = f(-π/8+x),得证
=2(1-cos2x)+2sin2x-2
=2(sin2x-cos2x)
=2根号2(sin2xcosπ/2-cos2xsinπ/2)
=2根号2sin(2x-π/4)
最小正周春陆期:2π/2=π
(2x+π/4)=2kπ-π/2,k∈Z 时,sin(2x-π/4)=1,f(x)有最大值2根号2
此时x=kπ-3π/8,k∈Z
f(-π/8-x)=2根号2sin[2(-π/8-x)-π/4]=2根号2sin[-2x-π/2]=2根号2sin[π-(-2x-π/2)]=2根号2sin[2x+3π/2]=2根号2sin[(2x+3π/2)-2π]=2根号谈帆2sin[2x-π/2]
f(-π/8+x)=2根号2sin[2(-π/8+x)-π/扒侍顷4]=2根号2sin[2x-π/2]
f(-π/8-x) = f(-π/8+x),得证
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