设向量a=(sinx,cosx),b=(cosx,cosx)x∈R,函数f(X)=a(a-b)
问题描述：
设向量a=(sinx,cosx),b=(cosx,cosx)x∈R,函数f(X)=a(a-b)
1 求函数f（x）的最小正周期
2.当x属于【-pai/4,pai/4】时,求函数f(x)的值域
3.求使不等式f(x)≥1成立的x的取值范围
不要复制f(x)=a(a+b)的那个题答案 我要的是a(a-b)
答
f(x)=a(a-b)
=(sinx,cosx)*(sinx-cosx,0)
=sin²x-sinxcosx
=1/2*(1-cos2x)-1/2*sin2x.正弦余弦二倍角公式
=-1/2*(sin2x-cos2x)+1/2
=-√2/2*sin(2x+π/4)+1/2
(1)最小正周期 =2π/2=π
(2)x∈[-π/4,π/4]
2x∈[-π/2,π/2]
2x+π/4∈[-π/4,3π/4]
sin(2x-π/4)∈[-√2/2,1]
-√2/2*sin(2x-π/4)+1/2∈[(1-√2)/2,1]
f(x)的值域是[(1-√2)/2,1]
(3)
-√2/2*sin(2x+π/4)+1/2≥1
-√2/2*sin(2x+π/4)≥1/2
sin(2x+π/4)≤-√2/2
5π/4+2kπ≤2x+π/4≤7π/4+2kπ,k∈Z
π+kπ≤x≤π+kπ,k∈Z
x的取值范围[π/2+kπ,3π/2+kπ],k∈Z