设向量a=(根号3sinx,cosx),向量b=(cosx,cosx),记f(x)=向量a×向量b(1)写出函数f(x)的最小正周期及递增区间；(2)若f(x0)=11/10,x0属于[π/4,π/2],求cos2x0的值
问题描述：
设向量a=(根号3sinx,cosx),向量b=(cosx,cosx),记f(x)=向量a×向量b
(1)写出函数f(x)的最小正周期及递增区间；
(2)若f(x0)=11/10,x0属于[π/4,π/2],求cos2x0的值
答
=√3/2sin(2α-π/3)+2
f(x)=a*b=√3sinxcosx+cos^2x
=1/2sin2x+1/2cos2x+1/2
=√2sin(2x+π/4)+1/2
T=2π/2=π
x在[kπ-3π/8,kπ+π/8]递增
f(x)==√2sin(2x+π/4)+1/2=11/10
sin(2x+π/4)=3√2/10
x属于[π/4，π/2]，
2x+π/4属于[π/4，π/2]，
cos(2x+π/4)= -2√22/10
cos2x=cos[(2x+π/4)-π/4]
=√2/2[cos(2x+π/4)+sin(2x+π/4)]
=√2/2(3√2/10-2√22/10)
=(3-√11)/10
答
设向量a = (√3sinx,cosx),向量b = (cosx,cosx),记ƒ(x) = 向量a • 向量b—————————————————————————————————写出函数ƒ(x)的最小正周期及递增区间.ƒ(x) = (√3...