1
(a+b)·(a-b)=|a|^2-|b|^2=1-1=0
即:(a+b)⊥(a-b)
2
a·b=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)
=cos(2x)
故:f(x)=cos(2x)+1
x∈[π/12,π/6],即:2x∈[π/6,π/3]
故:cos(2x)∈[1/2,√3/2]
即:f(x)∈[3/2,(√3+2)/2]
即:f(x)的最大值:(√3+2)/2
f(x)的最小值:3/2
1
(a+b)·(a-b)=|a|^2-|b|^2=1-1=0
即:(a+b)⊥(a-b)
2
a·b=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)
=cos(2x)
故:f(x)=cos(2x)+1
x∈[π/12,π/6],即:2x∈[π/6,π/3]
故:cos(2x)∈[1/2,√3/2]
即:f(x)∈[3/2,(√3+2)/2]
即:f(x)的最大值:(√3+2)/2
f(x)的最小值:3/2