已知向量a=(cos3/2x,sin3/2x),向量b=(cos1/2x,-sin1/2x),x∈[-π/8,π/4].（1）求向量a·向量b（内积）及丨向量a+向量b丨；（2）若f(x)=向量a·向量b-丨向量a+向量b丨,求f（x）的最值.
问题描述：
已知向量a=(cos3/2x,sin3/2x),向量b=(cos1/2x,-sin1/2x),x∈[-π/8,π/4].（1）求向量a·向量b（内积）及丨向量a+向量b丨；（2）若f(x)=向量a·向量b-丨向量a+向量b丨,求f（x）的最值.
答
(1)
a.b
=(cos(3x/2),sin(3x/2)).(cos(x/2),-sin(x/2))
=cos(3x/2).cos(x/2)- sin(3x/2).sin(x/2)
=cos2x
答
(1)
a.b
=(cos(3x/2),sin(3x/2)).(cos(x/2),-sin(x/2))
=cos(3x/2).cos(x/2)- sin(3x/2).sin(x/2)
=cos2x
(2)
|a+b|^2
=(cos(3x/2)+cos(x/2))^2+ (sin(3x/2)-sin(x/2))^2
= 2 + 2(cos(3x/2)cos(x/2)- sin(3x/2)sin(x/2) )
=2 + 2cos2x
= 2 + 2[2(cosx)^2 -1)]
|a+b| =2cosx
f(x) = a.b-|a+b|
= cos2x- 2cosx
= 2(cosx)^2-2cosx -1
= 2(cosx-1/2)^2 -3/2
min f(x) at cosx =1/2
min f(x) = -3/2