Question

已知向量a=[cos(3x/2),sin(3x/2)],b=[cos(x/2),-sin(x/2),]且x∈[0,π/2]

(1)求a·b及|a+b|;
(2)求函数f(x)=a·b-|a+b|的最小值及此时x的值

(过程全的另加分！！！)

Answer 1

a·b=cos(3x/2)*cos(x/2)-sin(3x/2)*)sin(x/2)
=cos(3x/2+x/2)
=cos2x,
向量a+b=[cos(3x/2)+cos(x/2),sin(3x/2)-sin(x/2)]
=(2cosxcosx/2,2cosxsinx/2),
|a+b|=√[(2cosxcosx/2)^2+(2cosxsinx/2)^2]
=2√{(cosx)^2[(sinx/2)^2+(cosx/2)^2]}
=2|cosx|,
x∈[0,π/2] ,
|a+b|=2cosx,
f(x)=cos2x-2cosx=2(cosx)^2-2cosx-1=2[(cosx-1/2)^2-5/4,
最小值为-5/4，此时cosx=1/2,x=π/3。