(1)因为向量a=(cos3x/2,sin3x/2),b=(cosx/2,—sinx/2),所以:
|a|=|b|=1
且a*b=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)
=cos(3x/2 +x/2)
=cos2x
则|a+b|²=|a|²+2a*b+|b|²
=2+2cos2x
=2(1+cos2x)
=4cos²x
因为x∈[0,π/2],所以:
|a+b|=2cosx
(2)由(1)可得:
f(x)=a·b-2λ│a+b│
=cos2x-2λ*2cosx
=2cos²x-4λcosx-1
=2(cosx-λ)²-2λ²-1
因为x∈[0,π/2],所以cosx∈[0,1]
若λ