已知向量a=(cos3x/2,sin3x/2),向量b=(cosx/2,-sinx/2),且x∈[0,π/2].
1个回答

1.ab=cos(3x/2)*cos(x/2)-sin(3x/2)*sin(x/2)

=cos[(3x+x)/2]

=cos(2x).

a+b=(cos(3x/2)+cos(x/2),sin(3x/2)-sin(x/2)),

|a+b|=√[(cos(3x/2)+cos(x/2))^2+(sin(3x/2)-sin(x/2))^2]

=√[2(1+cos2x)]

=2*|cosx|,

因为,x∈[-π/3,π/4].则有,cosx>0,

即,

|a+b|=2*|cosx|=2cosx.

2.f（x）=a*b-2λ|a+b|=2cos^2x-1+4λcosx=2(cosx+λ)^2-2λ^2-1

00f(x)递增

所以f(x)min=2(0+λ)^2-2λ^2-1=-1≠-3/2

1

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