已知向量a=（cos3/2x,sin3/2x),b=(cos1/2x,-sin1/2x),x属于[0,π/2]若f（x）=a·b-2t|a+b|的最小值为g（t）,求g（t）

a=(cos3/2x,sin3/2x),b=(cos1/2x,-sin1/2x)
a·b=cos3/2xcos1/2x-sin3/2xsin1/2x=cos(3x/2+x/2)=cos2x
a+b=(cos3/2x+cos1/2x,sin3/2x-sin1/2x)
(a+b)²=(cos3/2x+cos1/2x)²+(sin3/2x-sin1/2x)²
=cos²3/2x+cos²1/2x+2cos3/2xcos1/2x+sin²3/2x+sin²1/2x-2sin3/2xsin1/2x
=2+2cos3/2xcos1/2x-2sin3/2xsin1/2x
=2+2cos2x
=2+2(2cos²x-1)
=4cos²x
|a+b|=2cosx
f(x)=a·b-2t|a+b|=cos2x-4tcosx=2cos²x-4tcosx-1
f(x)=2x²-4tx-1 x∈[0,1]
对称轴x=t
1)tg(t)=f(0)=-1
2)0≤t≤1
g(t)=f(t)=2t²-4t²-1=-2t²-1
3)t>1
g(t)=f(1)=2-4t-1=1-4t
-1 ,tg(t) =-2t²-1,0≤t≤1
1-4t,t>1

已知向量a=（cos(3/2)x,sin(3/2)x),b=(cos(1/2)x,-sin(1/2)x),x属于[0,π/2],
若f（x）=a•b-2t|a+b|的最小值为g（t）,求g（t）
f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2t√{[cos(3x/2)+cos(x/2)]²+[sin(3x/2)-sin(x/2)]²}
=cos2x-2t√{2+2[cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)]}=cos2x-2t√(2+2cos2x)
x∈[0,π/2],当2x=π/2,即x=π/4时得g(t)=-2(√2)t