(1)f(x)=1-m*n
=1-(2a,1)*(1+cosx,1-cos2x)
=1-(2a+2acosx+1-cos2x)
=-2a-2acosx+cos2x
=-2a-2acosx+(2cos^2 x-1)
=2cos^2 x-2acosx-2a-1
(2)由f(x)=2(cosx-a/2)^2-1-a^2/2-2a
则x=a/2时,函数最小值=-a^2/2-2a-1
则g(a)=-a^2/2-1-2a
(3)g(a)=-a^2/2-1-2a=1/2
因为-12
则最小值为f(1)=-4a+1=1/2 a=1/8 矛盾
如果a/2