f(x)=(√3/2)sin2x-(1/2)cos2x-1=sin(2x-π/6)-1
左移π/6,
g(x)=sin2x-1
g(B)=0
B=π/4 tanB=1 cosB=√2/2
m*n=cosA+cosB*(sinA-cosA)
=cosBsinA+(1-cosB)cosA
=√[(cosB)^2+(1-cosB)^2] sin(A+u) cosu=cosB/√[(cosB)^2+(1-cosB)^2]=√2/2√(2-√2)
sinu=(1-cosB)/√[(cosB)^2+(1-cosB)^2]=(2-√2)/2√(2-√2)
=√[(1/2)+1+1/2-√2]sin(A+u)
=√(2-√2)sin(A-u)
A>0 sin(A-u)>sin(-u)=(√2-2)/[2√(2-√2)]
(2√2-3)/√(2-√2) < m*n