证明：利用倍角公式cos^2x-sin^2x=cos2x=1-2sin^2x,将原式移项有：sin^2A/2+sin^2C/2=cos^2B/2-sin^2B/2=cosB,因2(sin^2A/2+sin^2C/2)=2-cosA-cosC,即有2cosB=2-cosA-cosC,于是cosA+cosC=2(1-cosB)=4sin^2(B/2)……①,在△ABC中A+B+C=π,所以B/2=(π-A-C)/2,sinB/2=sin(π-A-C)/2=cos(A+C)/2,而cosA+cosC=2cos(A+C)/2cos(A-C)/2,代入①式化简即可得cos(A-C)/2=2cos(A+C)/2,即cosA/2cosC/2+sinA/2sinC/2=2cosA/2cosC/2-2sinA/2sinC/2,即cosA/2cosC/2=3sinA/2sinC/2,即tanA/2tanC/2=1/3=(tanπ/6)^2,即证!