在三角形ABC中,已知2sinB=sinA+sinC,且（根号3）*cos(B/2)=2cos(A/2)cos(C/2).
问题描述：
在三角形ABC中,已知2sinB=sinA+sinC,且（根号3）*cos(B/2)=2cos(A/2)cos(C/2).
(1)求证：tan(A/2)tan(C/2)=1/3
(2)求tan[(A+C)/2]的值.
答
(1)sinA +sinC =2sinB 左边和差化积,右边倍角公式,2sin[(A+C)/2]cos[(A-C)/2] =2*2sin(B/2)cos(B/2) 2sin[(π-B)/2]cos[(A-C)/2]=2*2sin(B/2)cos(B/2) cos(B/2)cos[(A-C)/2] =2sin(B/2)cos(B/2) cos[(A-C)/2] =2sin(...