设A+B+C=派,若sinA,sinB,sinC成等差数列,求证tanA/2×tanB/2=1/3
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sinA,sinC,sinB成等差数列

sinA+sinB=2sinC,

利用和差化积公式:2sin[(A+B)/2]cos(A-B)/2]= 4sin(C/2)cos(C/2)(1)

利用诱导公式 :COS(C/2)=2sin[(A+B)/2]

所以(1)可化为:cos(A-B)/2= 2sin(C/2)=2cos(A+B)/2

所以cos(A/2)cos(B/2)+sin(A/2)sin(B/2) =2cos(A/2)cos(B/2)-2sin(A/2)sin(B/2)

所以cos(A/2)cos(B/2)=3sin(A/2)sin(B/2)

所以tanA/2×tanB/2=sin(A/2)/cos(A/2)*sin(B/2)/cos(B/2)=1/3