在三角形ABC中,已知sinA、sinB、sinC成等差数列,证明cot（A/2）*cot(C/2)=3

题目

在三角形ABC中,已知sinA、sinB、sinC成等差数列,证明cot（A/2）*cot(C/2)=3

答案

sinA-sinC=sinC-sinB
--->2sinC=sinA+sinB 和差化积
--->4sin(C/2)cos(C/2)=2sin[(A+B)/2]cos(A-B)/2]
--->4sin(C/2)cos[(A+B)/2]=2sin(C/2)cos[(A-B)/2] 诱导公式
sin(C/2)0--->2cos[(A+B)/2]=cos[(A-B)/2]
--->2cos(A/2)cos(B/2)-2sin(A/2)sin(B/2)
=cos(A/2)cos(B/2)s+sin(A/2)sin(B/2)
--->cos(A/2)cos(B/2)=3sin(A/2)sin(B/2)
--->cos(A/2)/sin(A/2)*cos(B/2)/sin(B/2)=3
--->cot(A/2)cot(B/2)=3.证完