3cosA+cosC=2(1-cosB)
and A+B+C=pi
If in ΔABC ,cosA+2cosB+cosC=2
then prove that a,b,c are in AP
-
UP 0 DOWN 0 0 4
4 Answers
106cosA+cosC=2(1-cosB)
=> 2cos((A+C)/2)cos((A-C)/2) = 4sin2(B/2)
=> 2sin(B/2)cos((A-C)/2) = 4sin2(B/2)
=> cos((A-C)/2) = 2sin(B/2)
=> cos((A-C)/2) =2cos((A+C)/2)
=> cos((A-C)/2) - cos((A+C)/2) = cos((A+C)/2)
=> 2sin(A/2)sin(C/2) = sin(B/2)
=> 2√(s-b)(s-c)√bc√(s-a)(s-b)√ab = √(s-a)(s-c)√ac
=> 2(s-b) = b
=> a+b+c-2b = b
=> a+c=2b (proved)
- KhLuck I wonder if there"s any other proof?Upvote·0· Reply ·2015-01-18 15:23:55