if the side lengths a,b,c are in A.P. then prove that cos(A-C)/2 = 2sin B/2
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Answer:
cos(A-C)/2 = 2sin B/2
Step-by-step explanation:
a/SinA = b/SinB = c/SinC = k
a = kSinA
b =kSinB
c = kSinC
a,b,c are in A.P
=> 2kSinB = kSinA + k SinC
=> 2SinB = SinA + SinC
using Sin2x = 2SinxCosx
& Sinx + Siny = 2Sin((x + y)/2)Cos((x - y)/2)
=> 2 (2Sin(B/2)Cos(B/2)) = 2 Sin((A + C)/2) Cos((A-C)/2)
A+ B + C = 180° = π => A + C = π- B => (A + C)/2 = π/2 - B/2
=> Sin( (A + C)/2) = Sin(π/2 - B/2)= Cos (B/2)
=> 2Sin(B/2)Cos(B/2) = Cos (B/2) Cos((A-C)/2)
cancelling Cos(B/2) from both sides
=> 2Sin(B/2) = Cos((A-C)/2)
=> cos(A-C)/2 = 2sin B/2
QED
Proved
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