If in a triangle `A B C ,cosA+2cosB+cosC=2`prove that the sides of the triangle are in `AP`

If in a triangle `A B C ,cosA+2cosB+cosC=2`prove that the sides of the

1.	If in a triangle `A B C ,cosA+2cosB+cosC=2`prove that the sides of the triangle are in `AP`
Answer» `cosA+2cosB+cosC = 2` `=>cosA+cosC = 2(1-cosB)` `=>2cos((A+C)/2)cos((A-C)/2) = 2(2sin^2(B/2))` `=>2cos((pi-B)/2)cos((A-C)/2) = 2(2sin^2(B/2))` `=>sin(B/2)cos((A-C)/2) = 2sin^2(B/2)` `=>cos((A-C)/2) = 2sin(B/2)` `=>cos(B/2)cos((A-C)/2) = 2sin(B/2)cos(B/2)` `=>cos((pi-(A+C))/2)cos((A-C)/2) = sinB` `=>sin((A+C)/2)cos((A-C)/2) = sinB` `=>2sin((A+C)/2)cos((A-C)/2) = 2sinB` `=>sinA+sinC = 2sinB->(1)` Now, from sine rule, `sinA/a = sinB/b = sinC/c = k` `=>sinA = ka, sinB = kb,sinC = kc` So, (1) becomes, `=>ka +kc = 2kb` `=>(a+c)/2 = b` So, sides of the triangle are in A.P.