Question

In a triangle abc provw that acosa+bcosb+ccosc= 2asinbsinc

Answer 1

aCosA+bCosB+cCosC = 2aSinBCosC


aCosA+bCosB+cCosC 


=2R sinA cosA +2R SinB CosB + 2R Sin C CosC (From Sine rule)


=R(2 sinA cosA +2 SinB CosB + 2 Sin C CosC)


=R (Sin2A+Sin2B+Sin2C)


=R(2Sin (A+B)Cos(A-B)+2SinCCosC)


=R [ 2 SinC Cos(A-B) + 2SinC CosC]


=2RSinC[Cos(A-B)+CosC]


=2RSinC[Cos(A-B)-Cos(A+B)]


=2R SinC [ 2 SinA Sin B]


=(2R SinA) (2SinB SinC)


=a (2 SinB SinC)


=2a Sin B SinC