prove that aCosA+bCosB+cCosC = 2aSinBCosC - Maths - Trigonometric Functions

Question

prove that aCosA+bCosB+cCosC = 2aSinBCosC

S. Usha Rani · Accepted Answer

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