Question

2 (a cosA-bcosB) sinC = c ( sin2A - sin 2B)

Answer 1

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)

Step-by-step explanation:

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