Math, asked by B1hatTnivekar, 1 year ago

Prove that :sin2a+sin2b+sin2c= 4 sina sinb sinc

Answers

Answered by prmkulk1978

286

Given:
A+B+C=π
sin2A+sin2B+sin2C
2sinAcosA+(2sin(B+C)(cos(B-C)][∵sinC+sinD=Sin(C+D)cos(C-D)/2]
2sinAcosA+2sin(B+C)cos(B-C)
2sinACosA+2 SinAcos(B-C)[∵A+B+C=π]
2SinA[-cos(B+c)+cos(B-C)[∵A+B+C=π; CosA=cos[π-(B+C)]=-cosB+C
2sinA(2SinBSinc) [∵cos(A-B)-cos(A+B)=2SinASinB]

Answered by kashyapswati312

97

Answer:

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