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