Question

If A+B+C =0,then prove that sin2A+sin2B + sin2C = 4sinA.sunB.sinC

Answer 1

Answer:

Step-by-step explanation:

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

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Answer 2

HEY MATE,

FIND THE ANSWER BELOW !!

Answer:

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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