If A+B+C=180degree. Prove that sin2A+sin2B+sin2C=4×sinA×sinB×sinC.

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Solution

A+B+C=180
sin2A+sin2B+sin2C
=2sinAcosA+(2sin(B+C)(cos(B-C)][because sinC+sinD=Sin(C+D)cos(C-D)/2]
=2sinAcosA+2sin(B+C)cos(B-C)
=2sinACosA+2 SinAcos(B-C)[since B+C=180-A and sin(180-A)=sinA]
=2SinA[-cos(B+c)+cos(B-C)[since A+B+C=180; CosA=cos[180-(B+C)]=-cosB+C
=2sinA(2SinBSinc) [since cos(A-B)-cos(A+B)=2SinASinB]
=4×sinA×sinB×sinC

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