Prove sin2a+sin2b+sin2c/sina+sinb+sinc=8sina/2+sinb/2sinc/2
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check your question answer is 8sina/2sinb/2sinc/2.
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LHS = sin 2A + sin 2B + sin 2C = sin (pie – 2A) + sin (pie –2B) + sin (pie –2C) = 2sin(A+B)cos(A-B) + 2sinCcosC = 2sin(pie - C)cos(A-B) + 2sinCcosC = 2sinCcos(A - B) + 2sinCcosC = 2sinC (cos (A - B)+ cos C) = 2sinC (2cos (A - B + C)/2 cos (A - B - C)/2) = 2sinC (2cos (pie - 2B)/2 cos(2A - pie)/2 = 4sinAsinBsinC = RHS Hence Proved
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