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Question

If : a+b+c=pie

then prove that : sin2A+sin2B+sin2C= 4sinAsinBsinC


Solution

If A+B+C=180 degree, then A+B=180-C hence sin(A+B) = sinC, cos(A+B)=-cosC 

sin2A + sin 2B +sin2C 

= 2 sin(A+B)cos(A-B) + 2sinC cosC 

=2sinC cos(A-B)+2sinC cosC 

=2sinC (cos(A-b) + cos C) 

=2sin C(cos(A-B) - cos(A+B) ) 

= 2sinC . 2sin A sin B 

=4 sinA sin B sin C

Mathematics

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