Question

If A + B + C = 2S, then prove that sin(S -A) + sin(S - B) + sin C = 4 $cos\frac{S - A}{2}cos\frac{S - B}{2}sin\frac{C}{2}$

Answer 1

Solution :

A + B + C = 2s -----( 1 )



LHS = sin(s-A) + sin(s-B) + sinC

= 2cos[( 2s-A-B)/2]cos[(B-A)/2] + sinC

= 2sin(C/2)cos[(A-B)/2]+2sin(C/2)cos(C/2)

= 2sin(C/2){cos[(A-B)/2] + cos(C/2)}

= 2sin(C/2){2[(A-B+C)/4]×cos[(A-B-C)/4]}

=4sin(C/2){cos[(2s-B-B)/4]cos[(2s-A-A)/4]}

= 4cos[(s-A)/2]cos[(s-B)/2]sin(C/2)

= RHS

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Answer 2

Answer:

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