Question

If A + B + C = 2S, then prove that cos(S -A) + cos(S - B) + cos C = -1 + 4 $cos\frac{S - A}{2}cos\frac{S - B}{2}cos\frac{C}{2}$

Answer 1

Solution :

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

LHS = cos(s-A)+cos(s-B)+cos(s-C)

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

= 2cos(C/2)cos[(B-A)/2]+2cos²(C/2)-1

= -1+2cos(C/2)[cos[(B-A)/2]+cos(C/2)

= -1+2cos(C/2){2cos[(B-A+C)/4] ×

cos[(B-A-C)/4]}

= -1+4cos(C/2)cos[(B+C-A)/4]×

cos[(A+C-B)/4]

= -1 + 4cos(C/2)cos[(2s-A-A)/4]×

cos[(2s-B-B)/4]

= -1 +4cos(C/2)cos[(S-A)/2]cos[(s-B)/2]

= -1+4cos[(s-A)/2]cos[(s-B)/2]cos[(s-C)/2]×

cos(C/2)

= RHS

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