Question

If A + B + C = 2π, then prove that sin A - sin B + sin C - sin D = -4 $cos\frac{A + B}{2}sin\frac{A + C}{2}cos\frac{A + D}{2}$.

Answer 1

Solution :

i ) A+B+C+D = 2π

=> A/2 + B/2 + C/2 + D/2 = π

=> ( A+B)/2 = π - (C+D)/2

Now ,

LHS = sinA-sinB+sinC-sinD

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

+ 2cos[(C+D)/2]sin[(C-D)/2]

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

+ 2cos{π+(A+B)/2)}sin[(C-D)/2]

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

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

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

= 2cos[(A+B)/2]{2cos[(A+B+C-D)/4}×

sin{(A-B-C+D)/4}

= 4cos[(A+B)/2]cos{(A+B-2π+A+C)/4}×

sin[(A+D-2π+A+D)/4}

= 4cos[(A+B)/2]cos[(A+C)/2 - π/2]×

sin[(A+D)/2 - π/2]

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

= RHS

•••••

Answer 2

Answer:

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