Question

Prove that sinA+sinB+sinC= 4cos(A/2)cos(B/2)cos(C/2) where A, B and C stand for the angles of a triangle.

Answer 1

RHS
\[=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\]\[=4[\frac{1}{2}(cos\frac{A+B}{2}-cos\frac{A-B}{2})]cos\frac{C}{2}\]\[=2(cos\frac{A+B}{2}-cos\frac{A-B}{2})cos\frac{C}{2}\]\[=2(cos\frac{A+B}{2}cos\frac{C}{2}-cos\frac{A-B}{2}cos\frac{C}{2})\]\[=2(cos\frac{180-C}{2}cos\frac{C}{2}-cos\frac{A-B}{2}cos\frac{C}{2})\]\[=2(sin\frac{C}{2}cos\frac{C}{2}-cos\frac{A-B}{2}cos\frac{C}{2})\]\[=2sin\frac{C}{2}cos\frac{C}{2}-2cos\frac{A-B}{2}cos\frac{C}{2}\]\[=sinC-2cos\frac{A-B}{2}cos\frac{C}{2}\]\[=sinC-2(\frac{1}{2}cos\frac{A-B+C}{2}+cos\frac{A-B-C}{2})\]\[=sinC-(cos\frac{A-B+C}{2}+cos\frac{A-B-C}{2})\]\[=sinC-(cos\frac{A+C-B}{2}+cos\frac{A-(B+C)}{2})\]\[=sinC-(cos\frac{(180-B)-B}{2}+cos\frac{A-(180-A)}{2})\]\[=sinC-(cos\frac{180-2B}{2}+cos\frac{-180+2A}{2})\]\[=sinC-(sin(-B)-sinA)\]\[=sinC+sinB+sinA\]=LHS

Answer 2

\[A + B + C = \pi\]\[A + B = \pi - C\]\[C = \pi - (A + B)\]\[\sin A + \sin B + \sin C = 2 \sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) + \sin \left( \pi - (A + B) \right)\]\[= 2 \sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) + \sin \left(A + B\right)\]\[= 2 \sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) + 2 \sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A+B}{2} \right)\]\[= \left( 2 \sin \left( \frac{A+B}{2} \right) \right)\left( \cos \left( \frac{A+B}{2} \right) + \cos \left( \frac{A-B}{2} \right)\right)\]\[= \left( 2 \sin \left( \frac{A + B}{2} \right) \right)\left( 2 \cos \left( \frac{A}{2} \right) \cos \left( \frac{B}{2} \right) \right)\]\[= 4 \cos \left( \frac{A}{2} \right) \cos \left( \frac{B}{2} \right) \sin \left( \frac{\pi - C}{2} \right)\]\[= 4 \cos \left( \frac{A}{2} \right)\cos \left( \frac{B}{2} \right) \sin \left( \frac{\pi}{2} - \frac{C}{2} \right)\]\[= 4 \cos \left( \frac{A}{2} \right) \cos \left( \frac{B}{2} \right) \cos \left( \frac{C}{2} \right)\]

Answer 3

Thank you guys so much!