Answer: see proof below
Step-by-step explanation:
Given: A + B + C = π → A + B = π - C
A + C = π - B
B + C = π - A
Use the Cofunction Identity: sin A = cos (π/2 - A)
Use the following Sum to Product Identity:
sin A + sin B = 2 sin [(A + B)/2] · cos [(A + B)/2]
Use the Double Angle Identity: sin 2A = 2 sin A · cos A
Proof LHS → RHS
LHS: sin (B + C - A) + sin (C + A - B) + sin (A + B - C)
Given: sin[(π - A) - A) + sin [(π - B) - B] + sin [(π - C) - C]
= sin (π - 2A) + sin (π - 2B) + sin (π - 2C)
= sin 2A + sin 2B + sin 2C
= (sin 2A + sin 2B) + sin 2C
[tex]\text{Sum to Product:}\qquad 2\sin \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)+\sin 2C\\\\.\qquad \qquad \qquad \qquad =2\sin (A+B)\cdot \cos (A-B)+\sin 2C[/tex]
[tex]\text{Double Angle:}\qquad 2\sin (A+B)\cdot \cos (A-B)+2\sin C \cdot \cos C[/tex]
Given: 2 sin C · cos (A - B) + 2 sin C · cos C
Factor: 2 sin C [cos (A - B) + cos C]
[tex]\text{Sum Product:}\qquad 2\sin C\cdot 2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\\\\.\qquad \qquad \qquad =2\sin C\cdot 2\cos \bigg(\dfrac{(A+C)-B}{2}\bigg)\cdot \cos \bigg(\dfrac{A-(B+C)}{2}\bigg)[/tex]
[tex]\text{Given:}\qquad \qquad 4\sin C\cdot \cos \bigg(\dfrac{(\pi -B)-B}{2}\bigg)\cdot \cos \bigg(\dfrac{A-(\pi -A)}{2}\bigg)\\\\.\qquad \qquad \qquad =4\sin C\cdot \cos \bigg(\dfrac{\pi -2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-\pi}{2}\bigg)\\\\.\qquad \qquad \qquad =4\sin C\cdot \cos \bigg(\dfrac{\pi}{2}-B\bigg)\cdot \cos \bigg(\dfrac{\pi}{2}-A\bigg)[/tex]
Cofunction: 4 sin A · sin B · sin C
LHS = RHS: 4 sin A · sin B · sin C = 4 sin A · sin B · sin C [tex]\checkmark[/tex]
