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