Answer: see proof below
Step-by-step explanation:
Given: A + B + C = π → A + B = π - C
→ A + C = π - B
Use the Cofunction Identities: sin (π - A) = sin A
sin (A - π/2) = cos A
cos (π/2 - A) = - sin A
Use the Power Reducing Identity: sin ² A - sin² B = sin (A + B) · sin (A - B)
Use the Sum to Product Identity:
sin A - sin B = 2 [cos (A + B)/2] · [sin (A - B)/2]
Proof LHS → RHS:
LHS: sin² A - sin² B - sin² C
= (sin² A - sin² B) - sin² C
Power Reducing: sin (A + B) · sin (A - B) - sin² C
Given: sin (π - C) · sin (A - B) - sin² C
Cofunction: sin C · sin (A - B) - sin² C
Factor: sin C [sin (A - B) - sin C]
[tex]\text{Sum to Product:}\qquad \sin C\bigg[2\cos \bigg(\dfrac{(A-B)+C}{2}\bigg)\cdot \sin \bigg(\dfrac{(A-B)-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =2\sin C\bigg[\cos \bigg(\dfrac{(A+C)-B}{2}\bigg)\cdot \sin \bigg(\dfrac{A-(B+C)}{2}\bigg)\bigg][/tex]
[tex]\text{Given:}\qquad \qquad 2\sin C\bigg[\cos \bigg(\dfrac{(\pi - B)-B}{2}\bigg)\cdot \sin \bigg(\dfrac{A-(\pi - A)}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\sin C\bigg[\cos \bigg(\dfrac{\pi -2B}{2}\bigg)\cdot \sin \bigg(\dfrac{2A-\pi}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\sin C\bigg[\cos \bigg(\dfrac{\pi}{2} -B}\bigg)\cdot \sin \bigg(A-\dfrac{\pi}{2}\bigg)\bigg][/tex]
Cofunction: 2 sin C (-sin B) · (cos A)
= -2 sin C · sin B · cos A
= -2 cos A · sin B · sin C
LHS = RHS: -2 cos A · sin B · sin C = -2 cos A · sin B · sin C [tex]\checkmark[/tex]
