Answer: see proof below
Step-by-step explanation:
Given: A + B + C = π → A = π - (B + C)
→ B + C = π - A
Use the Pythagorean Identity: cos² A + sin² A = 1 → sin² A = 1 - cos² A
Use Double Angle Identities: cos 2A = 2 cos² A - 1 → cos² A = (cos 2A + 1)/2
→ cos A = 1 - 2 sin² (A/2)
Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]
Use Cofunction Identities: cos (π/2 - A) = sin (A)
sin (π/2 - A) = cos A
cos (-A) = cos (A)
Proof LHS → RHS:
[tex]\text{LHS:}\qquad \qquad \sin^2\bigg(\dfrac{B}{2}\bigg)+\sin^2 \bigg(\dfrac{C}{2}\bigg)-\sin^2\bigg(\dfrac{A}{2}\bigg)[/tex]
[tex]\text{Pythagorean:}\qquad 1-\cos^2 \bigg(\dfrac{B}{2}\bigg)+1-\cos^2 \bigg(\dfrac{C}{2}\bigg)-\bigg[1-\cos^2 \bigg(\dfrac{A}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1-\cos^2 \bigg(\dfrac{B}{2}\bigg)-\cos^2 \bigg(\dfrac{C}{2}\bigg)+\cos^2 \bigg(\dfrac{A}{2}\bigg)[/tex]
[tex]\text{Double Angle:}\quad 1-\bigg(\dfrac{\cos(2\cdot \frac{B}{2})+1}{2}\bigg)-\bigg(\dfrac{\cos (2\cdot \frac{C}{2})+1}{2}\bigg)+\bigg(\dfrac{\cos (2\cdot \frac{A}{2})+1}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1-\dfrac{\cos B}{2}-\dfrac{1}{2}-\dfrac{\cos C}{2}-\dfrac{1}{2}+\dfrac{\cos A}{2}+\dfrac{1}{2}\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}[1-(\cos B+\cos C)+\cos A][/tex]
[tex]\text{Sum to Product:}\qquad \dfrac{1}{2}\bigg(1-\bigg[2\cos \bigg(\dfrac{B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)\bigg]+\cos A\bigg)[/tex]
[tex]\text{Given:}\qquad \dfrac{1}{2}\bigg(1-\bigg[2\cos \bigg(\dfrac{\pi -A}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)\bigg]+\cos A\bigg)[/tex]
[tex]\text{Cofunction:}\qquad \dfrac{1}{2}\bigg(1-\bigg[2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)\bigg]+\cos A\bigg)[/tex]
[tex]\text{Double Angle:}\qquad \dfrac{1}{2}\bigg[1-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)+1-2\sin^2 \bigg(\dfrac{A}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg[2-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)-2\sin^2 \bigg(\dfrac{A}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1-\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B-C}{2}\bigg)-\sin^2 \bigg(\dfrac{A}{2}\bigg)[/tex]
[tex]\text{Factor:}\qquad \qquad 1-\sin \bigg(\dfrac{A}{2}\bigg)\bigg[ \cos \bigg(\dfrac{B-C}{2}\bigg)-\sin \bigg(\dfrac{A}{2}\bigg)\bigg][/tex]
[tex]\text{Given:}\qquad \qquad 1-\sin \bigg(\dfrac{A}{2}\bigg)\bigg[ \cos \bigg(\dfrac{B-C}{2}\bigg)-\sin \bigg(\dfrac{\pi -(B+C)}{2}\bigg)\bigg][/tex]
[tex]\text{Cofunction:}\qquad 1-\sin \bigg(\dfrac{A}{2}\bigg)\bigg[ \cos \bigg(\dfrac{B-C}{2}\bigg)+\cos \bigg(\dfrac{B+C}{2}\bigg)\bigg][/tex]
[tex]\text{Sum to Product:}\ 1-\sin \bigg(\dfrac{A}{2}\bigg)\cdot 2 \cos \bigg(\dfrac{(B-C)+(B-C)}{2\cdot 2}\bigg)\cdot \cos \bigg(\dfrac{(B-C)-(B+C)}{2\cdot 2}\bigg)\\\\\\.\qquad \qquad \qquad =1-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \cos \bigg(-\dfrac{C}{2}\bigg)[/tex][tex]\text{Cofunction:}\qquad =1-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \cos \bigg(\dfrac{C}{2}\bigg)[/tex]
[tex]\text{LHS = RHS:}\quad \checkmark\\\\\quad 1-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \cos \bigg(\dfrac{C}{2}\bigg)=1-2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \cos \bigg(\dfrac{C}{2}\bigg)\quad[/tex]
