Answer: see proof below
Step-by-step explanation:
Given: A + B + C = 90° → A + B = 90° - C
→ C = 90° - (A + B)
Use the Double Angle Identity: cos 2A = 1 - 2 sin² A
→ sin² A = (1 - cos 2A)/2
Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]
Use the Product to Sum Identity: cos (A - B) - cos (A + B) = 2 sin A · sin B
Use the Cofunction Identities: cos (90° - A) = sin A
sin (90° - A) = cos A
Proof LHS → RHS:
LHS: sin² A + sin² B + sin² C
[tex]\text{Double Angle:}\qquad \dfrac{1-\cos 2A}{2}+\dfrac{1-\cos 2B}{2}+\sin^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2-\cos 2A-\cos 2B\bigg)+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\sin^2 C[/tex]
[tex]\text{Sum to Product:}\quad 1-\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\cos (A+B)\cdot \cos (A-B)+\sin^2 C[/tex]
Given: 1 - cos (90° - C) · cos (A - B) + sin² C
Cofunction: 1 - sin C · cos (A - B) + sin² C
Factor: 1 - sin C [cos (A - B) + sin C]
Given: 1 - sin C[cos (A - B) - sin (90° - (A + B))]
Cofunction: 1 - sin C[cos (A - B) - cos (A + B)]
Sum to Product: 1 - sin C [2 sin A · sin B]
= 1 - 2 sin A · sin B · sin C
LHS = RHS: 1 - 2 sin A · sin B · sin C = 1 - 2 sin A · sin B · sin C [tex]\checkmark[/tex]
