Question

Please help me to prove this!!!
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Answer 1

Answer: see proof below

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]

`\text{Sum to Product:}\qquad \sin C\bigg[2\cos \bigg(((A-B)+C)/(2)\bigg)\cdot \sin \bigg(((A-B)-C)/(2)\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =2\sin C\bigg[\cos \bigg(((A+C)-B)/(2)\bigg)\cdot \sin \bigg((A-(B+C))/(2)\bigg)\bigg]`

`\text{Given:}\qquad \qquad 2\sin C\bigg[\cos \bigg(((\pi - B)-B)/(2)\bigg)\cdot \sin \bigg((A-(\pi - A))/(2)\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\sin C\bigg[\cos \bigg((\pi -2B)/(2)\bigg)\cdot \sin \bigg((2A-\pi)/(2)\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\sin C\bigg[\cos \bigg((\pi)/(2) -B}\bigg)\cdot \sin \bigg(A-(\pi)/(2)\bigg)\bigg]`

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
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