Question

Please help me to prove this!​

Answer 1

Answer: see proof below

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:

`\text{LHS:}\qquad \qquad \sin^2\bigg((B)/(2)\bigg)+\sin^2 \bigg((C)/(2)\bigg)-\sin^2\bigg((A)/(2)\bigg)`

`\text{Pythagorean:}\qquad 1-\cos^2 \bigg((B)/(2)\bigg)+1-\cos^2 \bigg((C)/(2)\bigg)-\bigg[1-\cos^2 \bigg((A)/(2)\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1-\cos^2 \bigg((B)/(2)\bigg)-\cos^2 \bigg((C)/(2)\bigg)+\cos^2 \bigg((A)/(2)\bigg)`

`\text{Double Angle:}\quad 1-\bigg((\cos(2\cdot (B)/(2))+1)/(2)\bigg)-\bigg((\cos (2\cdot (C)/(2))+1)/(2)\bigg)+\bigg((\cos (2\cdot (A)/(2))+1)/(2)\bigg)\\\\\\.\qquad \qquad \qquad =1-(\cos B)/(2)-(1)/(2)-(\cos C)/(2)-(1)/(2)+(\cos A)/(2)+(1)/(2)\\\\\\.\qquad \qquad \qquad =(1)/(2)[1-(\cos B+\cos C)+\cos A]`

`\text{Sum to Product:}\qquad (1)/(2)\bigg(1-\bigg[2\cos \bigg((B+C)/(2)\bigg)\cdot \cos \bigg((B-C)/(2)\bigg)\bigg]+\cos A\bigg)`

`\text{Given:}\qquad (1)/(2)\bigg(1-\bigg[2\cos \bigg((\pi -A)/(2)\bigg)\cdot \cos \bigg((B-C)/(2)\bigg)\bigg]+\cos A\bigg)`

`\text{Cofunction:}\qquad (1)/(2)\bigg(1-\bigg[2\sin \bigg((A)/(2)\bigg)\cdot \cos \bigg((B-C)/(2)\bigg)\bigg]+\cos A\bigg)`

`\text{Double Angle:}\qquad (1)/(2)\bigg[1-2\sin \bigg((A)/(2)\bigg)\cdot \cos \bigg((B-C)/(2)\bigg)+1-2\sin^2 \bigg((A)/(2)\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =(1)/(2)\bigg[2-2\sin \bigg((A)/(2)\bigg)\cdot \cos \bigg((B-C)/(2)\bigg)-2\sin^2 \bigg((A)/(2)\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1-\sin \bigg((A)/(2)\bigg)\cdot \cos \bigg((B-C)/(2)\bigg)-\sin^2 \bigg((A)/(2)\bigg)`

`\text{Factor:}\qquad \qquad 1-\sin \bigg((A)/(2)\bigg)\bigg[ \cos \bigg((B-C)/(2)\bigg)-\sin \bigg((A)/(2)\bigg)\bigg]`

`\text{Given:}\qquad \qquad 1-\sin \bigg((A)/(2)\bigg)\bigg[ \cos \bigg((B-C)/(2)\bigg)-\sin \bigg((\pi -(B+C))/(2)\bigg)\bigg]`

`\text{Cofunction:}\qquad 1-\sin \bigg((A)/(2)\bigg)\bigg[ \cos \bigg((B-C)/(2)\bigg)+\cos \bigg((B+C)/(2)\bigg)\bigg]`

`\text{Sum to Product:}\ 1-\sin \bigg((A)/(2)\bigg)\cdot 2 \cos \bigg(((B-C)+(B-C))/(2\cdot 2)\bigg)\cdot \cos \bigg(((B-C)-(B+C))/(2\cdot 2)\bigg)\\\\\\.\qquad \qquad \qquad =1-2\sin \bigg((A)/(2)\bigg)\cdot \cos \bigg((B)/(2)\bigg)\cdot \cos \bigg(-(C)/(2)\bigg)`
`\text{Cofunction:}\qquad =1-2\sin \bigg((A)/(2)\bigg)\cdot \cos \bigg((B)/(2)\bigg)\cdot \cos \bigg((C)/(2)\bigg)`

`\text{LHS = RHS:}\quad \checkmark\\\\\quad 1-2\sin \bigg((A)/(2)\bigg)\cdot \cos \bigg((B)/(2)\bigg)\cdot \cos \bigg((C)/(2)\bigg)=1-2\sin \bigg((A)/(2)\bigg)\cdot \cos \bigg((B)/(2)\bigg)\cdot \cos \bigg((C)/(2)\bigg)\quad`