Question

If A, B,C are the angles of a triangle then prove: (the following in picture)

Please help me to prove this. ​

Answer 1

The proof for this is simple. Let's say that A + B + C = π. From here on we require several trigonometric identities that must be applied.

`\cos \left(A\right)+\cos \left(B\right)+\cos \left(C\right) \\= 2 * cos((A + B) / 2) * cos((A - B) / 2) + \cos C \\= 2 * cos((\pi /2) - (C/2)) * cos((A - B) / 2) +\cos C \\= 2 * sin(C/2) * cos((A - B) / 2) + (1 - 2 * sin^2 (C/2)) \\= 1 + 2 sin (C/2) * cos((A - B) / 2) - sin (C/2) \\= 1 + 2 sin (C/2) * cos((A - B) / 2) - sin((\pi /2) - (A + B)/2 ))\\= 1 + 2 sin (C/2) * cos((A - B) / 2) - cos((A + B)/ 2)\\= 1 + 2 sin (C/2) * 2 sin (A/2) * sin(B/2) \\= 1 + 4 sin(A/2) sin(B/2) sin(C/2)`

Hope that helps!

Answer 2

Answer: see proof below

Explanation:

Given: A + B + C = π → A + B = π - C

→ C = π - (A + B)

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use Product to Sum Identity: 2 sin A · sin B = cos [(A + B)/2] - cos [(A - B)/2]

Use the Double Angle Identity: cos 2A = 1 - 2 sin² A

Use the Cofunction Identity: cos (π/2 - A) = sin A

Proof LHS → RHS:

LHS: cos A + cos B + cos C

= (cos A + cos B) + cos C

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

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

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

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

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

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

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

`\text{Product to Sum:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[2\sin \bigg((A)/(2)\bigg)\cdot \sin\bigg((B)/(2)\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =1+4\sin \bigg((C)/(2)\bigg)\bigg[\sin \bigg((A)/(2)\bigg)\cdot \sin\bigg((B)/(2)\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)`

`\text{LHS = RHS:}\ 1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)=1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)\quad \checkmark`