Respuesta :

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.

[tex]\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)[/tex]

Hope that helps!

tramserran

Answer:  see proof below

Step-by-step 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

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

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

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

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

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

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

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

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

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

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