Question

Please help me!!!!!​

Answer 1

Answer:  see proof below

Step-by-step explanation:

Given: A + B + C = π               → A = π - (B + C)

                                               → B = π - (A + C)

                                               → C = π - (A + B)

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

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

Proof LHS → RHS:

LHS:                        sin A - sin B + sin C

                             = (sin A - sin B) + sin C

$\text{Sum to Product:}\quad 2\cos \bigg(\dfrac{A+B}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

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

$.\qquad \qquad =2\cos \bigg(\dfrac{\pi}{2} -\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

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

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

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

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

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

$\text{LHS = RHS:}\quad 4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)=4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\quad \checkmark$