Question

If A, B ,C are the angles of a triangle then,Please help me to prove this!​

Answer 1

Answer:  see proof below

Step-by-step explanation:

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

                                                     → C = π - (A + B)

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

                                                       cos (π/2 - A) = sin A

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

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

Proof LHS → RHS:

$\text{LHS:}\qquad \qquad \sin \bigg(\dfrac{A}{2}\bigg)+\sin \bigg(\dfrac{B}{2}\bigg)+\sin \bigg(\dfrac{C}{2}\bigg)$

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

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

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

$\text{Double Angle:}\qquad 2\sin \bigg(\dfrac{A+B}{4}\bigg)\cdot\cos \bigg(\dfrac{A-B}{4}\bigg)+1-2\sin^2 \bigg(\dfrac{A+B}{2\cdot 2}\bigg)$

$\text{Factor:}\qquad \qquad 1+2\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg[\cos \bigg(\dfrac{A-B}{4}\bigg)-\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]$

$\text{Cofunction:}\qquad 1+2\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg[\cos \bigg(\dfrac{A-B}{4}\bigg)-\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg[\cos \bigg(\dfrac{A-B}{4}\bigg)-\sin \bigg(\dfrac{2\pi -(A+B)}{4}\bigg)\bigg]$

$\text{Sum to Product:}\quad 1+2\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \sin \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \sin \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+4\sin \bigg(\dfrac{A+B}{4}\bigg)\cdot \sin \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \sin \bigg(\dfrac{\pi -A}{4}\bigg)$

$\text{Given:}\qquad \qquad 1+4\sin \bigg(\dfrac{\pi-C}{4}\bigg)\cdot \sin \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \sin \bigg(\dfrac{\pi -A}{4}\bigg)\\\\\\.\qquad \qquad \qquad =1+4\sin \bigg(\dfrac{\pi-A}{4}\bigg)\cdot \sin \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \sin \bigg(\dfrac{\pi -C}{4}\bigg)$

LHS = RHS $\checkmark$