If A + B + C = π, prove that

Solution:

$=\sin 2 A+\sin 2 B-\sin 2 C$

$=2 \sin (B+C) \cos A+2 \sin (A+C) \cos B-2 \sin (A+B) \cos C$

Using formula, $\sin (A+B)=\sin A \cos B+\cos A \sin B$

$=\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}-\sin 2 \mathrm{C}$

Using formula

sin2A = 2sinAcosA

= 2sinAcosA + 2sinBcosB - 2sinCcosC

Since A + B + C = π

$\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$

And $\sin (\pi-A)=\sin A$

$=2 \sin (B+C) \cos A+2 \sin (A+C) \cos B-2 \sin (A+B) \cos C$

$=2(\sin B \cos C+\cos B \sin C) \cos A+2(\sin A \cos C+\cos A \sin C) \cos B-2(\sin A \cos B+$

$\cos A \sin B) \cos C$

$=2 \cos A \sin B \cos C+2 \cos A \cos B \sin C+2 \sin A \cos B \cos C+2 \cos A \cos B \sin C-$

$2 \sin A \cos B \cos C-2 \cos A \sin B \cos C$

$=2 \cos A \cos B \sin C+2 \cos A \cos B \sin C$

$=4 \cos A \cos B \sin C$

= R.H.S