If A + B + C = π, prove that

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

Using formula,

$\frac{1-\cos 2 A}{2}=\sin ^{2} A$

$=\frac{1-\cos 2 \mathrm{~A}}{2}-\frac{1-\cos 2 \mathrm{~B}}{2}+\frac{1-\cos 2 \mathrm{C}}{2}$

$=\frac{1-\cos 2 \mathrm{~A}-1+\cos 2 \mathrm{~B}+1-\cos 2 \mathrm{C}}{2}$

$=\frac{1-\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}-\cos 2 \mathrm{C}}{2}$

Using ,

$\cos A-\cos B=2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)$

$=\frac{1-\cos 2 \mathrm{~A}+\left\{2 \sin \left(\frac{2 \mathrm{~B}+2 \mathrm{C}}{2}\right) \sin \left(\frac{2 \mathrm{C}-2 \mathrm{~B}}{2}\right)\right\}}{2}$

$=\frac{1-\cos 2 \mathrm{~A}+2 \sin (\mathrm{B}+\mathrm{C}) \sin (\mathrm{C}-\mathrm{B})}{2}$

since A + B + C = π

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

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

$=\frac{1-\cos 2 A+2 \sin (\pi-A) \sin (C-B)}{2}$

$=\frac{1-\cos 2 A+2 \sin A \sin (C-B)}{2}$

Using, $\cos 2 A=1-2 \sin ^{2} A$

$=\frac{1-1+2 \sin ^{2} A+2 \sin A \sin (C-B)}{2}$

$=\frac{2 \sin A\{\sin A+\sin (C-B)\}}{2}$

$=\frac{2 \sin A\{\sin A+\sin (C-B)\}}{2}$

$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$

$=\frac{2 \sin A\left\{2 \sin \left(\frac{A+C-B}{2}\right) \cos \left(\frac{A-C+B}{2}\right)\right\}}{2}$

$=\frac{1-2 \sin A\left\{2 \sin \left(\frac{\pi-B-B}{2}\right) \cos \left(\frac{\pi-C-C}{2}\right)\right\}}{2}$

$=\frac{2 \sin A\left\{2 \sin \left(\frac{\pi}{2}-\frac{2 B}{2}\right) \cos \left(\frac{\pi}{2}-\frac{2 C}{2}\right)\right\}}{2}$

As, $\sin \left(\frac{\pi}{2}-\mathrm{A}\right)=\cos \mathrm{A}$

$=\frac{2 \sin A\{2 \cos B \sin C\}}{2}$

= 2sinAcosBsinC

= R.H.S