If A + B + C = π, prove that

Solution:

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

Using formula

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

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

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

Since $A+B+C=\pi$

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

$=\cos 2 \mathrm{~A}-\{2 \cos (\pi-\mathrm{A}) \cos (\mathrm{B}-\mathrm{C})\}$

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

$=\cos 2 A-\{-2 \cos A \cos (B-C)\}$

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

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

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

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

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

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

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

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

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

$=2 \cos A\{2 \sin C \sin B\}-1$

$=4 \cos A \sin B \sin C-1$

= R.H.S