If A + B + C = π, prove that

Solution:

= cosA + cosB + cosC

Using

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

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

since A + B + C = π

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

And,

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

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

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

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

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

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

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

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

$=4 \sin \frac{A}{2} \cos \left(\frac{B}{2}\right) \cos \left(\frac{C}{2}\right)+1$

= R.H.S