Question

In $△ABC$ prove that $\cos A+\cos B+\cos C=1+\frac{r}{R}$

Answer 1

$\cos A+\cos B+\cos C=1+\frac{r}{R}$

Consider, $\cos A+\cos B+\cos C$

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

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

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

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

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

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

$=1+2\sin \frac{C}{2}\left[2\sin \frac{A}{2}\sin \frac{B}{2}\right]$

$=1+4\sin \frac{C}{2}\sin \frac{A}{2}\sin \frac{B}{2}$

$=1+4\left[\sqrt{\frac{(s-a)(s-b)}{ab}}\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-c)}{ac}}\right]$ {multiple and divide with $s^2$}

$1+4\left[\sqrt{\frac{\left(s\right(s-a\left)\right(s-b\left)\right(s-c){)}^2}{a^2{b}^2{c}^2{s}^2}}\right]$

$=\frac{1+4\left(s\right(s-a\left)\right(s-b\left)\right(s-c)}{sabc}$

$=1+4\frac{{△}^2}{sabc}$

$=1+\left(\frac{△}{s}\right)\left(\frac{4△}{abc}\right)$

$=1+\frac{r}{R}$.