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  4. In any triangle A B C cos 2 (A/2)+ cos 2 (B/2)+ cos 2 (C/2)=

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Q. In any triangle $A B C$ $\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2}=$

TS EAMCET 2020

Solution:

We have,
$\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2} $
$=\frac{1}{2}\{(1+\cos A)+(1+\cos B)+(1+\cos C)\} $
$=\frac{1}{2}\{3+\cos A+\cos B+\cos C\} $
$=\frac{1}{2}\left\{3+1+\frac{r}{R}\right\}\left[\because \cos A+\cos B+\cos C=1+\frac{r}{R}\right] $
$=\frac{1}{2}\left(4+\frac{r}{R}\right)=2+\frac{r}{2 R} \cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2} $
$=\frac{1}{2}\{(1+\cos A)+(1+\cos B)+(1+\cos C)\}$
$=\frac{1}{2}\{3+\cos A+\cos B+\cos C\} $
$=\frac{1}{2}\left\{3+1+\frac{r}{R}\right\}\left[\because \cos A+\cos B+\cos C=1+\frac{r}{R}\right]$
$=\frac{1}{2}\left(4+\frac{r}{R}\right)=2+\frac{r}{2 R}$