Question

In ΔABC, show that $b\ cos^{2}(\frac{C}{2})\ +\ c\ cos^{2}(\frac{B}{2}) = \ s}$

Answer 1

Answer:

Step-by-step explanation:

Formula used:

Projection formula:

In triangle ABC,

a = b cosC + c cosB

Half perimeter of triangle ABC is

s =(a+b+c)/2

$b{cos}^2(\frac{C}{2})+c{cos}^2(\frac{B}{2})$

$=b[\frac{1+cosC}{2}]+c[\frac{1+cosB}{2}]$

$=\frac{b+bcosC+c+ccosB}{2}$

$=\frac{(bcosCccosB)+b+c}{2}$

$=\frac{a+b+c}{2}$

$=s$