Question

Prove that $bc{\cos }^2\frac{A}{2}+ca{\cos }^2\frac{B}{2}+ab{\cos }^2\frac{C}{2}=s^2$.

Answer 1

L.H.S $=bc{\cos }^2\frac{A}{2}+ca{\cos }^2\frac{B}{2}+ab{\cos }^2\frac{C}{2}$
$=\frac{1}{2}(bc{\cos }^2\frac{A}{2}+ca{\cos }^2\frac{B}{2}+ab{\cos }^2\frac{C}{2})$
$=\frac{1}{2}\left[bc\right(1+\cos A)+ca(1+\cos B)+ab(\cos C\left)\right]$
$=\frac{1}{2}[bc+ca+ab+b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2]$ ...[ Cosine rule: $\cos A=\frac{b^2+c^{-}{a}^2}{2bc}$]
$=\frac{1}{4}[a^2+b^2+c^2+2ab+2bc+2ca]$
$=\frac{1}{4}{(a+b+c)}^2=s^2=$R.H.S