Question

In any $\Delta ABC$, prove that $4(bc{\cos }^2\frac{A}{2}+ca{\cos }^2\frac{B}{2}+ab{\cos }^2\frac{C}{2})=(a+b+c{)}^2$.

Answer 1

Given that:- $ABC$ is a triangle.

To prove:- $4(bc{\cos }^2\frac{A}{2}+ca{\cos }^2\frac{B}{2}+ab{\cos }^2\frac{C}{2})={(a+b+c)}^2$

Proof:-

Taking L.H.S.-

$4(bc{\cos }^2\frac{A}{2}+ca{\cos }^2\frac{B}{2}+ab{\cos }^2\frac{C}{2})$

Using identity $\cos 2x=2{\cos }^{2}x-1\Rightarrow {\cos }^{2}x=\frac{1}{2}(1+\cos 2x)$

$4(bc{\cos }^2\frac{A}{2}+ca{\cos }^2\frac{B}{2}+ab{\cos }^2\frac{C}{2})=4(\frac{1}{2}bc(1-\cos A)+\frac{1}{2}ca(1-\cos B)+\frac{1}{2}ab(1-\cos C))$

$=2bc+2bc\cos A+2ca+2ca\cos B+2ab+2ab\cos C$

Using the property-

$2bc\cos A=b^2+c^2-a^2$

$2ac\cos B=a^2+c^2-b^2$

$2ab\cos C=a^2+b^2-c^2$

$\therefore 4(bc{\cos }^2\frac{A}{2}+ca{\cos }^2\frac{B}{2}+ab{\cos }^2\frac{C}{2})=2bc+(b^2+c^2-a^2)+2ca+(a^2+c^2-b^2)+2ab+(a^2+b^2-c^2)$

$=2(ab+bc+ca)+(a^2+b^2+c^2)$

$={(a+b+c)}^2$

$=$ R.H.S.

Hence proved that $4(bc{\cos }^2\frac{A}{2}+ca{\cos }^2\frac{B}{2}+ab{\cos }^2\frac{C}{2})={(a+b+c)}^2$.