Question

In ∆ABC, prove the following:
$a^2={\left(b+c\right)}^2-4bc{\cos }^2\frac{A}{2}$

Answer 1

$$\begin{gathered} \mathrm{RHS}={\left(b+c\right)}^2-4bc{\cos }^2\frac{A}{2} \\ =b^2+c^2+2bc-4bc\left(\frac{1+\cos A}{2}\right) \\ =b^2+c^2+2bc-2bc\left(1+\cos A\right) \\ =b^2+c^2+2bc\left(1-1-\cos A\right) \\ =b^2+c^2-2bc\cos A \\ =b^2+c^2-2bc\left(\frac{b^2+c^2-a^2}{2bc}\right)\left(\because \cos A=\frac{b^2+c^2-a^2}{2bc}\right) \\ =b^2+c^2-b^2-c^2+a^2 \\ =a^2=\mathrm{LHS} \end{gathered}$$

Hence proved.