Question

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

Answer 1

Hint: We will be using the concept of trigonometry and solution of triangles to solve the problem. We will be using cosine rules to relate the side of the triangles with the cosine of the angle of the triangle then we will use these values in the left hand side of the equation to prove it to be equal to the right hand side of the equation.

Complete Step-by-Step solution:
We have been given a$\Delta ABC$ and we have to prove that,$4\left( bc{{\cos }^{2}}\dfrac{A}{2}+ca{{\cos }^{2}}\dfrac{B}{2}+ab{{\cos }^{2}}\dfrac{C}{2} \right)={{\left( a+b+c \right)}^{2}}$.
Now, we will draw a $\Delta ABC$ and label its sides as a, b, c.

We will now take L.H.S of the question and prove it to be equal to the R.H.S. In L.H.S we have,
$4\left( bc{{\cos }^{2}}\dfrac{A}{2}+ca{{\cos }^{2}}\dfrac{B}{2}+ab{{\cos }^{2}}\dfrac{C}{2} \right)$
Now, we know the trigonometry formula that,
$\cos 2A=2{{\cos }^{2}}A-1$
So, we have,
\[\begin{align}
  & {{\cos }^{2}}\left( \dfrac{A}{2} \right)=\dfrac{1+\cos A}{2}..........\left( 1 \right) \\
 & {{\cos }^{2}}\left( \dfrac{B}{2} \right)=\dfrac{1+\cos B}{2}..........\left( 2 \right) \\
 & {{\cos }^{2}}\left( \dfrac{C}{2} \right)=\dfrac{1+\cos C}{2}..........\left( 3 \right) \\
\end{align}\]
Now, we will substitute the value of \[{{\cos }^{2}}\left( \dfrac{A}{2} \right),{{\cos }^{2}}\left( \dfrac{B}{2} \right)\ and\ {{\cos }^{2}}\left( \dfrac{C}{2} \right)\] from (1), (2) and (3) we have,
$4\left( bc{{\cos }^{2}}\dfrac{A}{2}+ca{{\cos }^{2}}\dfrac{B}{2}+ab{{\cos }^{2}}\dfrac{C}{2} \right)=4\left( bc\dfrac{\left( 1+\cos A \right)}{2}+ca\dfrac{\left( 1+\cos B \right)}{2}+ab\dfrac{\left( 1+\cos C \right)}{2} \right)$
Now, we will take 2 as L.C.M and expand the numerator,
$\begin{align}
  & 4\left( bc{{\cos }^{2}}\dfrac{A}{2}+ca{{\cos }^{2}}\dfrac{B}{2}+ab{{\cos }^{2}}\dfrac{C}{2} \right)=\dfrac{4}{2}\left( bc+bc\cos A+ca+ca\cos B+ab+ab\cos c \right) \\
 & =2\left( ab+bc+ca+bc\cos A+ca\cos B+ab\cos c \right) \\
\end{align}$
Now, we know that according to cosine rule,
$\begin{align}
  & \cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}............\left( 4 \right) \\
 & \cos B=\dfrac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ac}............\left( 5 \right) \\
 & \cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}............\left( 6 \right) \\
\end{align}$
Now, we will substitute the value of \[\cos A,\cos B\ and\ \cos C\] from (1), (2) and (3) we have,
$\begin{align}
  & =2\left( ab+bc+ca+bc\dfrac{\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)}{2bc}+ca\dfrac{\left( {{c}^{2}}+{{a}^{2}}-{{b}^{2}} \right)}{2ac}+ab\dfrac{\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}{2ab} \right) \\
 & =2\left( ab+bc+ca+\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2}+\dfrac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2}+\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2} \right) \\
 & =2\left( ab+bc+ca+\dfrac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{2} \right) \\
\end{align}$
Now, we will expand it, we have,
$\begin{align}
  & =2ab+2bc+2ca+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca \\
\end{align}$
Now, we know that,
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$
Therefore, we have,
$4\left( bc{{\cos }^{2}}\dfrac{A}{2}+ca{{\cos }^{2}}\dfrac{B}{2}+ab{{\cos }^{2}}\dfrac{C}{2} \right)={{\left( a+b+c \right)}^{2}}$
Since, L.H.S = R.H.S, Hence proved.

Note: To solve these types of questions it is important to remember that we have used the cosine rule to relate the side of the triangle with the cosine of the angle of the triangle. It is also advised that one must remember the cosine rule and trigonometric identities like $\cos 2A=2{{\cos }^{2}}A-1$.