Question

If the value of the sum of the angles A, B and C is given by $A+B+C=\pi $, then prove that
${{\cos }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{B}{2} \right)+{{\cos }^{2}}\left( \dfrac{C}{2} \right)=2\left( 1+\sin \left( \dfrac{A}{2} \right)\sin \left( \dfrac{B}{2} \right)\sin \left( \dfrac{C}{2} \right) \right)$

Answer 1

Hint: In this question, we should use the relation between ${{\sin }^{2}}\left( \theta \right)$ and \[{{\cos }^{2}}\left( \theta \right)\] and the formula for the ${{\cos }^{2}}(\theta )-{{\sin }^{2}}(\theta )$ to obtain the solution to this question.

Complete step-by-step answer:
It is given that $A+B+C=\pi $. Also we know that for any angle $\theta $,
\[\begin{align}
  & {{\cos }^{2}}\left( \theta \right)+{{\sin }^{2}}\left( \theta \right)=1 \\
 & \Rightarrow {{\cos }^{2}}\left( \theta \right)=1-{{\sin }^{2}}\left( \theta \right)..............(1.1) \\
\end{align}\]
We can use equation (1.1) in the expression given in the question to obtain
$\begin{align}
  & {{\cos }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{B}{2} \right)+{{\cos }^{2}}\left( \dfrac{C}{2} \right)={{\cos }^{2}}\left( \dfrac{A}{2} \right)+\left( 1-{{\sin }^{2}}\left( \dfrac{B}{2} \right) \right)+{{\cos }^{2}}\left( \dfrac{C}{2} \right) \\
 & =1+\left( {{\cos }^{2}}\left( \dfrac{A}{2} \right)-{{\sin }^{2}}\left( \dfrac{B}{2} \right) \right)+{{\cos }^{2}}\left( \dfrac{C}{2} \right)................(1.2) \\
\end{align}$
Now, we can use the formula
${{\cos }^{2}}\left( x \right)-{{\sin }^{2}}\left( y \right)=\cos (x+y)\cos (x-y).....(1.2a)$
In equation (1.2) to obtain
$\begin{align}
  & {{\cos }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{B}{2} \right)+{{\cos }^{2}}\left( \dfrac{C}{2} \right)=1+\left( {{\cos }^{2}}\left( \dfrac{A}{2} \right)-{{\sin }^{2}}\left( \dfrac{B}{2} \right) \right)+{{\cos }^{2}}\left( \dfrac{C}{2} \right) \\
 & =1+\left( \cos \left( \dfrac{A}{2}+\dfrac{B}{2} \right)\cos \left( \dfrac{A}{2}-\dfrac{B}{2} \right) \right)+1-{{\sin }^{2}}\left( \dfrac{C}{2} \right)................(1.3) \\
\end{align}$
Now, it is given that
$A+B+C=\pi \Rightarrow \dfrac{A}{2}+\dfrac{B}{2}=\dfrac{\pi }{2}-\dfrac{C}{2}.........(1.4)$
Also, we can use the formula that
$\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \left( \theta \right)$
In equation (1.4) to obtain
$\cos \left( \dfrac{A}{2}+\dfrac{B}{2} \right)=\sin \left( \dfrac{C}{2} \right)..........(1.5)$
Now, using equation (1.5), we can simplify equation 1.3 as
$\begin{align}
  & {{\cos }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{B}{2} \right)+{{\cos }^{2}}\left( \dfrac{C}{2} \right)=1+\left( \sin \left( \dfrac{C}{2} \right)\cos \left( \dfrac{A}{2}-\dfrac{B}{2} \right) \right)+1-{{\sin }^{2}}\left( \dfrac{C}{2} \right) \\
 & =2+\sin \left( \dfrac{C}{2} \right)\left( \cos \left( \dfrac{A}{2}-\dfrac{B}{2} \right)-\sin \left( \dfrac{C}{2} \right) \right) \\
 & =2+\sin \left( \dfrac{C}{2} \right)\left( \cos \left( \dfrac{A}{2}-\dfrac{B}{2} \right)-\cos \left( \dfrac{A+B}{2} \right) \right)................(1.4) \\
\end{align}$
Now, we can use the sum of the cosines of sum and difference of the angles as
$\cos (a-b)-\cos (a+b)=2\sin (a)\sin (b).......(1.5)$
and taking $b=\dfrac{B}{2}$$a=\dfrac{A}{2}$ and using equation (1.5) in equation (1.4), we obtain
 $\begin{align}
  & {{\cos }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{B}{2} \right)+{{\cos }^{2}}\left( \dfrac{C}{2} \right)=2+\sin \left( \dfrac{C}{2} \right)\left( \cos \left( \dfrac{A}{2}-\dfrac{B}{2} \right)-\cos \left( \dfrac{A+B}{2} \right) \right) \\
 & =2+\sin \left( \dfrac{C}{2} \right)\left( 2\sin \left( \dfrac{A}{2} \right)\sin \left( \dfrac{B}{2} \right) \right) \\
 & =2\left( 1+\sin \left( \dfrac{A}{2} \right)\sin \left( \dfrac{B}{2} \right)\sin \left( \dfrac{C}{2} \right) \right)................(1.6) \\
\end{align}$
We see that equation (1.6) is exactly the equation we wanted to prove. Therefore, we have successfully proved the given equation.

Note: We should be careful to use the equations for the cosine and sine with the proper signs in equation (1.2a) as the difference should be between ${{\cos }^{2}}(\theta )$ and \[{{\sin }^{2}}(\theta )\] and not in the opposite order as it will result in an overall negative sign even if the cos function is an even function. Also, in equation (1.3), we have used equation (1.1) again to simplify the last term as then the sin term would come out as a common factor in the subsequent steps.