Question

If $A+B+C=\pi $, prove that ${{\cos }^{2}}\dfrac{A}{2}+{{\cos }^{2}}\dfrac{B}{2}-{{\cos }^{2}}\dfrac{C}{2}=2\cos \dfrac{A}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2}$.

Answer 1

Hint: For solving this question as there are squares of trigonometric ratios so, we will trigonometric formulas like formula for $\cos 2\theta $ and then further for simplifying the term written on the left-hand side we will use formulas for $\cos C+\cos D$ and $\cos \left( \dfrac{\pi }{2}-\theta \right)$. After that, we will prove the term on the left-hand side is equal to the term on the right-hand side.

Complete step-by-step answer:

Given:

It is given that if $A+B+C=\pi $ and we have to prove the following equation:

${{\cos }^{2}}\dfrac{A}{2}+{{\cos }^{2}}\dfrac{B}{2}-{{\cos }^{2}}\dfrac{C}{2}=2\cos \dfrac{A}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2}$

Now, before we proceed we should know the following four formulas:

$\begin{align}

  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1..............................\left( 1 \right) \\

 & \cos 2\theta =2{{\cos }^{2}}\theta -1 \\

 & \Rightarrow 2{{\cos }^{2}}\theta =\cos 2\theta +1 \\

 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{\cos 2\theta +1}{2}..........................\left( 2 \right)

\\

 & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2}

\right)................\left( 3 \right) \\

 & A+B+C=\pi \\

 & \Rightarrow A+B=\pi -C \\

 & \Rightarrow \dfrac{A+B}{2}=\dfrac{\pi }{2}-\dfrac{C}{2} \\

 & \Rightarrow \cos \left( \dfrac{A+B}{2} \right)=\cos \left( \dfrac{\pi }{2}-\dfrac{C}{2}

\right)=\sin \dfrac{C}{2}........................\left( 4 \right) \\

\end{align}$

Now, we will be using the above four formulas to simplify the term on the left-hand side to prove the desired result.

Now, L.H.S is equal to ${{\cos }^{2}}\dfrac{A}{2}+{{\cos }^{2}}\dfrac{B}{2}-{{\cos }^{2}}\dfrac{C}{2}$ so, using the formula from equation (2) and equation (1). Then,

$\begin{align}

  & {{\cos }^{2}}\dfrac{A}{2}+{{\cos }^{2}}\dfrac{B}{2}-{{\cos }^{2}}\dfrac{C}{2} \\

 & \Rightarrow \dfrac{\cos A+1}{2}+\dfrac{\cos B+1}{2}-\left( 1-{{\sin }^{2}}\dfrac{C}{2} \right)

\\

 & \Rightarrow \dfrac{\cos A+\cos B}{2}+\dfrac{1}{2}+\dfrac{1}{2}-1+{{\sin }^{2}}\dfrac{C}{2}


\\
 & \Rightarrow \dfrac{\cos A+\cos B}{2}+{{\sin }^{2}}\dfrac{C}{2} \\

\end{align}$

Now, using the formula from equation (3) in the above equation. Then,

$\begin{align}

  & \dfrac{\cos A+\cos B}{2}+{{\sin }^{2}}\dfrac{C}{2} \\

 & \Rightarrow \dfrac{1}{2}\left( 2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2}

\right) \right)+{{\sin }^{2}}\dfrac{C}{2} \\

 & \Rightarrow \cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)+{{\sin

}^{2}}\dfrac{C}{2} \\

\end{align}$

Now, using the formula from the equation (4) in the above equation. Then,

$\begin{align}

  & \cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)+{{\sin }^{2}}\dfrac{C}{2} \\

 & \Rightarrow \sin \dfrac{C}{2}\cos \left( \dfrac{A-B}{2} \right)+\sin \dfrac{C}{2}\times \cos \left( \dfrac{A+B}{2} \right) \\

 & \Rightarrow \sin \dfrac{C}{2}\left( \cos \left( \dfrac{A+B}{2} \right)+\cos \left( \dfrac{A-B}{2} \right) \right) \\

\end{align}$

Now, using the formula from the equation (3) in the above equation. Then,

$\begin{align}

  & \sin \dfrac{C}{2}\left( \cos \left( \dfrac{A+B}{2} \right)+\cos \left( \dfrac{A-B}{2} \right) \right) \\

 & \Rightarrow \sin \dfrac{C}{2}\left( 2\cos \left( \dfrac{{}^{A}/{}_{2}+{}^{B}/{}_{2}+{}^{A}/{}_{2}-{}^{B}/{}_{2}}{2} \right)\cos \left( \dfrac{{}^{A}/{}_{2}+{}^{B}/{}_{2}-{}^{A}/{}_{2}+{}^{B}/{}_{2}}{2} \right) \right) \\

 & \Rightarrow \sin \dfrac{C}{2}\left( 2\cos \dfrac{A}{2}\cos \dfrac{B}{2} \right) \\

 & \Rightarrow 2\cos \dfrac{A}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2} \\

\end{align}$

Now, from the above result, we can say that ${{\cos }^{2}}\dfrac{A}{2}+{{\cos }^{2}}\dfrac{B}{2}-{{\cos }^{2}}\dfrac{C}{2}=2\cos \dfrac{A}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2}$ .

Thus, $L.H.S=R.H.S$ .

Hence Proved.

Note: Here, the student should first understand what we have to prove in the question and then proceed in a stepwise manner while solving. For making the simplification process smooth, we should also try to make use of trigonometric results like $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta $ for making equations that will help us further in the solution. Moreover, the formulas like $\cos C+\cos D$ and $\cos 2\theta $ should be applied correctly with proper values and avoid making calculation mistakes while solving the problem.