Question

If $A+B+C=\pi $, prove that $\cos A+\cos B-\cos C=\left( 4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2} \right)-1$.

Answer 1

Hint: For solving this question we will use some trigonometric formulas like formula for$\cos C+\cos D$, $\cos 2\theta $, $\cos \left( \dfrac{\pi }{2}-\theta \right)$ for simplifying the term written on the left-hand side. After that, we will prove it 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:

$\cos A+\cos B-\cos C=\left( 4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2} \right)-1$

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

$\begin{align}

  & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)...........................\left( 1 \right) \\

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

 & A+B+C=\pi \\

 & \Rightarrow A+B=\pi -C \\

 & \Rightarrow \left( \dfrac{A+B}{2} \right)=\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( 3 \right) \\

\end{align}$

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

Now, L.H.S is equal to $\cos A+\cos B-\cos C$ so, using the formula form equation (1). Then,

$\begin{align}

  & \cos A+\cos B-\cos C \\

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

\end{align}$

Now, using the formula from equation (3) and equation (2) in the above equation. Then,

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

$\begin{align}

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

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

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

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

\end{align}$

Now, using the formula from equation (1) in the above equation. Then,

$\begin{align}

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

 & \Rightarrow 2\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)-1 \\

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

 & \Rightarrow 4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2}-1 \\

\end{align}$

Now, from the above result, we can say that $\cos A+\cos B-\cos C=\left( 4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2} \right)-1$ .

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 less calculative, 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.