Question

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

Answer 1

Hint: For solving this question we will use some trigonometric formula like formula for $\sin C+\sin D$ , $\cos C-\cos D$ and $\sin 2\theta $ 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:

$\sin A+\sin B-\sin C=4\sin \dfrac{A}{2}\sin \dfrac{B}{2}\cos \dfrac{C}{2}$

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

$\begin{align}

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

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

 & \sin 2\theta =2\sin \theta \cos \theta ...........................................................\left( 3 \right) \\

 & A+B+C=\pi \\

 & \Rightarrow A+B=\pi -C \\

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

 & \Rightarrow \sin \left( \dfrac{A+B}{2} \right)=\sin \left( \dfrac{\pi }{2}-\dfrac{C}{2} \right)=\cos \dfrac{C}{2}..............................\left( 4 \right) \\

 & \Rightarrow \cos \left( \dfrac{A+B}{2} \right)=\cos \left( \dfrac{\pi }{2}-\dfrac{C}{2} \right)=\sin \dfrac{C}{2}..............................\left( 5 \right) \\

\end{align}$

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

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

Then,

$\begin{align}

  & \sin A+\sin B-\sin C \\

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

\end{align}$

Now, using the formula from equation (4) and equation (3). Then,

$\begin{align}

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

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

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

\end{align}$

Now, using the formula form equation (5) and equation (2). Then,

$\begin{align}

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

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

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


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

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

\end{align}$

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

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 easier, 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 $\sin C+\sin D$ , $\cos C-\cos D$ and $\sin 2\theta $ should be applied correctly with proper values and avoid making calculation mistakes while solving the problem.