Question

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

Answer 1

Hint: For solving this question we will use some trigonometric formula like formula for $\sin C-\sin D$ and $\sin \left( \pi -\theta \right)$ 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:

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

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

$\begin{align}

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

 & A+B+C=\pi \\

 & \Rightarrow A+B=\pi -C \\

 & \Rightarrow \sin \left( A+B \right)=\sin \left( \pi -C \right)=\sin C..................................\left( 2 \right) \\

 & \Rightarrow \cos \left( A+B \right)=\cos \left( \pi -C \right)=-\cos C..............................\left( 3 \right) \\

 & \sin 2\theta =2\sin \theta \cos \theta ..................................................\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 $\sin 2A-\sin 2B+\sin 2C$ so, using the formula from equation (1).

Then,

$\begin{align}

  & \sin 2A-\sin 2B+\sin 2C \\

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

\\

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

\end{align}$

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

$\begin{align}

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

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

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

\end{align}$

Now, using the formula from equation (2) and equation (1) in the above equation. Then,

$\begin{align}

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

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

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

\dfrac{A+B-A+B}{2} \right) \right) \\

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

 & \Rightarrow 4\cos A\sin B\cos C \\

\end{align}$

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

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. We should also try to make use of trigonometric results like $\sin \left( \pi -\theta \right)=\sin \theta $ for making equations that will help us further in the solution. Moreover, the formula $\sin C-\sin D$ should be applied correctly with proper values and avoid making calculation mistakes while solving the problem.