Question

If$A,B,C$ are angles of a triangle, then $\sin 2A+\sin 2B-\sin 2C$ is equal to

• A. $4\sin A\cos BcosC$
• B. $4\cos A$
• C. $4\sin A\cos A$
• D. $4\cos A\cos B\sin C$

Answer 1

The correct option is D

$4\cos A\cos B\sin C$

Finding the value of :

Given that, $A,B,C$ are the angle of triangle.

$$\begin{gathered} \Rightarrow A+B+C=\pi \\ \Rightarrow 2A+2B+2C=2\pi \dots .\left(i\right) \end{gathered}$$

Now,

$\begin{array}{rcl}& & \sin 2A+\sin 2B-\sin 2C\left[\begin{array}{rcl}\mathbf{sinx}\mathbf{+}\mathbf{siny}& \mathbf{=}& \mathbf{2}\mathbf{sin}\frac{\mathbf{(}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{y}\mathbf{)}}{\mathbf{2}}\mathbf{cos}\frac{\mathbf{(}\mathbf{x}\mathbf{}\mathbf{–}\mathbf{}\mathbf{y}\mathbf{)}}{\mathbf{2}}\mathbf{,}\end{array}\right]\\ & =& 2sin\frac{(2A+2B)}{2}cos\frac{(2A–2B)}{2}–sin[2\pi –2(A+B\left)\right]from\left(i\right)\\ & =& 2sin(A+B)cos(A–B)+sin2(A+B)\\ & =& 2sin(A+B)cos(A–B)+2sin(A+B)cos(A+B)\\ since,sin2x& =& 2sinxcosx\\ & =& 2\sin (A+B)\left[\cos \right(A–B)+\cos (A+B\left)\right]\\ & =& 2\sin (\pi –C)\left(2\cos A\cos B\right)\\ & =& 4\cos A\cos B\sin C\end{array}$

Hence, correct option is $\left(D\right)$.