Question

In a triangle $ABC,$ the value of $\sin \left(2A\right)+\sin \left(2B\right)+\sin \left(2C\right)$ is equal to:

• A. $4\sin \left(A\right)\sin \left(B\right)\sin \left(C\right)$
• B. $4\cos \left(A\right)\cos \left(B\right)\cos \left(C\right)$
• C. $2\cos \left(A\right)\cos \left(B\right)\cos \left(C\right)$
• D. $2\sin \left(A\right)\sin \left(B\right)\sin \left(C\right)$

Answer 1

The correct option is A

$4\sin \left(A\right)\sin \left(B\right)\sin \left(C\right)$

Finding the value of $\sin \left(2A\right)+\sin \left(2B\right)+\sin \left(2C\right)$ in a triangle $ABC.$

In triangle $ABC,$ $A+B+C=180°$ ( sum of all the interior angles in any triangle is $180°$)

So, $2A+2B+2C=360°$

Now, solving for $\sin \left(2A\right)+\sin \left(2B\right)+\sin \left(2C\right)$:

$$\begin{gathered} \Rightarrow \sin \left(2A\right)+\sin \left(2B\right)+\sin \left(2C\right)=2\sin \frac{\left(2A+2B\right)}{2}\cos \frac{\left(2A-2B\right)}{2}+\sin \left(2C\right) \\ \Rightarrow =2\sin \left(A+B\right)\cos \left(A-B\right)+\sin \left(2C\right) \\ \Rightarrow =2\sin \left(A+B\right)\cos \left(A-B\right)+\sin \left(360°-2(A+B)\right) \\ \Rightarrow =2\sin \left(A+B\right)\cos \left(A-B\right)-\sin 2\left(A+B\right) \\ \Rightarrow =2\sin \left(A+B\right)\cos \left(A-B\right)-2\sin \left(A+B\right)\cos \left(A+B\right) \\ \Rightarrow =2\sin \left(A+B\right)(\cos \left(A-B\right)-\cos \left(A+B\right)) \\ \Rightarrow =2\sin \left(A+B\right)(2\sin \left(A\right)\sin \left(B\right)) \\ \Rightarrow =4\sin \left(180°-C\right)2\sin \left(A\right)\sin \left(B\right) \\ \Rightarrow =4\sin \left(C\right)\sin \left(A\right)\sin \left(B\right) \\ \Rightarrow \sin \left(2A\right)+\sin \left(2B\right)+\sin \left(2C\right)=4\sin \left(A\right)\sin \left(B\right)\sin \left(C\right) \end{gathered}$$

Hence, Option (C) is correct.