Search any question & find its solution
Question:
Answered & Verified by Expert
In a $\triangle A B C, \sin 2 A+\sin 2 B+\sin 2 C=$
Options:
Solution:
2846 Upvotes
Verified Answer
The correct answer is:
$4 \sin A \sin B \sin C$
In $\triangle A B C, A+B+C=\pi \Rightarrow A+B=\pi-C$
$\begin{aligned} & \text { Now, } \sin 2 A+\sin 2 B+\sin 2 C \\ & =2 \sin \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)+2 \sin C \cdot \cos C \\ & =2 \sin (A+B) \cdot \cos (A-B)+2 \sin C \cdot \cos C\end{aligned}$
$\begin{aligned} & =2 \sin (\pi-C) \cdot \cos (A-B)+2 \sin C \cdot \cos C \\ & =2 \sin C[\cos (A-B)+\cos C] \\ & =2 \sin C[\cos (A-B)+\cos \{\pi-(A+B)\}] \\ & =2 \sin C[\cos (A-B)-\cos (A+B)] \\ & =2 \sin C \times 2 \sin A \sin B=4 \sin A \cdot \sin B \cdot \sin C\end{aligned}$
$\begin{aligned} & \text { Now, } \sin 2 A+\sin 2 B+\sin 2 C \\ & =2 \sin \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)+2 \sin C \cdot \cos C \\ & =2 \sin (A+B) \cdot \cos (A-B)+2 \sin C \cdot \cos C\end{aligned}$
$\begin{aligned} & =2 \sin (\pi-C) \cdot \cos (A-B)+2 \sin C \cdot \cos C \\ & =2 \sin C[\cos (A-B)+\cos C] \\ & =2 \sin C[\cos (A-B)+\cos \{\pi-(A+B)\}] \\ & =2 \sin C[\cos (A-B)-\cos (A+B)] \\ & =2 \sin C \times 2 \sin A \sin B=4 \sin A \cdot \sin B \cdot \sin C\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.