Q61 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics

In a $\triangle A B C, \sin 2 A+\sin 2 B+\sin 2 C=$

  • Correct answer
    A
    $4 \sin A \sin B \sin C$
  • B
    $2 \sin A \sin B \sin C$
  • C
    $4 \cos A \cos B \cos C$
  • D
    $2 \sin A \cos B \cos C$

Solution

In $\triangle A B C, A+B+C=\pi \Rightarrow A+B=\pi-C$ $\begin{aligned} & \text { Now, } \sin 2 A+\sin 2 B+\sin 2 C \\ & =2 \sin \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)+2 \sin C \cdot \cos C \\ & =2 \sin (A+B) \cdot \cos (A-B)+2 \sin C \cdot \cos C\end{aligned}$ $\begin{aligned} & =2 \sin (\pi-C) \cdot \cos (A-B)+2 \sin C \cdot \cos C \\ & =2 \sin C[\cos (A-B)+\cos C] \\ & =2 \sin C[\cos (A-B)+\cos \{\pi-(A+B)\}] \\ & =2 \sin C[\cos (A-B)-\cos (A+B)] \\ & =2 \sin C \times 2 \sin A \sin B=4 \sin A \cdot \sin B \cdot \sin C\end{aligned}$

Hence, the correct answer is option: 'A'.

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