If $A, B, C$ are the angles of triangle then $\sin 2 A-\sin 2 B+$ $\sin 2 C=$

If $A, B, C$ are the angles of triangle then $\sin 2 A-\sin 2 B+$ $\sin 2 C=$
  1. $4 \cos A \cos B \sin C$
  2. $4 \cos A \sin B \cos C$
  3. $4 \cos A \sin B \sin C-1$
  4. $4 \sin A \cos B \sin C$

Solution

Given that $A, B, C$ are the angle of triangle $\Rightarrow A+B+C=\pi \Rightarrow 2 A+2 B+2 C=2 \pi$...(i) Now, $\begin{aligned} & \sin 2 A-\sin 2 B+\sin 2 C=2 \sin \frac{2 A+2 C}{2} \cos \left(\frac{2 A-2 C}{2}\right) \\ & -\quad \sin [2 \pi-2(A+C)] \\ & =2 \sin (A+C) \cos (A-C)+\sin 2(A+C) \\ & =2 \sin (A+C) \cos (A-C)+2 \sin (A+C) \cos (A+C) \\ & =2 \sin (A+C)[\cos (A-C)+\cos (A+C)] \\ & =2 \sin (\pi-B)[2 \cos A \cos C]=4 \cos A \sin B \cos C . \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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