For \(A, B\) and \(C\), if \(A+B+C=0\), then \(\sin (2 A)+\sin (2 B)+\sin (2 C)\) is equal to

For \(A, B\) and \(C\), if \(A+B+C=0\), then \(\sin (2 A)+\sin (2 B)+\sin (2 C)\) is equal to
  1. \(4 \sin (A) \cdot \sin (B) \cdot \sin (C)\)
  2. \(2 \sin (A) \cdot \sin (B) \cdot \sin (C)\)
  3. \(-4 \sin (A) \cdot \sin (B) \cdot \sin (C)\)
  4. \(-2 \sin (A) \cdot \sin (B) \cdot \sin (C)\)

Solution

For \(A+B+C=0\) \(\begin{aligned} \sin 2 A+ & \sin 2 B+\sin 2 C \\ & =2 \sin (A+B) \cos (A-B)+2 \sin C \cos C \\ & =-2 \sin C \cos (A-B)+2 \sin C \cos (A+B) \\ & =-2 \sin C[\cos (A-B)-\cos (A+B)] \\ & =-2 \sin C[2 \sin A \sin B] \\ & =-4 \sin A \sin B \sin C. \end{aligned}\)

Asked in: AP EAMCET 2020 (18 Sep Shift 1)

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