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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2018 · Mathematics · Trigonometric Ratios & Identities

If A+B+C=180^ ^ , then what is 2 A- 2 B- 2 C equal to?

If A+B+C=180∘\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{{}^{\circ}}, then what is sin⁡2A−sin⁡2B−sin⁡2C\sin 2 \mathrm{ A}-\sin 2 \mathrm{ B}-\sin 2 \mathrm{C} equal to?

Options

  1. A−4sin⁡Asin⁡Bsin⁡C-4 \sin \mathrm{ A} \sin \mathrm{ B} \sin \mathrm{C}
  2. B−4cos⁡Asin⁡Bcos⁡C-4 \cos \mathrm{ A} \sin \mathrm{ B} \cos \mathrm{C}
  3. C−4cos⁡Acos⁡Bsin⁡C-4 \cos \mathrm{ A} \cos \mathrm{ B} \sin \mathrm{C}
  4. D−4sin⁡Acos⁡Bcos⁡C-4 \sin A \cos B \cos C

Correct answer

D. −4sin⁡Acos⁡Bcos⁡C-4 \sin A \cos B \cos C

Step-by-step solution

∵sin⁡2A−sin⁡2B−sin⁡2C=2cos⁡(A+B)⋅sin⁡(A−B)+sin⁡(2A+2B)=2cos⁡(A+B)sin⁡(A−B)+2sin⁡(A+B)cos⁡(A+B)=2cos⁡(A+B)[sin⁡(A−B)+sin⁡(A+B)]=−2cos⁡C2sin⁡A⋅cos⁡B]=−4sin⁡Acos⁡Bcos⁡C\begin{aligned} & \because \sin 2 A-\sin 2 B-\sin 2 C \\ & =2 \cos (A+B) \cdot \sin (A-B)+\sin (2 A+2 B) \\ & =2 \cos (A+B) \sin (A-B)+2 \sin (A+B) \cos (A+B) \\ & =2 \cos (A+B)[\sin (A-B)+\sin (A+B)] \\ & =-2 \cos C 2 \sin A \cdot \cos B] \\ & =-4 \sin A \cos B \cos C\end{aligned}
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