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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2015 · Mathematics · Trigonometric Ratios & Identities

Let α be the root of the equation 25 ^2 θ+5 θ-12=0 , where (π)/(2)...

Let α\alpha be the root of the equation 25cos⁡2θ+5cos⁡θ−12=025 \cos ^2 \theta+5 \cos \theta-12=0, where π2<α<π\frac{\pi}{2} < \alpha < \pi. What is sin⁡2α\sin 2 \alpha equal to?

Options

  1. A2425\frac{24}{25}
  2. B−2425\frac{-24}{25}
  3. C−512\frac{-5}{12}
  4. D−2125\frac{-21}{25}

Correct answer

B. −2425\frac{-24}{25}

Step-by-step solution

sin⁡2α=2sin⁡α⋅cos⁡α=2(35)(−45)=65×−45=−2425\begin{aligned} & \sin 2 \alpha=2 \sin \alpha \cdot \cos \alpha \\ & =2\left(\frac{3}{5}\right)\left(\frac{-4}{5}\right) \\ & =\frac{6}{5} \times \frac{-4}{5}=\frac{-24}{25}\end{aligned}\therefore \quad$ Option (B) is correct.
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