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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

If A=(3)/(4) , then what is the value of ((A)/(2)) ((3 A)/(2))?

If cos⁡A=34\cos A=\frac{3}{4}, then what is the value of sin⁡\sin\left(\frac{A}{2}\right) \sin (3A2)?\left(\frac{3 A}{2}\right)?

Options

  1. A58\frac{5}{8}
  2. B516\frac{5}{16}
  3. C524\frac{5}{24}
  4. D732\frac{7}{32}

Correct answer

B. 516\frac{5}{16}

Step-by-step solution

cos⁡A=34\cos A=\frac{3}{4}sin⁡(A2)⋅sin⁡(3A2)=12[2sin⁡A2sin⁡3A2]=12[cos⁡(3A2−A2)−cos⁡(3A2+A2)]=12[cos⁡A−cos⁡2A]=12[cos⁡A−(2cos⁡2A−1)]=12[cos⁡A−2cos⁡2A+1]\begin{aligned} & \sin \left(\frac{A}{2}\right) \cdot \sin \left(\frac{3 A}{2}\right)=\frac{1}{2}\left[2 \sin \frac{A}{2} \sin \frac{3 A}{2}\right] \\ & =\frac{1}{2}\left[\cos \left(\frac{3 A}{2}-\frac{A}{2}\right)-\cos \left(\frac{3 A}{2}+\frac{A}{2}\right)\right] \\ & =\frac{1}{2}[\cos A-\cos 2 A]=\frac{1}{2}\left[\cos A-\left(2 \cos ^2 A-1\right)\right] \\ & =\frac{1}{2}\left[\cos A-2 \cos ^2 A+1\right]\end{aligned}$\begin{aligned} & =\frac{1}{2}\left[\left(\frac{3}{4}\right)-2\left(\frac{3}{4}\right)^2+1\right] \\ & =\frac{1}{2}\left[\frac{3}{4}-\frac{18}{16}+1\right]=\frac{1}{2}\left[\frac{12-18+16}{16}\right]=\frac{1}{2}\left[\frac{10}{16}\right]=\frac{5}{16}\end{aligned}
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