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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

The value of sin (9π)/(14)sin (11π)/(14)sin (13π)/(14) is equal to

The value of sin9π14sin11π14sin13π14sin \dfrac{9\pi}{14}sin \dfrac{11\pi}{14}sin \dfrac{13\pi}{14} is equal to

Options

  1. A164\dfrac{1}{64}
  2. B−164- \dfrac{1}{64}
  3. C18\dfrac{1}{8}
  4. D−18- \dfrac{1}{8}

Correct answer

C. 18\dfrac{1}{8}

Step-by-step solution

sin9π14sin11π14sin13π14=sin⁡(π2+π7)sin⁡(π2+2π7)sin⁡(π2+3π7)sin \dfrac{9\pi}{14}sin \dfrac{11\pi}{14}sin \dfrac{13\pi}{14} = \sin \left(\dfrac{\pi}{2} + \dfrac{\pi}{7}\right)\sin \left(\dfrac{\pi}{2} + \dfrac{2\pi}{7}\right)\sin \left(\dfrac{\pi}{2} + \dfrac{3\pi}{7}\right)
=cosπ7cos2π7cos3π7= cos \dfrac{\pi}{7}cos \dfrac{2\pi}{7}cos \dfrac{3\pi}{7}
=cosπ7cos2π7cos⁡(π−4π7)= cos \dfrac{\pi}{7}cos \dfrac{2\pi}{7}\cos \left(\pi - \dfrac{4\pi}{7}\right)
=−cosπ7cos2π7cos4π7= - cos \dfrac{\pi}{7}cos \dfrac{2\pi}{7}cos \dfrac{4\pi}{7}
=−sin8π78sinπ7=−sin⁡(π+π7)8sinπ7= \dfrac{- sin \dfrac{8\pi}{7}}{8sin \dfrac{\pi}{7}} = \dfrac{- \sin \left(\pi + \dfrac{\pi}{7}\right)}{8sin \dfrac{\pi}{7}}
=18= \dfrac{1}{8}
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