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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 cos (π)/(11)cos (2π)/(11)cos (3π)/(11) cos (11π)/(11)...

The value of cosπ11cos2π11cos3π11……cos11π11cos \dfrac{\pi}{11}cos \dfrac{2\pi}{11}cos \dfrac{3\pi}{11} \ldots \ldots cos \dfrac{11\pi}{11} is equal to

Options

  1. A−132- \dfrac{1}{32}
  2. B1512\dfrac{1}{512}
  3. C11024\dfrac{1}{1024}
  4. D−12048- \dfrac{1}{2048}

Correct answer

C. 11024\dfrac{1}{1024}

Step-by-step solution

Required expression =(cosπ11cos2π11cos3π11cos4π11cos5π11)2= {\left(cos \dfrac{\pi}{11}cos \dfrac{2\pi}{11}cos \dfrac{3\pi}{11}cos \dfrac{4\pi}{11}cos \dfrac{5\pi}{11}\right)}^{2}
=(cosπ11cos2π11cos8π11cos4π11cos5π11)2= {\left(cos \dfrac{\pi}{11}cos \dfrac{2\pi}{11}cos \dfrac{8\pi}{11}cos \dfrac{4\pi}{11}cos \dfrac{5\pi}{11}\right)}^{2}
=(sin16π1116sinπ11×cos5π11)2= {\left(\dfrac{sin \dfrac{16\pi}{11}}{16sin \dfrac{\pi}{11}} \times cos \dfrac{5\pi}{11}\right)}^{2}
=(2sin5π11cos5π1132sinπ11)2= {\left(\dfrac{2sin \dfrac{5\pi}{11}cos \dfrac{5\pi}{11}}{32sin \dfrac{\pi}{11}}\right)}^{2}
=(sin10π1132×sinπ11)2=1322=11024= \left(\dfrac{sin \dfrac{10\pi}{11}}{32 \times sin \dfrac{\pi}{11}}\right)^{2} = \dfrac{1}{322} = \dfrac{1}{1024}
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