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JEE Main 202225 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
25 July
Shift
Evening Shift
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2022 · 25 July · Evening Shift · Mathematics · Trigonometric Ratios & Identities

The value of...

The value of 2sinπ22sin3π22sin5π22sin7π22sin9π222sin\dfrac{\pi}{22}sin\dfrac{3\pi}{22}sin\dfrac{5\pi}{22}sin\dfrac{7\pi}{22}sin\dfrac{9\pi}{22} is equal to:

Options

  1. A116\dfrac{1}{16}
  2. B516\dfrac{5}{16}
  3. C716\dfrac{7}{16}
  4. D316\dfrac{3}{16}

Correct answer

A. 116\dfrac{1}{16}

Step-by-step solution

Given, 2sinπ22sin3π22sin5π22sin7π22sin9π222sin\dfrac{\pi}{22}sin\dfrac{3\pi}{22}sin\dfrac{5\pi}{22}sin\dfrac{7\pi}{22}sin\dfrac{9\pi}{22}

Now by we know that sin⁡(θ)=cos⁡(π2−θ)\sin \left(\theta\right) = \cos \left(\dfrac{\pi}{2} - \theta\right), so by using this we get sinπ22=cos⁡(π2−π22)=cos5π11sin\dfrac{\pi}{22} = \cos \left(\dfrac{\pi}{2} - \dfrac{\pi}{22}\right) = cos\dfrac{5\pi}{11} and similarly sin3π22=cos4π11,sin5π22=cos3π11,sin7π22=2π11,sin9π22=cosπ11sin\dfrac{3\pi}{22} = cos\dfrac{4\pi}{11}, sin\dfrac{5\pi}{22} = cos\dfrac{3\pi}{11}, sin\dfrac{7\pi}{22} = \dfrac{2\pi}{11}, sin\dfrac{9\pi}{22} = cos\dfrac{\pi}{11}

So, 2sinπ22sin3π22sin5π22sin7π22sin9π222sin\dfrac{\pi}{22}sin\dfrac{3\pi}{22}sin\dfrac{5\pi}{22}sin\dfrac{7\pi}{22}sin\dfrac{9\pi}{22}

=2cos5π11cos4π11cos3π11cos2π11cos1π11= 2cos\dfrac{5\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{1\pi}{11}

=2cos1π11cos2π11cos4π11cos8π11cos16π11= 2cos\dfrac{1\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{8\pi}{11}cos\dfrac{16\pi}{11}

{As cos3π22=−cos⁡(π−3π11)=−cos8π11cos\dfrac{3\pi}{22} = - \cos \left(\pi - \dfrac{3\pi}{11}\right) = - cos\dfrac{8\pi}{11} and cos5π11=−cos16π11cos\dfrac{5\pi}{11} = - cos\dfrac{16\pi}{11}}

=2sinπ112sinπ11cosπ11cos2π11cos4π11cos8π11cos16π11= \dfrac{2sin\dfrac{\pi}{11}}{2sin\dfrac{\pi}{11}}cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{8\pi}{11}cos\dfrac{16\pi}{11}

=2sin4π11cos4π11cos8π11cos16π1122sinπ11= \dfrac{2sin\dfrac{4\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{8\pi}{11}cos\dfrac{16\pi}{11}}{{2}^{2}sin\dfrac{\pi}{11}}

=2sin8π11cos8π11cos16π1123sinπ11= \dfrac{2sin\dfrac{8\pi}{11}cos\dfrac{8\pi}{11}cos\dfrac{16\pi}{11}}{{2}^{3}sin\dfrac{\pi}{11}}

=2sin16π11cos16π1124sinπ11= \dfrac{2sin\dfrac{16\pi}{11}cos\dfrac{16\pi}{11}}{{2}^{4}sin\dfrac{\pi}{11}}

=sin32π1124sinπ11= \dfrac{sin\dfrac{32\pi}{11}}{{2}^{4}sin\dfrac{\pi}{11}}

=116sin⁡(3π−π11)sinπ11=116= \dfrac{1}{16}\dfrac{\sin \left(3\pi - \dfrac{\pi}{11}\right)}{sin\dfrac{\pi}{11}} = \dfrac{1}{16}

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