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JEE Main 20238 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2023 · 8 April · Evening Shift · Mathematics · Trigonometric Ratios & Identities

The value of 36 (4 ^ 2 9 ^ ∘ - 1) (4 ^ 2 27 ^ ∘ - 1)(4 ^ 2 81 ^ ∘ -...

The value of 3636\left(4\cos^{2}{9}^{∘} - 1\right)(4cos⁡227∘−1)(4cos⁡281∘−1)(4cos⁡2243∘−1)\left(4\cos^{2}{27}^{∘} - 1\right)\left(4\cos^{2}{81}^{∘} - 1\right)\left(4\cos^{2}{243}^{∘} - 1\right) is

Options

  1. A5454
  2. B1818
  3. C2727
  4. D3636

Correct answer

D. 3636

Step-by-step solution

We need to find the value of

36(4cos⁡29∘−1)(4cos⁡227∘−1)(4cos⁡281∘−1)(4cos⁡2243∘−1)36\left(4\cos^{2}{9}^{∘} - 1\right)\left(4\cos^{2}{27}^{∘} - 1\right)\left(4\cos^{2}{81}^{∘} - 1\right)\left(4\cos^{2}{243}^{∘} - 1\right)

=36[(4cos⁡29∘−1)(4sin⁡29∘−1)(4cos⁡227∘−1)(4sin⁡227∘−1)]= 36\left[\left(4\cos^{2}{9}^{∘} - 1\right)\left(4\sin^{2}{9}^{∘} - 1\right)\left(4\cos^{2}{27}^{∘} - 1\right)\left(4\sin^{2}{27}^{∘} - 1\right)\right]

=36[(4(2cos9∘sin⁡9°)2−4cos⁡29°−4sin⁡29°+1)(4(2sin27∘cos⁡9°)2−4cos⁡227°−4sin⁡227°+1)]= 36\left[\left(4{\left(2cos{9}^{∘}\sin 9 °\right)}^{2} - 4\cos^{2}9 ° - 4\sin^{2}9 ° + 1\right)\left(4{\left(2sin{27}^{∘}\cos 9 °\right)}^{2} - 4\cos^{2}27 ° - 4\sin^{2}27 ° + 1\right)\right]

=36[(4sin⁡218∘−4+1)(4sin⁡254∘−4+1)]= 36\left[\left(4\sin^{2}{18}^{∘} - 4 + 1\right)\left(4\sin^{2}{54}^{∘} - 4 + 1\right)\right]

=36[(4sin⁡218∘−3)(4sin⁡254∘−3)]= 36\left[\left(4\sin^{2}{18}^{∘} - 3\right)\left(4\sin^{2}{54}^{∘} - 3\right)\right]
=36[{(5−1)24−3}{(5+1)24−3}]= 36\left[\left\{\dfrac{{\left(\sqrt{5} - 1\right)}^{2}}{4} - 3\right\}\left\{\dfrac{{\left(\sqrt{5} + 1\right)}^{2}}{4} - 3\right\}\right]
=36[{(6−25−12)4(6+25−12)4}]= 36\left[\left\{\dfrac{\left(6 - 2\sqrt{5} - 12\right)}{4}\dfrac{\left(6 + 2\sqrt{5} - 12\right)}{4}\right\}\right]
=−36[{(25)2−3616}]=36= - 36\left[\left\{\dfrac{{\left(2\sqrt{5}\right)}^{2} - 36}{16}\right\}\right] = 36

Hence this is the correct option.

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