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QJEEM23Mac2e8: JEE Main 2023 Mathematics - Class 11 JEEM23Mac2e8 Solution

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

Multiple Choice Options

  • A 18
  • B 36
  • C 54
  • D 27

Step-by-Step Verified Solution

amp;4cos⁡2θ−1=4(1−sin⁡2θ)−1amp;=3−4sin⁡2θamp;=3sin⁡θ−4sin⁡3θsin⁡θamp;=sin⁡3θsin⁡θ\begin{aligned}& 4 \cos ^2 \theta-1=4\left(1-\sin ^2 \theta\right)-1 \\\\& =3-4 \sin ^2 \theta \\\\& =\frac{3 \sin \theta-4 \sin ^3 \theta}{\sin \theta} \\\\& =\frac{\sin 3 \theta}{\sin \theta}\end{aligned}

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

amp;=36[sin⁡27∘sin⁡9∘×sin⁡81∘sin⁡27∘×sin⁡243∘sin⁡81∘×sin⁡729∘sin⁡243∘]amp;=36[sin⁡729∘sin⁡9∘]=36×1=36\begin{aligned}& =36\left[\frac{\sin 27^{\circ}}{\sin 9^{\circ}} \times \frac{\sin 81^{\circ}}{\sin 27^{\circ}} \times \frac{\sin 243^{\circ}}{\sin 81^{\circ}} \times \frac{\sin 729^{\circ}}{\sin 243^{\circ}}\right] \\\\& =36\left[\frac{\sin 729^{\circ}}{\sin 9^{\circ}}\right]=36 \times 1=36\end{aligned}

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