JEE MAIN - Mathematics (2023 - 8th April Evening Shift - No. 12)
The value of $$36\left(4 \cos ^{2} 9^{\circ}-1\right)\left(4 \cos ^{2} 27^{\circ}-1\right)\left(4 \cos ^{2} 81^{\circ}-1\right)\left(4 \cos ^{2} 243^{\circ}-1\right)$$ is :
18
36
54
27
Explanation
$$
\begin{aligned}
& 4 \cos ^2 \theta-1=4\left(1-\sin ^2 \theta\right)-1 \\\\
& =3-4 \sin ^2 \theta \\\\
& =\frac{3 \sin \theta-4 \sin ^3 \theta}{\sin \theta} \\\\
& =\frac{\sin 3 \theta}{\sin \theta}
\end{aligned}
$$
$$36\left(4 \cos ^{2} 9^{\circ}-1\right)\left(4 \cos ^{2} 27^{\circ}-1\right)\left(4 \cos ^{2} 81^{\circ}-1\right)\left(4 \cos ^{2} 243^{\circ}-1\right)$$
$$ \begin{aligned} & =36\left[\frac{\sin 27^{\circ}}{\sin 9^{\circ}} \times \frac{\sin 81^{\circ}}{\sin 27^{\circ}} \times \frac{\sin 243^{\circ}}{\sin 81^{\circ}} \times \frac{\sin 729^{\circ}}{\sin 243^{\circ}}\right] \\\\ & =36\left[\frac{\sin 729^{\circ}}{\sin 9^{\circ}}\right]=36 \times 1=36 \end{aligned} $$
$$36\left(4 \cos ^{2} 9^{\circ}-1\right)\left(4 \cos ^{2} 27^{\circ}-1\right)\left(4 \cos ^{2} 81^{\circ}-1\right)\left(4 \cos ^{2} 243^{\circ}-1\right)$$
$$ \begin{aligned} & =36\left[\frac{\sin 27^{\circ}}{\sin 9^{\circ}} \times \frac{\sin 81^{\circ}}{\sin 27^{\circ}} \times \frac{\sin 243^{\circ}}{\sin 81^{\circ}} \times \frac{\sin 729^{\circ}}{\sin 243^{\circ}}\right] \\\\ & =36\left[\frac{\sin 729^{\circ}}{\sin 9^{\circ}}\right]=36 \times 1=36 \end{aligned} $$
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