Which of the following quantities are rational?

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$\sin (\frac{11 π}{12}) \sin (\frac{5 π}{12})$

$cosec (\frac{9 π}{10}) \sec (\frac{4 π}{5})$

$\sin^{4} (\frac{π}{8}) + \cos^{4} (\frac{π}{8})$

$(1 + \cos \frac{2 π}{9}) (1 + \cos \frac{4 π}{9}) (1 + \cos \frac{8 π}{9})$

answer is A.

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Detailed Solution

$\begin{aligned} (1) \sin (\frac{11 π}{12}) \sin (\frac{5 π}{12}) = \sin (\frac{π}{12}) \cos (\frac{π}{12}) \\ = \frac{1}{2} \sin (\frac{π}{6}) = \frac{1}{4} \in Q \end{aligned}$

$\begin{aligned} (2) cosec (\frac{9 π}{10}) \sec (\frac{4 π}{5}) = - cosec (\frac{π}{10}) \sec (\frac{π}{5}) \\ = \frac{1}{- \sin 18^{\circ} \cos 36^{\circ}} \\ = - \frac{16}{(\sqrt{5} - 1) (\sqrt{5} + 1)} \\ = - 4 \in Q \end{aligned}$

$\begin{aligned} (3) \sin^{4} (\frac{π}{8}) + \cos^{4} (\frac{π}{8}) = 1 - 2 \sin^{2} (\frac{π}{8}) \cos^{2} (\frac{π}{8}) \\ = 1 - \frac{1}{2} \sin^{2} (\frac{π}{4}) \\ = 1 - \frac{1}{4} = \frac{3}{4} \in Q \end{aligned}$

$\begin{array}{l} (4) 2 \cos^{2} \frac{π}{9} 2 \cos^{2} \frac{2 π}{9} 2 \cos^{2} \frac{4 π}{9} \\ = 8 {(\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ})}^{2} = \frac{1}{8} \in Q \end{array}$