Which of the following quantities are rational ?

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

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c o s e c (\frac{9 π}{10}) sec (\frac{4 π}{5})

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

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

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Solution

The correct options are
A $sin (\frac{11 π}{12}) sin (\frac{5 π}{12})$
B $c o s e c (\frac{9 π}{10}) sec (\frac{4 π}{5})$
C ${sin}^{4} (\frac{π}{8}) + {cos}^{4} (\frac{π}{8})$
D $(1 + cos \frac{2 π}{9}) (1 + cos \frac{4 π}{9}) (1 + cos \frac{8 π}{9})$
(A) $sin (\frac{11 π}{12}) sin (\frac{5 π}{12}) = sin (\frac{π}{12}) cos (\frac{π}{12})$
$= \frac{1}{2} sin (\frac{π}{6}) = \frac{1}{4} \in Q$
(B) $c o s e c (\frac{9 π}{10}) sec (\frac{4 π}{5}) = - c o s e c (\frac{π}{10}) sec (\frac{π}{5})$
$= \frac{- 1}{sin 18^{\circ} cos 36^{\circ}} = \frac{- 16}{(\sqrt{5} - 1) (\sqrt{5} + 1)} = - 4 \in Q$
(C) ${sin}^{4} (\frac{π}{8}) + {cos}^{4} (\frac{π}{8}) = 1 - 2 {sin}^{2} (\frac{π}{8}) {cos}^{2} (\frac{π}{8})$
$= 1 - \frac{1}{2} {sin}^{2} \frac{π}{4} = 1 - \frac{1}{4} = \frac{3}{4} \in Q$
(D) $2 {cos}^{2} \frac{π}{9} 2 {cos}^{2} \frac{2 π}{9} 2 {cos}^{2} \frac{4 π}{9} = 8 {(cos 20^{\circ} cos 40^{\circ} cos 80^{\circ})}^{2} = \frac{1}{8} \in Q$

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Similar questions

cos \frac{2 π}{7} . cos \frac{4 π}{7} . cos \frac{8 π}{7} = \frac{1}{8}

x = sin \frac{2 π}{7} + sin \frac{4 π}{7} + sin \frac{8 π}{7} a n d y = cos \frac{2 π}{7} + cos \frac{4 π}{7} + cos \frac{8 π}{7} t h e n x^{2} + y^{2} =

c o s \frac{2 π}{7}, c o s \frac{4 π}{7} a n d c o s \frac{8 π}{7}

c o s \frac{2 π}{7}, c o s \frac{4 π}{7} a n d c o s \frac{6 π}{7}

c o s \frac{2 π}{4} c o s \frac{4 π}{7} c o s \frac{8 π}{7} = S_{3} = \frac{1}{8}

c o s \frac{2 π}{4} + c o s \frac{4 π}{7} + c o s \frac{8 π}{7} = S_{1} = - \frac{1}{2}

s e c \frac{2 π}{7}, s e c \frac{4 π}{7}, s e c \frac{8 π}{7} (= s e c \frac{6 π}{7})

A = sin \frac{2 π}{7} + sin \frac{4 π}{7} + sin \frac{8 π}{7}

B = cos \frac{2 π}{7} + cos \frac{4 π}{7} + cos \frac{8 π}{7}

A^{2} + B^{2}

A = sin \frac{2 π}{7} + sin \frac{4 π}{7} + sin \frac{8 π}{7}

B = cos \frac{2 π}{7} + cos \frac{4 π}{7} + cos \frac{8 π}{7}

\sqrt{A^{2} + B^{2}}

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