Question

Which of the following quantities are rational?
Option A: $\sin (\dfrac{{11\pi }}{{12}})\sin (\dfrac{{5\pi }}{{12}})$
Option B: ${\sin ^4}(\dfrac{\pi }{8}) + {\cos ^4}(\dfrac{\pi }{8})$
Option C: $(1 + \cos (\dfrac{{2\pi }}{9}))(1 + \cos (\dfrac{{4\pi }}{9}))(1 + \cos (\dfrac{{8\pi }}{9}))$

Answer 1

By definition, rational number is a real number which is in the form $\dfrac{p}{q}$ and $q \ne 0$ . Any number which follows the above criteria, is a rational number. To solve this question, we solve step by step every option and try to simplify the equation using trigonometric identities:
$\sin 2\theta = 2\sin \theta \cos \theta$
${\sin ^2}(\dfrac{\pi }{8}) + {\cos ^2}(\dfrac{\pi }{8}) = 1$
${\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}$
and few trigonometric ratios:
$\sin (\dfrac{\pi }{4}) = \dfrac{1}{{\sqrt 2 }}$ and $\sin (\dfrac{\pi }{6}) = \dfrac{1}{2}$

Complete step-by-step answer:
Let’s start by solving every option.
Option A: $\sin (\dfrac{{11\pi }}{{12}})\sin (\dfrac{{5\pi }}{{12}})$
We know that:
$\sin (\dfrac{{5\pi }}{{12}}) = \cos (\dfrac{\pi }{{12}})$ because $\dfrac{{5\pi }}{{12}} + \dfrac{\pi }{{12}} = \dfrac{\pi }{2}$
$\sin (\dfrac{{11\pi }}{{12}}) = \sin (\dfrac{\pi }{{12}})$ because $\dfrac{{11\pi }}{{12}} + \dfrac{\pi }{{12}} = \pi$
Substituting them in the given equation, we get:
$\sin (\dfrac{\pi }{{12}})\cos (\dfrac{\pi }{{12}}) = \dfrac{1}{2}\sin (\dfrac{\pi }{6})$ because $\sin 2\theta = 2\sin \theta \cos \theta$
Using trigonometric ratios, we get: $\sin (\dfrac{\pi }{6}) = \dfrac{1}{2}$
After further simplification, we get: $\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$

Therefore, it is a rational number.

Option B: ${\sin ^4}(\dfrac{\pi }{8}) + {\cos ^4}(\dfrac{\pi }{8})$
We know that ${\sin ^2}(\dfrac{\pi }{8}) + {\cos ^2}(\dfrac{\pi }{8}) = 1$
Squaring on both the sides, we get: $[{\sin ^2}(\dfrac{\pi }{8}) + {\cos ^2}(\dfrac{\pi }{8})] = {\sin ^4}(\dfrac{\pi }{8}) + {\cos ^4}(\dfrac{\pi }{8}) + 2{\sin ^2}(\dfrac{\pi }{8}){\cos ^2}(\dfrac{\pi }{8}) = 1$
Further simplifying the equation, we get:
${\sin ^4}(\dfrac{\pi }{8}) + {\cos ^4}(\dfrac{\pi }{8}) = 1 - 2{\sin ^2}(\dfrac{\pi }{8}){\cos ^2}(\dfrac{\pi }{8})$
We know that $\sin 2\theta = 2\sin \theta \cos \theta$
Therefore, $\sin (\dfrac{\pi }{8}) = 2\sin (\dfrac{\pi }{8})\cos (\dfrac{\pi }{8})$
Squaring on both the sides, we get: ${\sin ^2}(\dfrac{\pi }{8}) = 4{\sin ^2}(\dfrac{\pi }{8}){\cos ^2}(\dfrac{\pi }{8})$
Substituting it in our previous equation, we get: ${\sin ^4}(\dfrac{\pi }{8}) + {\cos ^4}(\dfrac{\pi }{8}) = 1 - 4{\sin ^2}(\dfrac{\pi }{8}){\cos ^2}(\dfrac{\pi }{8}) = 1 - \dfrac{1}{2}{\sin ^2}(\dfrac{\pi }{4})$
By using trigonometric ratios, we get: $\sin (\dfrac{\pi }{4}) = \dfrac{1}{{\sqrt 2 }}$
Substituting the value, we get: ${\sin ^4}(\dfrac{\pi }{8}) + {\cos ^4}(\dfrac{\pi }{8}) = 1 - \dfrac{1}{2}{\sin ^2}(\dfrac{\pi }{4}) = 1 - \dfrac{1}{2}{(\dfrac{1}{{\sqrt 2 }})^2} = 1 - \dfrac{1}{4} = \dfrac{3}{4}$
This is also a rational number.

Option C: $(1 + \cos (\dfrac{{2\pi }}{9}))(1 + \cos (\dfrac{{4\pi }}{9}))(1 + \cos (\dfrac{{8\pi }}{9}))$
We know that ${\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}$
Using this formula, we get: $(1 + \cos (\dfrac{{2\pi }}{9}))(1 + \cos (\dfrac{{4\pi }}{9}))(1 + \cos (\dfrac{{8\pi }}{9})) = 8({\cos ^2}(\dfrac{\pi }{9}))({\cos ^2}(\dfrac{{2\pi }}{9}))({\cos ^2}(\dfrac{{4\pi }}{9}))$
$8{[(\cos (\dfrac{\pi }{9}))(\cos (\dfrac{{2\pi }}{9}))(\cos (\dfrac{{4\pi }}{9}))]^2}$
Multiplying and dividing by $\sin \left( {\dfrac{\pi }{9}} \right)$ we get:
$8{\left( {\dfrac{{(\cos (\dfrac{\pi }{9}))(\cos (\dfrac{{2\pi }}{9}))(\cos (\dfrac{{4\pi }}{9}))}}{{\sin (\dfrac{\pi }{9})}} \times \sin (\dfrac{\pi }{9})} \right)^2}$
Using the formula $\sin 2\theta = 2\sin \theta \cos \theta$ , we can simplify we get:
$8{\left( {\dfrac{{\sin (\dfrac{\pi }{9})(\cos (\dfrac{\pi }{9}))(\cos (\dfrac{{2\pi }}{9}))(\cos (\dfrac{{4\pi }}{9}))}}{{\sin (\dfrac{\pi }{9})}}} \right)^2} = 8{\left( {\dfrac{{\sin (\dfrac{{2\pi }}{9}))(\cos (\dfrac{{2\pi }}{9}))(\cos (\dfrac{{4\pi }}{9}))}}{{\sin (\dfrac{\pi }{9})}}} \right)^2}$
$8{\left( {\dfrac{{2\sin (\dfrac{{2\pi }}{9})(\cos (\dfrac{{2\pi }}{9}))(\cos (\dfrac{{4\pi }}{9}))}}{{2 \times 2\sin (\dfrac{\pi }{9})}}} \right)^2} = 8{\left( {\dfrac{{\sin (\dfrac{{4\pi }}{9})\cos (\dfrac{{4\pi }}{9})}}{{4\sin (\dfrac{\pi }{9})}}} \right)^2}$
$8{\left( {\dfrac{{2\sin (\dfrac{{4\pi }}{9})\cos (\dfrac{{4\pi }}{9})}}{{8\sin (\dfrac{\pi }{9})}}} \right)^2} = 8{\left( {\dfrac{{\sin (\dfrac{{8\pi }}{9})}}{{8\sin (\dfrac{\pi }{9})}}} \right)^2}$
We know that $\dfrac{{8\pi }}{9} + \dfrac{\pi }{9} = \pi$
The equation simplifies to: $8 \times {\left( {\dfrac{1}{8}} \right)^2} = \dfrac{1}{8}$
This is also a rational number.
All the options are rational numbers.

So, the correct answer is “Option A,B and C”.

Note: This question becomes easy to solve if one remembers the trigonometric formulae used in the question above. This is not the only solution to solve the above question, one can use many methods. We can also directly substitute the trigonometric ratios, but this is not a suggestible method.