Question

Which of the following numbers is/or are rational
1) \[\sin {15^ \circ }\]
2) \[\cos {15^ \circ }\]
3) \[\sin {15^ \circ }\cos {15^ \circ }\]
4) \[\sin {15^ \circ }\cos {75^ \circ }\]

Answer 1

Hint: Here in this question we have to determine the value of the trigonometric function and then we have to say whether it belongs to rational or not. Since the function is a trigonometric function, by considering the formulas of trigonometric function and table of trigonometric function we determine the value.

Complete step by step answer:
Consider every option and simplify.
1) \[\sin {15^ \circ }\]
As we know that the double angle formula \[\cos 2x = 1 - 2{\sin ^2}x\]
\[ \Rightarrow \cos ({30^ \circ }) = 1 - {\sin ^2}{15^ \circ }\]
By the table of trigonometric ratios we have
\[ \Rightarrow \dfrac{{\sqrt 3 }}{2} = 1 - 2{\sin ^2}{15^ \circ }\]
On simplifying we have
\[ \Rightarrow 2{\sin ^2}{15^ \circ } = 1 - \dfrac{{\sqrt 3 }}{2}\]
\[ \Rightarrow {\sin ^2}{15^ \circ } = \dfrac{{2 - \sqrt 3 }}{4}\]
On taking square root on both sides
\[ \Rightarrow \sin {15^ \circ } = \sqrt {\dfrac{{2 - \sqrt 3 }}{4}} \]
\[ \Rightarrow \sin {15^ \circ } = \dfrac{{\sqrt {2 - \sqrt 3 } }}{4}\]
This is not a rational number.

2) \[\cos {15^ \circ }\]
As we know that the double angle formula \[\cos 2x = 2{\sin ^2}x - 1\]
\[ \Rightarrow \cos ({30^ \circ }) = 2{\cos ^2}{15^ \circ } - 1\]
By the table of trigonometric ratios we have
\[ \Rightarrow \dfrac{{\sqrt 3 }}{2} = 2{\cos ^2}{15^ \circ } - 1\]
On simplifying we have
\[ \Rightarrow 2{\cos ^2}{15^ \circ } = 1 + \dfrac{{\sqrt 3 }}{2}\]
\[ \Rightarrow {\sin ^2}{15^ \circ } = \dfrac{{2 + \sqrt 3 }}{4}\]
On taking square root on both sides
\[ \Rightarrow \sin {15^ \circ } = \sqrt {\dfrac{{2 + \sqrt 3 }}{4}} \]
\[ \Rightarrow \sin {15^ \circ } = \dfrac{{\sqrt {2 + \sqrt 3 } }}{2}\]
This is not a rational number.

3) \[\sin {15^ \circ }\cos {15^ \circ }\]
Multiply and divide by 2 we have
\[ \Rightarrow \dfrac{1}{2} \times 2\sin {15^ \circ }\cos {15^ \circ }\]
As we know that the double angle formula \[\sin 2x = 2\sin x\cos x\]
\[ \Rightarrow \dfrac{1}{2}\sin ({30^ \circ })\]
By the table of trigonometric ratios we have
\[ \Rightarrow \sin {15^ \circ }\cos {15^ \circ } = \dfrac{1}{2} \times \dfrac{1}{2}\]
On simplifying we have
\[ \Rightarrow \sin {15^ \circ }\cos {15^ \circ } = \dfrac{1}{4}\]
This is a rational number.

4) \[\sin {15^ \circ }\cos {75^ \circ }\]
On considering the trigonometric ratio for the allied angles we have
\[ \Rightarrow \sin {15^ \circ }\cos ({90^ \circ } - {15^ \circ })\]
\[ \Rightarrow \sin {15^ \circ }\sin {15^ \circ }\]
On multiplying we get
\[ \Rightarrow {\sin ^2}{15^ \circ }\]
From the first problem we know the value of \[\sin {15^ \circ }\], on squaring the first problem we get
\[ \Rightarrow {\sin ^2}{15^ \circ } = {\left( {\dfrac{{\sqrt {2 - \sqrt 3 } }}{2}} \right)^2}\]
\[ \Rightarrow {\sin ^2}{15^ \circ } = \dfrac{{2 - \sqrt 3 }}{4}\]
This is not a rational number.

Note:
Rational Numbers: Numbers that can be written in the form of \[\dfrac{p}{q}\] where \[q \ne 0\].
Irrational Numbers: All the numbers which are not rational and cannot be written in the form of \[\dfrac{p}{q}\].
The solution of four questions is in the form of \[\dfrac{p}{q}\], but the numerator value is recurring; there is no end of the numbers. So we are considering those numbers as irrational.