
How do you evaluate $\cos \left( {\dfrac{\pi }{8}} \right)$?
Answer
542.7k+ views
Hint: Here, we are required to find the value of $\cos \left( {\dfrac{\pi }{8}} \right)$. Thus, we will use the half-angle formula of cosine and determine the value of the given function. With the help of this, we will be able to further solve the question, and using the trigonometric table, we will be able to find the exact value of the given trigonometric function.
Formula Used:
$\cos 2\theta = 2{\cos ^2}\theta - 1$
Complete step by step solution:
In order to find the value of $\cos \left( {\dfrac{\pi }{8}} \right)$,
We will use the half-angle formulas.
According to which, $\cos 2\theta = 2{\cos ^2}\theta - 1$
Hence, substituting $\theta = \dfrac{\pi }{8}$ and hence, $2\theta = \dfrac{\pi }{4}$, we get,
$\cos \left( {\dfrac{\pi }{4}} \right) = 2{\cos ^2}\left( {\dfrac{\pi }{8}} \right) - 1$
Using trigonometric tables, we know that $\cos \left( {\dfrac{\pi }{4}} \right) = \cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow \dfrac{1}{{\sqrt 2 }} = 2{\cos ^2}\left( {\dfrac{\pi }{8}} \right) - 1$
Adding 1 on both sides and dividing both sides by 2, we get,
$ \Rightarrow \dfrac{{\dfrac{1}{{\sqrt 2 }} + 1}}{2} = {\cos ^2}\left( {\dfrac{\pi }{8}} \right)$
$ \Rightarrow \dfrac{{1 + \sqrt 2 }}{{2\sqrt 2 }} = {\cos ^2}\left( {\dfrac{\pi }{8}} \right)$
Taking square root on both the sides,
$ \Rightarrow \cos \left( {\dfrac{\pi }{8}} \right) = \pm \sqrt {\dfrac{{1 + \sqrt 2 }}{{2\sqrt 2 }}} $
But, we will reject the negative value because the given angle lies in the first quadrant.
Since there is a square root in the denominator, thus, rationalizing the RHS, we get,
$ \Rightarrow \cos \left( {\dfrac{\pi }{8}} \right) = \sqrt {\dfrac{{1 + \sqrt 2 }}{{2\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}} = \sqrt {\dfrac{{\sqrt 2 + 2}}{4}} $
Splitting the denominator,
$ \Rightarrow \cos \left( {\dfrac{\pi }{8}} \right) = \sqrt {\dfrac{{\sqrt 2 }}{4} + \dfrac{2}{4}} = \sqrt {\dfrac{{\sqrt 2 }}{4} + \dfrac{1}{2}} $
Therefore, the required value of $\cos \left( {\dfrac{\pi }{8}} \right)$ is $\sqrt {\dfrac{{\sqrt 2 }}{4} + \dfrac{1}{2}} $.
Note:
In this question, we have used trigonometry. Trigonometry is a branch of mathematics that helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine, and the cosine function. In simple terms, they are written as ‘sin’, ‘cos’, and ‘tan’. Hence, trigonometry is not just a chapter to study, in fact, it is being used in everyday life.
Formula Used:
$\cos 2\theta = 2{\cos ^2}\theta - 1$
Complete step by step solution:
In order to find the value of $\cos \left( {\dfrac{\pi }{8}} \right)$,
We will use the half-angle formulas.
According to which, $\cos 2\theta = 2{\cos ^2}\theta - 1$
Hence, substituting $\theta = \dfrac{\pi }{8}$ and hence, $2\theta = \dfrac{\pi }{4}$, we get,
$\cos \left( {\dfrac{\pi }{4}} \right) = 2{\cos ^2}\left( {\dfrac{\pi }{8}} \right) - 1$
Using trigonometric tables, we know that $\cos \left( {\dfrac{\pi }{4}} \right) = \cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow \dfrac{1}{{\sqrt 2 }} = 2{\cos ^2}\left( {\dfrac{\pi }{8}} \right) - 1$
Adding 1 on both sides and dividing both sides by 2, we get,
$ \Rightarrow \dfrac{{\dfrac{1}{{\sqrt 2 }} + 1}}{2} = {\cos ^2}\left( {\dfrac{\pi }{8}} \right)$
$ \Rightarrow \dfrac{{1 + \sqrt 2 }}{{2\sqrt 2 }} = {\cos ^2}\left( {\dfrac{\pi }{8}} \right)$
Taking square root on both the sides,
$ \Rightarrow \cos \left( {\dfrac{\pi }{8}} \right) = \pm \sqrt {\dfrac{{1 + \sqrt 2 }}{{2\sqrt 2 }}} $
But, we will reject the negative value because the given angle lies in the first quadrant.
Since there is a square root in the denominator, thus, rationalizing the RHS, we get,
$ \Rightarrow \cos \left( {\dfrac{\pi }{8}} \right) = \sqrt {\dfrac{{1 + \sqrt 2 }}{{2\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}} = \sqrt {\dfrac{{\sqrt 2 + 2}}{4}} $
Splitting the denominator,
$ \Rightarrow \cos \left( {\dfrac{\pi }{8}} \right) = \sqrt {\dfrac{{\sqrt 2 }}{4} + \dfrac{2}{4}} = \sqrt {\dfrac{{\sqrt 2 }}{4} + \dfrac{1}{2}} $
Therefore, the required value of $\cos \left( {\dfrac{\pi }{8}} \right)$ is $\sqrt {\dfrac{{\sqrt 2 }}{4} + \dfrac{1}{2}} $.
Note:
In this question, we have used trigonometry. Trigonometry is a branch of mathematics that helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine, and the cosine function. In simple terms, they are written as ‘sin’, ‘cos’, and ‘tan’. Hence, trigonometry is not just a chapter to study, in fact, it is being used in everyday life.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

What happens to glucose which enters nephron along class 10 biology CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Write a dialogue with at least ten utterances between class 10 english CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

A circle is inscribed in an equilateral triangle and class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Which of the following does not have a fundamental class 10 physics CBSE

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

