
How do you evaluate $\cos \left( {\dfrac{\pi }{8}} \right)$?
Answer
556.5k+ 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.
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