Answer
398.4k+ views
Hint: In the given problem, we are required to find the cosine of a given angle using some simple and basic trigonometric compound angle formulae and trigonometric identities. Such questions require basic knowledge of compound angle formulae and their applications in this type of questions. Unit circle is a circle with a radius of one unit drawn on a graph paper with its centre at origin.
Complete step by step solution:
Consider a unit circle (a circle of radius of 1 unit centered at origin).
We need to find out the value of $\cos \dfrac{{53\pi }}{6}$ using the unit circle.
So, we have, $\cos \dfrac{{53\pi }}{6}$$ = \cos \left( {\dfrac{{48\pi + 5\pi }}{6}} \right)$
Separating the numerator into two parts and distributing the denominator underneath both the parts, we get,
$ \Rightarrow \cos \dfrac{{53\pi }}{6}$$ = \cos \left( {8\pi + \dfrac{{5\pi }}{6}} \right)$
Since cosine and sine function are periodic functions with period of $2\pi $, so the value of cosine and sine gets repeated after intervals in multiples of $2\pi $. Hence, we can eliminating the $\left( {8\pi } \right)$ term from the angle, we get,
\[ \Rightarrow \cos \dfrac{{53\pi }}{6}\]$ = \cos \left( {\dfrac{{5\pi }}{6}} \right)$
Since, the angle $\left( {\dfrac{{5\pi }}{6}} \right)$ lies in the second quadrant and cosine ratio is negative in the second quadrant. So, we get,
$ \Rightarrow $$\cos \left( {\pi - \dfrac{\pi }{6}} \right)$
$ \Rightarrow $$ - \cos \left( {\dfrac{\pi }{6}} \right)$
We know the value of $\cos \left( {\dfrac{\pi }{6}} \right)$ is $\left( {\dfrac{{\sqrt 3 }}{2}} \right)$.
Substituting the same, we get,
$ \Rightarrow $$ - \dfrac{{\sqrt 3 }}{2}$
Hence, the value of $\cos \dfrac{{53\pi }}{6}$ is $\left( { - \dfrac{{\sqrt 3 }}{2}} \right)$
Note: Periodic Function is a function that repeats its value after a certain interval. For a real number $T > 0$, $f\left( {x + T} \right) = f\left( x \right)$ for all x. If T is the smallest positive real number such that $f\left( {x + T} \right) = f\left( x \right)$ for all x, then T is called the fundamental period.
Complete step by step solution:
Consider a unit circle (a circle of radius of 1 unit centered at origin).
We need to find out the value of $\cos \dfrac{{53\pi }}{6}$ using the unit circle.
So, we have, $\cos \dfrac{{53\pi }}{6}$$ = \cos \left( {\dfrac{{48\pi + 5\pi }}{6}} \right)$
Separating the numerator into two parts and distributing the denominator underneath both the parts, we get,
$ \Rightarrow \cos \dfrac{{53\pi }}{6}$$ = \cos \left( {8\pi + \dfrac{{5\pi }}{6}} \right)$
Since cosine and sine function are periodic functions with period of $2\pi $, so the value of cosine and sine gets repeated after intervals in multiples of $2\pi $. Hence, we can eliminating the $\left( {8\pi } \right)$ term from the angle, we get,
\[ \Rightarrow \cos \dfrac{{53\pi }}{6}\]$ = \cos \left( {\dfrac{{5\pi }}{6}} \right)$
Since, the angle $\left( {\dfrac{{5\pi }}{6}} \right)$ lies in the second quadrant and cosine ratio is negative in the second quadrant. So, we get,
$ \Rightarrow $$\cos \left( {\pi - \dfrac{\pi }{6}} \right)$
$ \Rightarrow $$ - \cos \left( {\dfrac{\pi }{6}} \right)$
We know the value of $\cos \left( {\dfrac{\pi }{6}} \right)$ is $\left( {\dfrac{{\sqrt 3 }}{2}} \right)$.
Substituting the same, we get,
$ \Rightarrow $$ - \dfrac{{\sqrt 3 }}{2}$
Hence, the value of $\cos \dfrac{{53\pi }}{6}$ is $\left( { - \dfrac{{\sqrt 3 }}{2}} \right)$
Note: Periodic Function is a function that repeats its value after a certain interval. For a real number $T > 0$, $f\left( {x + T} \right) = f\left( x \right)$ for all x. If T is the smallest positive real number such that $f\left( {x + T} \right) = f\left( x \right)$ for all x, then T is called the fundamental period.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)