Question

How do you find the exact value of the following using the unit circle: $\cos \dfrac{{53\pi }}{6}$?

Answer 1

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.