Question

How do you find the exact value of $\cos \left( {\dfrac{{11\pi }}{3}} \right)$?

Answer 1

For solving this very question we will first write the expression in such a way that it will follow the formula given by cos(A+B)=cosA.cosB−sinA.sinBcos⁡(A+B)=cos⁡A.cos⁡B−sin⁡A.sin⁡B . And then substituting the values we will get to the result.

Formula used:
The formula in terms of cosine,
$\cos \left( {A + B} \right) = \cos A.\cos B - \sin A.\sin B$

Complete step by step answer:
So we have the expression given as $\cos \left( {\dfrac{{11\pi }}{3}} \right)$.
And for solving it we will first expand the expression and it can be written as
$\Rightarrow \cos \left( {2\pi + \dfrac{{5\pi }}{3}} \right)$
[Since, $\dfrac{{11\pi }}{3}$ can be written as, $\dfrac{{11\pi }}{3} = \dfrac{{6\pi }}{3} + \dfrac{{5\pi }}{3} = 2\pi + \dfrac{{5\pi }}{3}$]
And as we know the formula ,
$\cos \left( {A + B} \right) = \cos A.\cos B - \sin A.\sin B$
So, by comparing the LHS of the equation with the above expression we have the values of constants will be as
$A = 2\pi ,B = \pi$
Therefore, on substituting the values, we will get the equation as
$\Rightarrow \cos \left( {2\pi + \dfrac{{5\pi }}{3}} \right) = \cos 2\pi .\cos \dfrac{{5\pi }}{3} - \sin 2\pi .\sin \dfrac{{5\pi }}{3}$
And as we know the value of $\sin \pi = 0$ and their even multiple will also be the same.
Whereas the value of $\cos \pi = 1$ and for the even multiple the sign will keep changing at the interval.
Therefore, on substituting the values, we will get the equation as
$\Rightarrow \cos \left( {2\pi + \dfrac{{5\pi }}{3}} \right) = \left( 1 \right).\cos \dfrac{{5\pi }}{3} - \left( 0 \right).\sin \dfrac{{5\pi }}{3}$
$\Rightarrow \cos \left( {2\pi + \dfrac{{5\pi }}{3}} \right) = \cos \dfrac{{5\pi }}{3} - 0$
$\Rightarrow \cos \left( {2\pi + \dfrac{{5\pi }}{3}} \right) = \cos \dfrac{{5\pi }}{3}$
Now, we know that, the value of $\cos \dfrac{{5\pi }}{3} = \dfrac{1}{2}$.
Therefore, we can write,
$\Rightarrow \cos \left( {2\pi + \dfrac{{5\pi }}{3}} \right) = \cos \dfrac{{5\pi }}{3} = \dfrac{1}{2}$
And on solving it we will get
$\Rightarrow \cos \left( {2\pi + \dfrac{{5\pi }}{3}} \right) = \dfrac{1}{2}$

Therefore, the exact value of $\cos \left( {\dfrac{{11\pi }}{3}} \right)$ will be equal to $\dfrac{1}{2}$.

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. Trigonometric functions are periodic functions. Sine and cosine functions have the fundamental period as $2\pi$ radians. The compound angle formula for cosine is $\cos \left( {A + B} \right) = \cos A.\cos B - \sin A.\sin B$.