Question

How do you evaluate $\cot \left( {\dfrac{{2\pi }}{3}} \right)$?

Answer 1

Hint: In order to determine the value of the above question, first rewriting the angle in the form of $\pi \pm \theta $, we get $\pi - \dfrac{\pi }{3}$ where $\theta = \dfrac{\pi }{3}$. As we know that $\cot (\pi - \theta ) = - \cot (\theta )$ because $\pi - \dfrac{\pi }{3}$ is the angle in the second quadrant and cotangent is always negative in 2nd quadrant. Rewriting the cotangent and put the exact value of $\cot \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{{\sqrt 3 }}$ in it. At the end multiply and divide with $\sqrt 3 $ to get your desired value.

Complete step by step answer:
We are given a $\cot \left( {\dfrac{{2\pi }}{3}} \right)$, and we have to evaluate its value.
Let’s write the angle $\dfrac{{2\pi }}{3}$ in the form of $\pi \pm n$. We get that $\dfrac{{2\pi }}{3}$ can be written as $\pi - \dfrac{\pi }{3}$
$ = \cot \left( {\pi - \dfrac{\pi }{3}} \right)$-----(1)
Note that $\cot (\pi - \theta ) = - \cot (\theta )$
As we can see that $\pi - \dfrac{\pi }{3}$ is the angle in the second quadrant and cotangent is always negative in 2nd quadrant, that’s by $\cot (\pi - \theta ) = - \cot (\theta )$.
We can write
$
   = \cot \left( {\pi - \dfrac{\pi }{3}} \right) \\
   = - \cot \left( {\dfrac{\pi }{3}} \right) \\
 $
The exact value of $\cot \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{{\sqrt 3 }}$, putting this value, we get
$ = - \dfrac{1}{{\sqrt 3 }}$
Now, multiplying and dividing with the value $\sqrt 3 $ in above
$
   = - \dfrac{1}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} \\
   = - \dfrac{{\sqrt 3 }}{3} \\
 $
Therefore, the value of $\cot \left( {\dfrac{{2\pi }}{3}} \right)$ is equal to $ - \dfrac{{\sqrt 3 }}{3}$.

Note:
1. Periodic Function = A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $\theta $ and n$ \in $N.
2. Even Function – A function $f(x)$ is said to be an even function, if $f( - x) = f(x)$ for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function, if $f( - x) = - f(x)$ for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta $
Therefore, $\sin \theta $ and $\tan \theta $ and their reciprocals, $\cos ec\theta $ and $\cot \theta $ are odd functions whereas \[\cos \theta \] and its reciprocal \[\sec \theta \] are even functions.
3. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.