Question

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

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 $\pi + \dfrac{\pi }{2}$ where $\theta = \dfrac{\pi }{2}$. As we know that $\cot \left( {\pi + \dfrac{\pi }{2}} \right) = - \cot \left( {\dfrac{\pi }{2}} \right)$ because $\pi + \dfrac{\pi }{2}$ is the angle in the fourth quadrant and cotangent is always negative in 4th quadrant. Rewrite the cotangent and put the exact value of $\cot \left( {\dfrac{\pi }{2}} \right) = 0$ in it to get your desired value.

Complete step by step answer:
We are given a $\cot \left( {\dfrac{{3\pi }}{2}} \right)$, and we have to evaluate its value.
Let’s write the angle $\dfrac{{3\pi }}{2}$ in the form of $\pi \pm n$. We get that $\dfrac{{3\pi }}{2}$ can be written as $\pi + \dfrac{\pi }{2}$
$ = \cot \left( {\pi + \dfrac{\pi }{2}} \right)$-----(1)
As we can see that $\pi + \dfrac{\pi }{2}$ is the angle in the fourth quadrant and cotangent is always negative in 4th quadrant, that’s by $\cot \left( {\pi + \dfrac{\pi }{2}} \right) = - \cot (\dfrac{\pi }{2})$.
We can write
$
   = \cot \left( {\pi + \dfrac{\pi }{2}} \right) \\
   = - \cot \left( {\dfrac{\pi }{2}} \right) \\
 $
The exact value of $\cot \left( {\dfrac{\pi }{2}} \right) = 0$, putting this value, we get
$
   = - 0 \\
   = 0 \\
 $
Therefore, the value of $\cot \left( {\dfrac{{3\pi }}{2}} \right)$ is equal to $0$.

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.