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

How do you evaluate cot⁡(3π2)\cot \left( {\dfrac{{3\pi }}{2}} \right)?

Explanation

Solution

In order to determine the value of the above question, first rewriting the angle in the form of π±θ\pi \pm \theta , we get π+π2\pi + \dfrac{\pi }{2} where θ=π2\theta = \dfrac{\pi }{2}. As we know that cot⁡(π+π2)=−cot⁡(π2)\cot \left( {\pi + \dfrac{\pi }{2}} \right) = - \cot \left( {\dfrac{\pi }{2}} \right) because π+π2\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⁡(π2)=0\cot \left( {\dfrac{\pi }{2}} \right) = 0 in it to get your desired value.

Complete step by step answer:
We are given a cot⁡(3π2)\cot \left( {\dfrac{{3\pi }}{2}} \right), and we have to evaluate its value.
Let’s write the angle 3π2\dfrac{{3\pi }}{2} in the form of π±n\pi \pm n. We get that 3π2\dfrac{{3\pi }}{2} can be written as π+π2\pi + \dfrac{\pi }{2}
=cot⁡(π+π2)= \cot \left( {\pi + \dfrac{\pi }{2}} \right)-----(1)
As we can see that π+π2\pi + \dfrac{\pi }{2} is the angle in the fourth quadrant and cotangent is always negative in 4th quadrant, that’s by cot⁡(π+π2)=−cot⁡(π2)\cot \left( {\pi + \dfrac{\pi }{2}} \right) = - \cot (\dfrac{\pi }{2}).
We can write
=cot⁡(π+π2) =−cot⁡(π2)  = \cot \left( {\pi + \dfrac{\pi }{2}} \right) \\\ = - \cot \left( {\dfrac{\pi }{2}} \right) \\\
The exact value of cot⁡(π2)=0\cot \left( {\dfrac{\pi }{2}} \right) = 0, putting this value, we get
=−0 =0  = - 0 \\\ = 0 \\\
Therefore, the value of cot⁡(3π2)\cot \left( {\dfrac{{3\pi }}{2}} \right) is equal to 00.

Note: 1. Periodic Function= A function f(x)f(x) is said to be a periodic function if there exists a real number T > 0 such that f(x+T)=f(x)f(x + T) = f(x) for all x.
If T is the smallest positive real number such that f(x+T)=f(x)f(x + T) = f(x) for all x, then T is called the fundamental period of f(x)f(x) .
Since sin⁡ (2nπ+θ)=sin⁡θ\sin \,(2n\pi + \theta ) = \sin \theta for all values of θ\theta and n∈\inN.
2. Even Function – A function f(x)f(x) is said to be an even function, if f(−x)=f(x)f( - x) = f(x) for all x in its domain.
Odd Function – A function f(x)f(x) is said to be an even function, if f(−x)=−f(x)f( - x) = - f(x) for all x in its domain.
We know that sin⁡(−θ)=−sin⁡θcos⁡(−θ)=cos⁡θ and tan⁡(−θ)=−tan⁡θ\sin ( - \theta ) = - \sin \theta \cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta
Therefore, sin⁡θ\sin \theta and tan⁡θ\tan \theta and their reciprocals,cos⁡ecθ\cos ec\theta and cot⁡θ\cot \theta are odd functions whereas cos⁡θ\cos \theta and its reciprocal sec⁡θ\sec \theta are even functions.