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Question: How do you evaluate \(\cot \left( {\dfrac{{5\pi }}{6}} \right)\)?...

How do you evaluate cot⁡(5π6)\cot \left( {\dfrac{{5\pi }}{6}} \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−π6\pi - \dfrac{\pi }{6} where θ=π6\theta = \dfrac{\pi }{6}. As we know that cot⁡(π−θ)=−cot⁡(θ)\cot (\pi - \theta ) = - \cot (\theta ) because π−π6\pi - \dfrac{\pi }{6} is the angle in the second quadrant and cotangent is always negative in 2nd2^{nd} quadrant. Rewrite the cotangent and put the exact value of cot⁡(π6)=3\cot \left( {\dfrac{\pi }{6}} \right) = \sqrt 3 to get your required result.

Complete step by step answer:
We are given a cot⁡(5π6)\cot \left( {\dfrac{{5\pi }}{6}} \right), and we have to evaluate its value.
Let’s write the angle 5π6\dfrac{{5\pi }}{6}in the form of π±n\pi \pm n. We get that 5π6\dfrac{{5\pi }}{6} can be written as π−π6\pi - \dfrac{\pi }{6}
=cot⁡(π−π6)= \cot \left( {\pi - \dfrac{\pi }{6}} \right)-----(1)
Note that cot⁡(π−θ)=−cot⁡(θ)\cot (\pi - \theta ) = - \cot (\theta )
As we can see that π−π6\pi - \dfrac{\pi }{6} is the angle in the second quadrant and cotangent is always negative in 2nd2^{nd} quadrant, that’s by cot⁡(π−θ)=−cot⁡(θ)\cot (\pi - \theta ) = - \cot (\theta ).
We can write
=cot⁡(π−π6) =−cot⁡(π6)  = \cot \left( {\pi - \dfrac{\pi }{6}} \right) \\\ = - \cot \left( {\dfrac{\pi }{6}} \right) \\\
The exact value of cot⁡(π6)=3\cot \left( {\dfrac{\pi }{6}} \right) = \sqrt 3 , substituting this value, we get
=−3= - \sqrt 3

Therefore, the value of cot⁡(5π6)\cot \left( {\dfrac{{5\pi }}{6}} \right) is equal to −3- \sqrt 3.

Additional information:
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⁡θ.cos⁡(−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.
3. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.

Note: 1. One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2. Cotangent and tangent are always negative in 2nd2^{nd} quadrant and positive in 1st1^{st} and 3rd3^{rd} quadrant.