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Question: How do you find the value of \(\tan \left( {\dfrac{\pi }{3}} \right)\)?...

How do you find the value of tan⁡(π3)\tan \left( {\dfrac{\pi }{3}} \right)?

Explanation

Solution

In order to solve this question ,calculate it as tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} taking θ  as  π3\theta \,\,as\,\,\dfrac{\pi }{3}.

Complete step-by-step answer:
In this we can use the fundamental trigonometric identity that is
tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
In our question, value of θ=π3\theta = \dfrac{\pi }{3}
Now putting θ=π3\theta = \dfrac{\pi }{3}
tan⁡(π3)=sin⁡(π3)cos⁡(π3)   \tan \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sin \left( {\dfrac{\pi }{3}} \right)}}{{\cos \left( {\dfrac{\pi }{3}} \right)}} \\\ \\\
From the trigonometric table we know the value of sin⁡(π3)=32\sin \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}and cos⁡(π3)=12\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}

tan⁡(π3)=3212 tan⁡(π3)=32×21 tan⁡(π3)=32×21 tan⁡(π3)=3   \tan \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}} \\\ \tan \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2} \times \dfrac{2}{1} \\\ \tan \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{{{2}}} \times \dfrac{{{2}}}{1} \\\ \tan \left( {\dfrac{\pi }{3}} \right) = \sqrt 3 \\\ \\\

Therefore ,value of tan⁡(π3)\tan \left( {\dfrac{\pi }{3}} \right) is 3\sqrt 3

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⁡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.
4.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.