Question

How do you evaluate \[arctan\left( \dfrac{\surd 3}{3} \right)\]?

Answer 1

Hint: We already know that tan or tangent function is well-defined and one to one. So its inverse exists and its inverse is called arctangent or shortly as \[arctan\]. Thus the domain of \[arctan\] is the range of tan which is the set of real numbers \[\left( -\infty ,\infty \right)\] and its range is equal to the domain of tangent function, that is the set \[\left( -\pi /2,\pi /2 \right)\]. So, we can evaluate the value of \[arctan\] easily if we know the value of tan. Let us suppose \[tan\text{ }x=y\].
Then $arctan(y)=ta{{n}^{(-1)}}(y)=ta{{n}^{(-1)}}(tanx)$
              \[=x\]
Hence this way is the most suitable to find the value of arctan. Although there exists a complete list of values of the functions for various values of the domain which could be used to determine the value of Trigonometric functions.

Complete step by step solution:
We have to evaluate the value of \[arctan\left( \dfrac{\surd 3}{3} \right)\]
To find this , we need to first simplify the value \[\left( \dfrac{\surd 3}{3} \right)\]
Let us consider,
$\dfrac{\sqrt{3}}{3}=\dfrac{\sqrt{3}}{{{(\sqrt{3})}^{2}}}=\dfrac{1}{\sqrt{3}}$
Now we can use this value in place of \[\left( \dfrac{\surd 3}{3} \right)\]
So \[arctan\left( \dfrac{\surd 3}{3} \right)=\arctan \dfrac{1}{\sqrt{3}}\]
We already know that the value of \[tan\left( \dfrac{\pi }{6} \right)\] is \[1/\surd 3\]
Then $arctan(\dfrac{1}{\sqrt{3}})=ta{{n}^{-1}}(\dfrac{1}{\sqrt{3}})=ta{{n}^{-1}}(tan\dfrac{\pi }{6})=\pi /6$.

Therefore , we have calculated the value of \[arctan\left( \dfrac{\surd 3}{3} \right)\] which is \[\pi /6\].

Note:
Arctan, arcsin, and arccos functions return angles either in degree or radians as the answer. It is because these angles are the same whose tan, sin, and cos respectively would give the given number.