Question

How do you simplify the expression $\cos \left( {\arctan \left( {\dfrac{x}{5}} \right)} \right)$?

Answer 1

Hint: This problem deals with applying the basic and important trigonometric identities. We are given a tangent trigonometric expression inside of which there is an inverse of the cosine trigonometric expression of a particular value. So in order to proceed to get the exact value of the expression, first we need to assign the given inverse cosine trigonometric value to a variable, and then solve.

Complete step-by-step answer:
Given the expression of trigonometric and inverse trigonometric ratio which is $\cos \left( {ta{n^{ - 1}}\left( {\dfrac{x}{5}} \right)} \right)$.
Consider the given expression, as given below:
$ \Rightarrow \cos \left( {ta{n^{ - 1}}\left( {\dfrac{x}{5}} \right)} \right)$
Now consider the inside of cosine value which is ${\tan ^{ - 1}}\left( {\dfrac{x}{5}} \right)$, as given below:
Let the expression of ${\tan ^{ - 1}}\left( {\dfrac{x}{5}} \right)$ is equal to $\alpha $, which is mathematically expressed below:
$ \Rightarrow \alpha = {\tan ^{ - 1}}\left( {\dfrac{x}{5}} \right)$
Now take inverse tangent trigonometric function on both the sides of the above equation, as shown below:
$ \Rightarrow \tan \alpha = \tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{5}} \right)} \right)$
Here on the right hand side of the above equation, cosine and inverse tangent trigonometric function gets cancelled, as shown below:
$ \Rightarrow \tan \alpha = \dfrac{x}{5}$
Now as we considered $\alpha = {\tan ^{ - 1}}\left( {\dfrac{x}{5}} \right)$, hence the expression $\cos \left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{5}} \right)} \right)$ becomes as shown below:
\[ \Rightarrow \cos \left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{5}} \right)} \right) = \cos \alpha \]
So if we find the value of $\cos \alpha $, then it is the same as finding the value of\[\cos \left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{5}} \right)} \right)\].
Hence finding the value of $\cos \alpha $.
But we know the value of $\tan \alpha $, which is equal to $\dfrac{x}{5}$.
Hence to find the value of $\cos \alpha $, we can express $\cos \alpha $in terms of $\tan \alpha $, and then can get the value of $\cos \alpha $.
So expressing $\cos \alpha $ in terms of $\tan \alpha $, as given below:
\[ \Rightarrow \cos \alpha = \dfrac{1}{{\sec \alpha }}\]
We know that from the basic trigonometric identity ${\sec ^2}\alpha - {\tan ^2}\alpha = 1$, from here the value of $\sec \alpha $ can be written as:
$ \Rightarrow {\sec ^2}\alpha = 1 + {\tan ^2}\alpha $
$\therefore \sec \alpha = \sqrt {1 + {{\tan }^2}\alpha } $
Now substituting the above expression in the expression of $\cos \alpha $, as given below:
\[ \Rightarrow \cos \alpha = \dfrac{1}{{\sqrt {1 + {{\tan }^2}\alpha } }}\]
We obtained that the value of $\tan \alpha = \dfrac{x}{5}$, hence substituting it in the above expression, as shown:
\[ \Rightarrow \cos \alpha = \dfrac{1}{{\sqrt {1 + {{\left( {\dfrac{x}{5}} \right)}^2}} }}\]
Simplifying the above expression, as given below:
\[ \Rightarrow \cos \alpha = \dfrac{1}{{\sqrt {\dfrac{{25 + {x^2}}}{{25}}} }}\]
$ \Rightarrow \cos \alpha = \dfrac{5}{{\sqrt {{x^2} + 25} }}$
As we know that the value of under root of 25 is 5, $\sqrt {25} = 5$, as shown above.

Hence the value of $\cos \left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{5}} \right)} \right) = \dfrac{5}{{\sqrt {{x^2} + 25} }}$.

Note:
Please note that while solving any trigonometric based problems, we need to be through with all the important and basic trigonometric identities, few are given below:
$ \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = 1$
From which we can obtain $\sin \alpha = \sqrt {1 - {{\cos }^2}\alpha } $
$ \Rightarrow {\sec ^2}\alpha - {\tan ^2}\alpha = 1$
From which we can obtain \[\sec \alpha = \sqrt {1 + {{\tan }^2}\alpha } \]
$ \Rightarrow \cos e{c^2}\alpha - {\cot ^2}\alpha = 1$
From which we can obtain $\cot \alpha = \sqrt {\cos e{c^2}\alpha - 1} $