Question

How do you evaluate \[\cos \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)\] without a calculator?

Answer 1

Hint: This question is from the topic of trigonometry. In solving this question, we will equate \[{{\tan }^{-1}}\left( \dfrac{1}{2} \right)\] with \[x\] and solve the further question. After that, we will use a right angled triangle and find the value of \[\cos x\]. After solving the further question, we will get our answer.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to evaluate the given term without using a calculator. The given term is \[\cos \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)\]
Let us first use the term \[{{\tan }^{-1}}\left( \dfrac{1}{2} \right)\] and equate this with \[x\].
So, we can write
\[x={{\tan }^{-1}}\left( \dfrac{1}{2} \right)\]
The above equation can also be written as
\[\Rightarrow \tan x=\dfrac{1}{2}\]
We know that in a right angled triangle, the value of tan is \[\dfrac{\text{perpendicular side}}{\text{base side}}\]
We can understand this from the following figure:

We know that in a right angled triangle,
\[\text{hypotenuse side = }\sqrt{{{\left( \text{perpendicular side} \right)}^{2}}+{{\left( \text{base side} \right)}^{2}}}\]
Putting the values of sides, we can write the above equation as
\[\Rightarrow \text{hypotenuse side = }\sqrt{{{1}^{2}}+{{\text{2}}^{2}}}\]
\[\Rightarrow \text{hypotenuse side = }\sqrt{5}\]
We know that in a right angled triangle, the value of cos is
\[\cos x=\dfrac{\text{base side}}{\text{hypotenuse side}}\]
Putting the values, we can write the above as
\[\Rightarrow \cos x=\dfrac{2}{\sqrt{5}}\]
The above can also be written as
\[\Rightarrow x={{\cos }^{-1}}\left( \dfrac{2}{\sqrt{5}} \right)\]
Now, we can write \[\cos \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)\] as
\[\Rightarrow \cos \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)=\cos \left( x \right)\]
By putting the value of \[x\] as \[{{\cos }^{-1}}\left( \dfrac{2}{\sqrt{5}} \right)\], we can write
\[\Rightarrow \cos \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)=\cos \left( {{\cos }^{-1}}\left( \dfrac{2}{\sqrt{5}} \right) \right)\]
The above can also be written as
\[\Rightarrow \cos \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)=\dfrac{2}{\sqrt{5}}\]
As we know that the value of tan is positive from 0 degrees to 90 degrees and from 180 degrees to 270 degrees, but the value of cos is positive from 0 degrees to 90 degrees and negative from 180 degrees to 270 degrees. So, we will have to write
\[\Rightarrow \cos \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)=\pm \dfrac{2}{\sqrt{5}}\]

Note: As we can see that this question is from the trigonometry, so we should have a better knowledge in the topic of trigonometry.
We should know everything about the right angled triangle. Let us understand about the same from the following:

Remember that:
\[\text{hypotenuse side = }\sqrt{{{\left( \text{perpendicular side} \right)}^{2}}+{{\left( \text{base side} \right)}^{2}}}\]
\[\sin x=\dfrac{\text{perpendicular side}}{\text{hypotenuse side}}\]
\[\cos x=\dfrac{\text{base side}}{\text{hypotenuse side}}\]
\[\tan x=\dfrac{\text{perpendicular side}}{\text{base side}}\]