Question

Find the principal value of each of the following:
\[{{\tan }^{-1}}\left( 2\cos \dfrac{2\pi }{3} \right)\]

Answer 1

Hint: To solve the question given above, first we will draw the rough graph of \[y={{\tan }^{-1}}x\] and we will determine the nature of the graph. Then we will find the value of \[\cos \dfrac{2\pi }{3}\]. Then we will put this in the term \[{{\tan }^{-1}}\left( 2\cos \dfrac{2\pi }{3} \right)\] and then we will find its value.

Complete step-by-step solution-
Before solving the question, we must know what is the nature of the graph \[y={{\tan }^{-1}}x\]. For determining the nature of the graph, we will draw the graph. Thus, we have:

From the above graph, we can say that the inverse trigonometric function \[{{\tan }^{-1}}x\] as an odd function then, we have the following relation:
\[\Rightarrow f\left( -x \right)=-f\left( x \right)\]
Now, the term of which we have to find the value is \[{{\tan }^{-1}}\left( 2\cos \dfrac{2\pi }{3} \right)\]. Let its value by y. Thus, we have the following equation:
\[y={{\tan }^{-1}}\left( 2\cos \dfrac{2\pi }{3} \right)\] --------- (1)
We will first find out the value of \[\cos \dfrac{2\pi }{3}\]. Let its value be ‘p’. Thus, we have,
\[\Rightarrow y={{\tan }^{-1}}\left( p \right)\] -------- (2)
Now, we will find the value of p. Thus, we have:
\[\begin{align}
  & p=\cos \left( \dfrac{2\pi }{3} \right) \\
 & \Rightarrow p=\cos \left( \pi -\dfrac{\pi }{3} \right) \\
\end{align}\]
Now, we will use the identity \[\cos \left( \pi -x \right)=-\cos x\] in the above equation. In our case, \[x=\dfrac{\pi }{3}\]. So, we have:
\[\Rightarrow p=-\cos \dfrac{\pi }{3}\]
The value of \[\cos \dfrac{\pi }{3}\] is \[\dfrac{1}{2}\]. So, we will get:
\[\Rightarrow p=-\dfrac{1}{2}\] ----- (3)
Now, we will put the value of p from (3) into (1). Thus, because \[{{\tan }^{-1}}x\] is an odd function. So, we will get:
\[\Rightarrow y={{\tan }^{-1}}\left( 1 \right)\] --------- (4)
We know that \[\tan \dfrac{\pi }{4}=1\]. So, \[\dfrac{\pi }{4}={{\tan }^{-1}}\left( 1 \right)\]. Thus, we will get:
\[\Rightarrow y=-\left( \dfrac{\pi }{4} \right)\]
\[\Rightarrow y=-\dfrac{\pi }{4}\] ------- (5)
From (1) and (5), we have:
\[{{\tan }^{-1}}\left( 2\cos \dfrac{2\pi }{3} \right)=-\dfrac{\pi }{4}\].

Note: We can also solve the above question by following method, we know that:
\[y={{\tan }^{-1}}\left( 2\cos \dfrac{2\pi }{3} \right)\]
The value of \[\cos \dfrac{2\pi }{3}=\dfrac{-1}{2}\]. So, we have, \[y={{\tan }^{-1}}\left( -1 \right)\].
We know that, \[\tan \dfrac{\pi }{4}=1\]. So, we will get:
\[y={{\tan }^{-1}}\left( -\tan \dfrac{\pi }{4} \right)\]
Now, we know that \[-\tan \theta \] can be written as \[\left( -\tan \theta \right)\].
So, we will get:
\[y={{\tan }^{-1}}\left( \tan \left( \dfrac{-\pi }{4} \right) \right)\]
Now, we will use the following identity:
\[{{\tan }^{-1}}\left( \tan x \right)=x\] (if \[\dfrac{-\pi }{2}\le x\le \dfrac{\pi }{2}\])
Thus, we will get:
\[y=\dfrac{-\pi }{4}\]