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:

& p=\cos \left( \dfrac{2\pi }{3} \right) \\\ & \Rightarrow p=\cos \left( \pi -\dfrac{\pi }{3} \right) \\\ \end{aligned}$$ Now, we will use the identity $$\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}$$