VerifiedHint: The range of \[{\tan ^{ - 1}}\theta \] is between \[\left. {\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right.} \right)\]. From the trigonometric table find the value of\[\cos \dfrac{{2\pi }}{3}\] . Now substitute this back into our given expression and simplify it to get the principal value.
Complete step-by-step answer:
A principal value of a function is the value selected at a point in the domain of a multiple-valued function, chosen so that the function has a single value at the point. The principal value of \[{\tan ^{ - 1}}\theta \]
The principal value of \[{\tan ^{ - 1}}\theta \] branches to,
\[{\tan ^{ - 1}}\theta \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\].
Hence the principal value of the given function will be between the range\[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\] .
Now we have been given the function, \[{\tan ^{ - 1}}\left( {2\cos \left( {\dfrac{{2\pi }}{3}} \right)} \right)\]. Let us put this function as equal to \[\theta \].
Thus we get,
$ {\tan ^{ - 1}}\left( {2\cos \dfrac{{2\pi }}{3}} \right) = \theta $
$ \tan \theta = 2\cos \dfrac{{2\pi }}{3} $
Let us first find the value of \[\cos \dfrac{{2\pi }}{3}\] from the above expression. By using the trigonometric table we can find the value of \[\cos \dfrac{{2\pi }}{3}\]. The angle \[\dfrac{{2\pi }}{3}\]is in the second quadrant where the cosine ratio has negative value. Thus we can write the angle \[\dfrac{{2\pi }}{3}\] as,
\[\dfrac{{2\pi }}{3} = \left( {\pi - \dfrac{\pi }{3}} \right) = \dfrac{\pi }{3}\]
As the cosine function is in the second quadrant the value of the cosine function will be negative. Thus we can write \[\cos \dfrac{{2\pi }}{3}\] as,
\[\cos \dfrac{{2\pi }}{3} = - \left( {\cos \dfrac{\pi }{3}} \right)\]
Now from the trigonometric table we can take the value of the function, \[\cos \dfrac{\pi }{3}\].
Hence from the above table we got the value of \[\cos \dfrac{\pi }{3} = \dfrac{1}{2}\]. Now let us substitute this into our given expression and simplify it.
$\tan \theta = 2\cos \dfrac{{2\pi }}{3} = 2 \times \left( { - \left( {\cos \dfrac{\pi }{3}} \right)} \right) $
$ \tan \theta = 2 \times \left( { - \dfrac{1}{2}} \right) = - 1 $
$ \tan \theta = - 1 $
Now let's take the inverse of the tangent function.
$\theta = {\tan ^{ - 1}}( - 1) $
$ \theta = - \dfrac{\pi }{4} $
Thus we got our principal value as \[\left( { - \dfrac{\pi }{4}} \right)\].
\[\therefore {\tan ^{ - 1}}\left( {2\cos \dfrac{{2\pi }}{3}} \right) = \left( { - \dfrac{\pi }{4}} \right)\].
Note:
In the first quadrant all the functions are positive. In the second quadrant sine and cosine functions are positive and the other functions are negative. Similarly, in the third quadrant tangent and cotangent functions are positive. In the fourth quadrant cosine and secant functions are positive. You can understand this from the figure we have drawn above.Students should remember the important trigonometric ratios and standard angles to solve these types of questions.
