Question

Find the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$.

Answer 1

Hint: The principal value of an inverse trigonometric function at a point x is the value of the inverse function at the point x, which lies in the range of the principal branch.

Complete step-by-step answer:
The principal values of the trigonometric functions are-

Function	Principal Value
$sin^{-1}$	$\left[ { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right]$
$cos^{-1}$	$\left[ {0,\;{{\pi }}} \right]$
$tan^{-1}$	$\left( { - \dfrac{{{\pi }}}{2},\dfrac{{\;{{\pi }}}}{2}} \right)$
$cot^{-1}$	$\left( {0,\;{{\pi }}} \right)$
$sec^{-1}$	\left[ {0,\;{{\pi }}} \right] - \left\\{ {\dfrac{{{\pi }}}{2}} \right\\}
$cosec^{-1}$	\left[ { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right] - \left\\{ 0 \right\\}

In the given question we need to find the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$. We know that for cosine function to be negative, the angle should be obtuse, that is greater than $90^{o}$.

We know that $\cos {60^o} = \dfrac{1}{2}$. This means that the value of ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = {60^o}$. So the value of the given expression can be calculated as-
${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = {180^{\text{o}}} - {60^{\text{o}}} = {120^{\text{o}}}$
We know that ${{\pi }}$ rad = $180^{o}$, so
${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{{2{{\pi }}}}{3}$
This is the required value.

Note: In such types of questions, we need to strictly follow the range of the principal values that have been specified. This is because there can be infinite values of any inverse trigonometric functions. For example, the value of $sin^{-1}0$ can be $0^{o}, 180^{o}, 360^{o}$ and so on. Hence, we need to write only that value which is inside the principal range.