Inverse Trigonometric Functions — Class 12 Maths Solution

Let $x = {\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) \Rightarrow - \frac{1}{2} = \cos x$

As we know that the range of principal value branch of ${\cos ^{ - 1}}\;is\;[0,\;\pi ]$

Then, $\left( { - \frac{1}{2}} \right) = - \cos \frac{\pi }{3} = \cos \left( {\pi - \frac{\pi }{3}} \right) = \cos \left( {\frac{{2\pi }}{3}} \right),$

Where $\frac{{2\pi }}{3} \in [0,\;\pi ]$
Hence, the principal value of ${\cos ^{ - 1}}\left( { - \frac{1}{2}} \right)\;is\;\frac{{2\pi }}{3}.$