If the branch of \(\cos^{-1}\) with range \([\pi, 2\pi]\) is used instead of the principal value branch, then the value of \(\cos^{-1}\left(\displaystyle\frac{1}{2}\right)\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{5\pi}{3}\)
On \([\pi, 2\pi]\) the cosine rises from \(\cos\pi = -1\) to \(\cos 2\pi = 1\), so it takes each value in \([-1, 1]\) exactly once. In this branch, \(\cos^{-1}\left(\frac{1}{2}\right)\) is the one angle in \([\pi, 2\pi]\) whose cosine is \(\frac{1}{2}\).
Cosine does not change when the angle is replaced by \(2\pi\) minus the angle:
\( \displaystyle \cos\left(2\pi - \frac{\pi}{3}\right) = \cos\frac{\pi}{3} = \frac{1}{2} \)
and \(2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\) lies in \([\pi, 2\pi]\). Hence the value is \(\frac{5\pi}{3}\).
Why the other options are incorrect
Option (1) \(\frac{\pi}{3}\) — Incorrect. This is the principal value; \(\frac{\pi}{3}\) is not in \([\pi, 2\pi]\).
Option (2) \(\frac{4\pi}{3}\) — Incorrect. \(\frac{4\pi}{3}\) is in the branch, but \(\cos\frac{4\pi}{3} = -\frac{1}{2}\).
Option (3) \(-\frac{\pi}{3}\) — Incorrect. \(\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}\), but a negative angle is not in \([\pi, 2\pi]\).