The principal value of \(\cos^{-1}\left(-\displaystyle\frac{1}{\sqrt{2}}\right)\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3\pi}{4}\)
The principal value branch of \(\cos^{-1}\) is \([0, \pi]\). A negative cosine needs an angle in the second quadrant, between \(\frac{\pi}{2}\) and \(\pi\).
Since \(\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\),
\( \displaystyle \cos\left(\pi - \frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} \)
Hence \(\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4}\).
Why the other options are incorrect
Option (2) \(\frac{\pi}{4}\) — Incorrect. \(\cos\frac{\pi}{4} = +\frac{1}{\sqrt{2}}\), not \(-\frac{1}{\sqrt{2}}\).
Option (3) \(-\frac{\pi}{4}\) — Incorrect. Negative angles are outside \([0, \pi]\), and \(\cos\left(-\frac{\pi}{4}\right) = +\frac{1}{\sqrt{2}}\).
Option (4) \(\frac{5\pi}{4}\) — Incorrect. \(\cos\frac{5\pi}{4} = -\frac{1}{\sqrt{2}}\), but \(\frac{5\pi}{4}\) is greater than \(\pi\), so it is outside the principal branch.