Question

Find the principal value of \( \sec^{-1}(-\sqrt{2}) \).

Hint:

To find the principal value of \( \sec^{-1}(x) \), it is often easier to convert the problem to its cosine equivalent, \( \cos^{-1}(1/x) \), and find the angle within the range \( [0, \pi] \).

Answer 1

Step 1: Understanding the Concept:
We need to find the principal value of \( \sec^{-1}(-\sqrt{2}) \). The principal value is the unique angle \( \theta \) within the defined range of the inverse secant function such that \( \sec(\theta) = -\sqrt{2} \).
Step 2: Key Formula or Approach:
The principal value range for \( y = \sec^{-1}(x) \) is \( [0, \pi] - \{\frac{\pi}{2}\} \).
Let \( \theta = \sec^{-1}(-\sqrt{2}) \). This implies \( \sec(\theta) = -\sqrt{2} \), which is equivalent to \( \cos(\theta) = -\frac{1}{\sqrt{2}} \).
Step 3: Detailed Explanation or Calculation:
We need to find the angle \( \theta \) in the interval \( [0, \pi] \) for which \( \cos(\theta) = -\frac{1}{\sqrt{2}} \).
The reference angle, for which \( \cos(\alpha) = \frac{1}{\sqrt{2}} \), is \( \alpha = \frac{\pi}{4} \).
Since the cosine value is negative, the angle \( \theta \) must lie in the second quadrant.
The angle in the second quadrant with the reference angle \( \frac{\pi}{4} \) is given by \( \theta = \pi - \alpha \).
\[ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \] This value, \( \frac{3\pi}{4} \), lies within the principal value range of \( [0, \pi] \) and is not equal to \( \frac{\pi}{2} \).
Step 4: Final Answer:
The principal value of \( \sec^{-1}(-\sqrt{2}) \) is \( \frac{3\pi}{4} \).