Question

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

Hint:

Remember the principal value ranges for inverse trigonometric functions. For \(\cos^{-1}(x)\), it's \([0, \pi]\) (Quadrants I and II). For \(\sin^{-1}(x)\) and \(\tan^{-1}(x)\), it's \([-\frac{\pi}{2}, \frac{\pi}{2}]\) (Quadrants I and IV). This is crucial for finding the correct principal value.

Answer 1

Step 1: Understanding the Concept:
The principal value of an inverse trigonometric function, \(\cos^{-1}(x)\), is the unique angle \(\theta\) in the range \([0, \pi]\) such that \(\cos(\theta) = x\). For negative inputs, we use the identity \(\cos^{-1}(-x) = \pi - \cos^{-1}(x)\).
Step 2: Key Formula or Approach:
1. Use the identity: \(\cos^{-1}(-x) = \pi - \cos^{-1}(x)\).
2. Find the value of \(\cos^{-1}(x)\).
3. Substitute this value back into the identity to find the principal value.
Step 3: Detailed Explanation or Calculation:
We need to find the principal value of \(\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)\).
Using the identity, we have: \[ \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \pi - \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \] We know that the angle \(\theta\) in \([0, \pi]\) for which \(\cos(\theta) = \frac{1}{\sqrt{2}}\) is \(\frac{\pi}{4}\).
So, \(\cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\).
Substituting this back: \[ \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \pi - \frac{\pi}{4} \] \[ = \frac{4\pi - \pi}{4} = \frac{3\pi}{4} \] The value \(\frac{3\pi}{4}\) lies in the principal value range of \(\cos^{-1}(x)\), which is \([0, \pi]\).
Step 4: Final Answer:
The principal value of \(\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)\) is \(\frac{3\pi}{4}\).