Question

The correct expression for \( \cos^{-1}(-x) \) is

• A. \( \frac{\pi}{2} - \cos^{-1}x \)
• B. \( \pi - \cos^{-1}x \)
• C. \( \pi + \cos^{-1}x \)
• D. \( \frac{\pi}{2} + \cos^{-1}x \)

Hint:

Remember the key inverse trigonometric identities for negative arguments: \[\begin{array}{rl} \bullet & \text{\( \sin^{-1}(-x) = -\sin^{-1}(x) \)} \\ \bullet & \text{\( \tan^{-1}(-x) = -\tan^{-1}(x) \)} \\ \bullet & \text{\( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \)} \\ \bullet & \text{\( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \)} \\ \bullet & \text{\( \sec^{-1}(-x) = \pi - \sec^{-1}(x) \)} \\ \end{array}\]

Answer 1

The Correct Option is B

This is a standard identity for inverse trigonometric functions. The range of the principal value of \( \cos^{-1}(x) \) is \( [0, \pi] \).
Let \( y = \cos^{-1}(-x) \). By definition, \( \cos(y) = -x \), where \( 0 \leq y \leq \pi \).
We know the trigonometric identity \( \cos(\pi - \theta) = -\cos(\theta) \). Let \( \theta = \cos^{-1}(x) \). Then \( \cos(\theta) = x \). So, \( -x = -\cos(\theta) = \cos(\pi - \theta) \). Substituting this back into our first equation:
\[ \cos(y) = \cos(\pi - \theta) \] \[ y = \pi - \theta \] Substitute back \( \theta = \cos^{-1}(x) \): \[ y = \pi - \cos^{-1}(x) \] Thus, \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \).