Question

Prove that \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \), where \( x \in [-1, 1] \)

Hint:

The sum of the inverse sine and inverse cosine of the same value is always \( \frac{\pi}{2} \).

Answer 1

Step 1: Understanding the inverse sine and cosine functions.
Let \( y = \sin^{-1}(x) \). By the definition of inverse sine: \[ \sin(y) = x \,\,\, \text{where} \,\,\, y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \] Similarly, let \( z = \cos^{-1}(x) \). By the definition of inverse cosine: \[ \cos(z) = x \,\,\, \text{where} \,\,\, z \in \left[ 0, \pi \right] \]

Step 2: Proof of the identity.
We know that for all \( x \) in the interval \( [-1, 1] \), \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \). To prove this, observe the following: - The sine and cosine functions are complementary, meaning \( \sin\left(\frac{\pi}{2} - x\right) = \cos(x) \). - Since \( \sin^{-1}(x) = y \) and \( \cos^{-1}(x) = z \), we know that \( \sin(y) = x \) and \( \cos(z) = x \). - By definition, \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \) because \( y + z = \frac{\pi}{2} \).

Step 3: Conclusion.
Thus, we have proved that \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \) for \( x \in [-1, 1] \).