Question

Simplify : \( \sin^{-1} \sqrt{\frac{1 + \cos 2x}{2}} \), \( 0 < x < \frac{\pi}{2} \).

Hint:

Memorizing \( \sin^{-1}(\cos x) = \frac{\pi}{2} - x \) for small \( x \) saves time in calculus.
Always pay attention to the interval to correctly resolve square roots of squares.

Answer 1

Concept:
• Double Angle Formulas: \( 1 + \cos 2x = 2 \cos^2 x \).
• Relation between Sine and Cosine: \( \cos x = \sin \left( \frac{\pi}{2} - x \right) \).

Step 1: Substitute the identity for \( 1 + \cos 2x \)
\[ \sqrt{\frac{1 + \cos 2x}{2}} = \sqrt{\frac{2 \cos^2 x}{2}} \] \[ = \sqrt{\cos^2 x} = |\cos x| \]

Step 2: Determine the sign based on the given interval
Given \( 0 < x < \frac{\pi}{2} \), \( x \) is in the first quadrant. In the first quadrant, \( \cos x \) is positive. \[ |\cos x| = \cos x \]

Step 3: Apply the inverse sine transformation
The expression becomes: \[ \sin^{-1}(\cos x) \] Convert cosine to sine to match the inverse function: \[ \sin^{-1} \left[ \sin \left( \frac{\pi}{2} - x \right) \right] \] Check the range: Since \( 0 < x < \frac{\pi}{2} \), then \( 0 < \frac{\pi}{2} - x < \frac{\pi}{2} \). This is within the principal range of \( \sin^{-1} \). \[ = \frac{\pi}{2} - x \]