Question

sin\(^{-1} \left( \sin \left( \frac{5\pi}{6} \right) \right) \) is equal to:

• A. \( \frac{5\pi}{6} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{2\pi}{3} \)
• E. \( \frac{\pi}{2} \)

Hint:

For inverse sine functions, adjust angles outside the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) to an equivalent angle within the range.

Answer 1

The Correct Option is B

Step 1: We are given that \( \sin^{-1} (\sin \theta) = \theta \) for \( \theta \) in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
Since \( \frac{5\pi}{6} \) is outside this range, we need to adjust it to an equivalent angle within the range. 
The sine function is periodic, and \( \sin \left( \frac{5\pi}{6} \right) = \sin \left( \pi - \frac{5\pi}{6} \right) = \sin \left( \frac{\pi}{6} \right) \).
Thus, \( \sin^{-1} \left( \sin \left( \frac{5\pi}{6} \right) \right) = \frac{\pi}{6} \).