Question

Find \( \sin^{-1} \left( \sin \frac{23\pi}{6} \right) = \) __________.

Hint:

When simplifying \( \sin^{-1} (\sin \theta) \), reduce \( \theta \) to an equivalent angle within the range \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \).

Answer 1

Step 1: We are given \( \sin^{-1} \left( \sin \frac{23\pi}{6} \right) \), and we need to simplify the expression. 

Step 2: First, simplify \( \frac{23\pi}{6} \) by reducing it within the range of the sine inverse function, which is \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \). To do this, observe that: \[ \frac{23\pi}{6} = 2\pi + \frac{11\pi}{6}. \] Since \( \sin(\theta) \) is periodic with period \( 2\pi \), we have: \[ \sin \frac{23\pi}{6} = \sin \frac{11\pi}{6}. \] 

Step 3: Now, \( \frac{11\pi}{6} \) lies in the fourth quadrant. In the fourth quadrant, \( \sin \left( \frac{11\pi}{6} \right) = - \sin \left( \frac{\pi}{6} \right) = -\frac{1}{2} \). 

Step 4: Therefore, we have: \[ \sin^{-1} \left( \sin \frac{23\pi}{6} \right) = \sin^{-1} \left( -\frac{1}{2} \right). \]

 Step 5: The angle whose sine is \( -\frac{1}{2} \) in the range \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \) is \( -\frac{\pi}{6} \). 

Thus, \[ \sin^{-1} \left( \sin \frac{23\pi}{6} \right) = -\frac{\pi}{6}. \]