Question

find the exact value of the real number y.

$y = \sin^{-1}\left(\frac{1}{2}\
ight)$

a. $\frac{\pi}{6}$
b. $-\frac{\pi}{6}$
c. 0
d. $\frac{\pi}{4}$

Explanation:

Step1: Recall the definition of inverse sine function

The inverse sine function \(y = \sin^{-1}(x)\) has a range \([-\frac{\pi}{2},\frac{\pi}{2}]\) and \(\sin^{-1}(\sin\theta)=\theta\) when \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\). We need to find \(\theta\) such that \(\sin\theta=\frac{1}{2}\) and \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\).

Step2: Use the unit - circle or special angles

We know that \(\sin\frac{\pi}{6}=\frac{1}{2}\) and \(\frac{\pi}{6}\in[-\frac{\pi}{2},\frac{\pi}{2}]\).

Answer:

A. \(\frac{\pi}{6}\)