18. arcsin \\frac{1}{2}

Explanation:

Step1: Recall the definition of arcsin function

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

Step2: Find the value of (\theta)

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

Answer:

(\frac{\pi}{6})