The problem in the image asks to find the exact value of the inverse sine function, specifically:

$\sin^{-1} \left( -\frac{1}{2} \right)$

This requires finding the angle whose sine is $-\frac{1}{2}$. Let's break this down:

In Radians:

The sine of $-\frac{1}{2}$ corresponds to an angle in the third or fourth quadrant, as sine is negative in those quadrants. The reference angle where sine equals $\frac{1}{2}$ is $\frac{\pi}{6}$. Therefore, in radians, the angle is:

$\theta = -\frac{\pi}{6}$

In Degrees:

Similarly, in degrees, the reference angle is $30^\circ$. Thus, the angle is:

$\theta = -30^\circ$

So the exact values are:

• In radians: $-\frac{\pi}{6}$
• In degrees: $-30^\circ$

Would you like further details or have any questions about this solution? Here are some related questions to expand on the topic:

1. How is the range of the inverse sine function defined?
2. What are the key differences between sine and inverse sine?
3. Can you explain why the reference angle is important in this context?
4. How would the answer change if the problem was asking for cosine or tangent?
5. What are some practical applications of inverse trigonometric functions?

Tip: For inverse trigonometric functions, always consider the principal range in which the output lies!