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The question asks to solve for \( \sin^{-1}(0.5) \) using the given equality \( \sin 30^\circ = 0.5 \).

From the given information, we know:
\[
\sin 30^\circ = 0.5
\]
Thus, the inverse sine of 0.5, \( \sin^{-1}(0.5) \), will return the angle whose sine is 0.5. From the equation, we know that:
\[
\sin^{-1}(0.5) = 30^\circ
\]

Therefore, the completed expression is:
\[
\sin^{-1}(0.5) = 30^\circ
\]

Would you like further details on how inverse trigonometric functions work or any related concepts? Here are five related questions you might find interesting:

1. What is the general solution for \( \sin^{-1}(x) \) for values other than 0.5?
2. How do the sine and cosine functions differ when dealing with inverse values?
3. Can \( \sin^{-1}(x) \) return multiple answers depending on the context?
4. How is the unit circle used to understand inverse trigonometric functions?
5. What are the real-life applications of inverse trigonometric functions?

**Tip:** When working with inverse trigonometric functions, remember that they return principal values, typically within a specific range (e.g., for \( \sin^{-1}(x) \), the output is typically between \( -90^\circ \) and \( 90^\circ \) or \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \) in radians).

Using the equality sin 30° = 0.5, copy and complete the following: sin^(-1)(0.5) = ______.

Learn how to solve the inverse sine function sin^(-1)(0.5) using the given equality sin 30° = 0.5. This question covers key trigonometric concepts, including how to find the angle whose sine is 0.5. Suitable for students in Grades 9-11.

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