The expression you have provided is:

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

This asks for the angle $\theta$ such that:

$\sin(\theta) = -\frac{\sqrt{2}}{2}$

Step-by-Step Solution:

1. Understand the range of $\sin^{-1}(x)$: The range of $\sin^{-1}(x)$ (arcsine) is $[- \frac{\pi}{2}, \frac{\pi}{2}]$, meaning the result must be an angle in this interval.

2. Recall the sine value: The value $\frac{\sqrt{2}}{2}$ corresponds to an angle of $\frac{\pi}{4}$ (or 45°) in the first quadrant. Since the sine is negative, the angle must lie in the fourth quadrant (within the range of arcsine).

3. Determine the angle: In the fourth quadrant, the corresponding angle is: $\theta = -\frac{\pi}{4}$

Final Answer:

$\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}$

---

Let me know if you want a detailed explanation or have any follow-up questions.

Related Questions:

1. What is $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)$?
2. How does the sine function behave in different quadrants?
3. What is the difference between $\sin^{-1}(x)$ and $\arcsin(x)$?
4. How is the unit circle used to find arcsine values?
5. How can the range of $\sin^{-1}(x)$ affect its solutions?

Tip:

To solve inverse trigonometric functions quickly, always visualize or refer to the unit circle for clarity!