To find the value of $\cos\left(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)$, let's break it down step by step.

Step 1: Interpret the inverse sine

We are given $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$, which is the angle $\theta$ such that:

$\sin(\theta) = -\frac{\sqrt{3}}{2}$

We need to determine the angle $\theta$ that satisfies this equation. The sine function $\sin(\theta) = -\frac{\sqrt{3}}{2}$ occurs at specific angles, and we know that:

$\sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}$

Thus, the angle $\theta = -\frac{\pi}{3}$.

Step 2: Find $\cos(\theta)$

Next, we want to find $\cos\left(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right) = \cos\left(-\frac{\pi}{3}\right)$.

Using the property of cosine that $\cos(-\theta) = \cos(\theta)$, we have:

$\cos\left(-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right)$

The cosine of $\frac{\pi}{3}$ is:

$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$

Thus,

$\cos\left(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right) = \frac{1}{2}$

Final Answer:

$\cos\left(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right) = \frac{1}{2}$

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Would you like further details or have any questions about this process?

Here are five related questions:

1. How do you find $\cos\left(\sin^{-1}(x)\right)$ for any value of $x$?
2. How do inverse trigonometric functions work in general?
3. What is the significance of $\sin^{-1}(x)$ and its range?
4. Can you solve for $\cos\left(\sin^{-1}(x)\right)$ when $x$ is positive?
5. How does the unit circle help in evaluating inverse trigonometric functions?

Tip: When working with inverse trigonometric functions, remember that the range of $\sin^{-1}(x)$ is restricted to $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, which helps determine the correct quadrant for the angle.