We are tasked with evaluating $\cos\left(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)$.

Step-by-step Solution:

1. Interpret the problem:
   The expression $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$ represents an angle $\theta$, where: $\sin(\theta) = \frac{\sqrt{3}}{2}.$ From trigonometry, $\theta$ lies in the range $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ for the inverse sine function.

2. Identify the reference angle:
   From basic trigonometric values, $\sin(\theta) = \frac{\sqrt{3}}{2}$ corresponds to $\theta = \frac{\pi}{3}$.

3. Use the Pythagorean identity:
   We need to find $\cos(\theta)$. Using the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1.$ Substitute $\sin(\theta) = \frac{\sqrt{3}}{2}$: $\left(\frac{\sqrt{3}}{2}\right)^2 + \cos^2(\theta) = 1.$ $\frac{3}{4} + \cos^2(\theta) = 1.$ $\cos^2(\theta) = 1 - \frac{3}{4} = \frac{1}{4}.$ $\cos(\theta) = \pm\frac{1}{2}.$

4. Determine the sign of $\cos(\theta)$:
   Since $\sin^{-1}$ always gives angles in the first or fourth quadrant (where cosine is non-negative), we take: $\cos(\theta) = \frac{1}{2}.$

Final Answer:

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

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Would you like more details or have any questions? Here are some related topics you might find interesting:

1. What are the ranges of inverse trigonometric functions?
2. How is the Pythagorean identity derived?
3. Why is the cosine positive in the first and fourth quadrants?
4. How do you simplify other expressions involving inverse trig functions?
5. What are the exact values of all six trigonometric functions at common angles?

Tip: Always sketch a triangle when working with inverse trig functions to visualize the relationships easily!