The value of \( \arcsin\left(\frac{\sqrt{3}}{2}\right) \) is the angle whose sine is \( \frac{\sqrt{3}}{2} \). In the unit circle, the sine of \( \frac{\sqrt{3}}{2} \) corresponds to an angle of:

\[
\arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \text{ radians or } 60^\circ
\]

This is because \( \sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \), and the principal value of \( \arcsin \) is restricted to the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

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Here are 5 related questions:
1. What is \( \arccos\left(\frac{\sqrt{3}}{2}\right) \)?
2. How is the range of the arcsin function determined?
3. Can the sine of any angle ever be greater than 1 or less than -1?
4. How would you compute \( \arcsin\left(\frac{1}{2}\right) \)?
5. What is the general solution for \( \sin(\theta) = \frac{\sqrt{3}}{2} \)?

**Tip:** The \( \arcsin \) function always returns an angle in the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

Explore the calculation of \( \arcsin\left(\frac{\sqrt{3}}{2}\right) \), the angle whose sine is \( \frac{\sqrt{3}}{2} \), in both radians and degrees. This trigonometry problem involves the use of inverse trigonometric functions and the unit circle, suitable for Grades 10-12.

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