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The sine function, denoted as \( \sin \), is a fundamental trigonometric function. It relates the ratio of the opposite side of a right triangle to its hypotenuse for any given angle within the triangle. It takes an angle as an input and outputs a value between -1 and 1.

To evaluate expressions involving the sine function, it's helpful to remember some common sine values for well-known angles:

• \( \sin 0 = 0 \)
• \( \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \)
• \( \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
• \( \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)
• \( \sin \left(\frac{\pi}{2}\right) = 1 \)

Negative angles reflect across the origin in the unit circle, causing the sine value to be negative. For instance, \( \sin \left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \). Recognizing these key values and their symmetrical properties is crucial for solving problems without a calculator.

Angles can be measured in degrees or radians, but for trigonometry, radians are often more convenient. One complete revolution around a circle is \( 2\pi \) radians, equivalent to 360 degrees. Consequently, \( \pi \) radians is equal to 180 degrees.

Radians provide a direct connection between the arc length of a circle and the radius. Therefore:

• 90 degrees is \( \frac{\pi}{2} \) radians
• 60 degrees is \( \frac{\pi}{3} \) radians
• 45 degrees is \( \frac{\pi}{4} \) radians

Understanding this conversion: \( \text{degrees} = \text{radians} \times \frac{180}{\pi} \), helps when transferring angles between systems. For trigonometric identities and inverse functions like \( \arcsin \), mastering angle measurements in radians is essential.

The range of a function represents all possible output values it can produce. For the \( \arcsin \) or inverse sine function, this range is restricted. It outputs angles \( \theta \) where \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \). This restriction ensures that each input has a unique output, making \( \arcsin \) a well-defined function.

When evaluating \( \arcsin \left(-\frac{\sqrt{3}}{2}\right) \), the resulting angle \( \theta \) must lie within this range. Since \( \sin \left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \), the solution to \( \arcsin \left(-\frac{\sqrt{3}}{2}\right) \) is \( -\frac{\pi}{3} \). This value satisfies the range condition perfectly:

• It stays within \( -\frac{\pi}{2} \leq -\frac{\pi}{3} \leq \frac{\pi}{2} \).

Understanding the range restriction is key to solving inverse function problems accurately.