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The inverse sine function, also known as arcsine, is denoted by \( \sin^{-1} \) or \( \arcsin \). It is used to find the angle when the sine value is given. For instance, if we say \( \sin(x) = y \), then \( x = \arcsin(y) \) finds the angle \( x \). In practical terms, this means that if you know the sine of an angle, the inverse sine function will tell you what that angle is.
Understanding the inverse sine is crucial when working with angles and trigonometric equations because it allows us to solve for angles using known sine values. The inverse sine function has a distinct range of values (which we'll discuss in the "Principal Range of Inverse Functions" section), making it easy to find angles that fit within common trigonometric scenarios. It's important to remember that the sine of an angle can only have values between -1 and 1, which limits how the inverse sine function operates.

Special right triangles are specific triangles with angles and side ratios that are easily recognizable and commonly used in trigonometry. These known ratios make them particularly valuable when using inverse trigonometric functions.
One of the most notable examples is the 45°-45°-90° triangle, where the sides are in the ratio 1:1:\( \sqrt{2} \). This means that both legs are equal and the hypotenuse is \( \sqrt{2} \) times the length of one leg. The sine of a 45-degree angle is \( \frac{1}{\sqrt{2}} \), which is essential to our problem.
Additionally, there are 30°-60°-90° triangles, which also have set ratios. However, in this problem, understanding the 45°-45°-90° triangle helps identify the angle corresponding to our given sine value of \( \frac{1}{\sqrt{2}} \). Recognizing these key triangles helps in quickly solving trigonometric rates without needing extensive calculations.

Inverse trigonometric functions, like \( \sin^{-1} \), have specific ranges where they output their values. This is known as the principal range. For \( \sin^{-1}(x) \), the principal range is \([-90^{\circ}, 90^{\circ}]\) or \([-\frac{\pi}{2}, \frac{\pi}{2}]\) in radians.
The principal range is essentially the set of angles that the inverse function will return, ensuring a one-to-one mapping of sine values to angles within that range. This is crucial because it allows us to predict exactly what angle the inverse function will yield without ambiguity.
In our particular example, since \( \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \) is within this range, we confirm that the angle is 45 degrees. Other angles might also share the same sine value, but only those within the principal range are considered by \( \sin^{-1} \). Thus, it simplifies solving the equation by focusing on a well-defined and constrained angle range.