Problem 4 - Precalculus Mathematics for Calculus

Question

Find the exact value of each expression, if it is defined. (a) \(\sin ^{-1}(-1)\) (b) \(\sin ^{-1} \frac{\sqrt{2}}{2}\) (c) \(\sin ^{-1}(-2)\)

Step-by-Step Solution

Answer

(a) \(-\frac{\pi}{2}\), (b) \(\frac{\pi}{4}\), (c) Undefined

Step 1: Understand the inverse sine function

The inverse sine function, written as \(\sin^{-1}(x)\), is the angle whose sine is \(x\). The range of \(\sin^{-1}\) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), meaning the output will be an angle in this interval.

Step 2: Evaluate \(\sin^{-1}(-1)\)

\(\sin^{-1}(-1)\) is the angle whose sine value is \(-1\). The sine of \(-\frac{\pi}{2}\) is \(-1\), so \(\sin^{-1}(-1) = -\frac{\pi}{2}\).

Step 3: Evaluate \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\)

\(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\) is the angle whose sine value is \(\frac{\sqrt{2}}{2}\). The sine of \(\frac{\pi}{4}\) is \(\frac{\sqrt{2}}{2}\), so \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}\).

Step 4: Determine if \(\sin^{-1}(-2)\) is defined

The sine function can only take values between \(-1\) and \(1\). \(-2\) is outside this range, so \(\sin^{-1}(-2)\) is undefined.

Key Concepts

Range of Inverse Sine Function

The inverse sine function, often denoted as \(\sin^{-1}(x)\) or \(\arcsin(x)\), maps a value from its domain, which ranges between \(-1\) to \(1\), to an angle. This angle is always within its specific range, \[-\frac{\pi}{2}, \frac{\pi}{2}\]. This range is crucial when working on problems involving inverse sine because it tells us where the resulting angle must lie. For example, when you evaluate \(\sin^{-1}(-1)\), you expect the angle to be at the edge of this range, precisely \(-\frac{\pi}{2}\), since the sine of \(-\frac{\pi}{2}\) is \(-1\). However, if you try to find \(\sin^{-1}(-2)\), it is undefined because \(-2\) lies outside the permissible domain of the sine function.

Sine Function Properties

The sine function is a fundamental trigonometric function with specific properties. It is periodic and oscillates between \(-1\) and \(1\), which defines its range.
This alternating pattern is critical because it dictates the input values that the inverse sine function can accept.

• Sine of an angle remains the same for supplementary angles (e.g., \(\sin(\pi - x) = \sin(x)\)).
• Its maximum value of 1 occurs at \(\frac{\pi}{2}\) (90°) and the minimum value of -1 at \(-\frac{\pi}{2}\) (-90°).

Understanding these properties helps when evaluating inverse sine expressions because they determine which angles in the primary interval produce certain sine values. For example, knowing that \(\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\) lets us easily determine that \(\sin^{-1}(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}\).

Evaluation of Inverse Trigonometric Expressions

Evaluating inverse trigonometric expressions involves understanding the relationships between angles and their sine values. Consider the expression \(\sin^{-1}(x)\):

• If \(x\) is within the range of \(-1\) to \(1\), the output is an angle within \[-\frac{\pi}{2}, \frac{\pi}{2}\].
• For example, \(\sin^{-1}(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}\) because \(\frac{\sqrt{2}}{2}\) corresponds to a standard angle \(\frac{\pi}{4}\) where the sine value matches.
• If \(x\) is outside the range, such as \(\sin^{-1}(-2)\), the expression is undefined, because no angle exists where the sine is \(-2\).

Understanding these principles is key for solving various trigonometric equations and expressions efficiently.