Problem 42 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 33-42: Sketch the graph of the equation. $$ y=\sin \left(\sin ^{-1} x\right) $$

Step-by-Step Solution

Answer

The graph is a line segment from (-1, -1) to (1, 1) along \( y = x \).

Step 1: Understand Inverse Functions

The function \( y = \sin(\sin^{-1}(x)) \) involves an inverse trigonometric function. The inverse sine function, \( \sin^{-1}(x) \), returns the angle whose sine is \( x \). Thus, \( \sin(\sin^{-1}(x)) = x \) for all \( x \) in the range of \( \sin^{-1}(x) \).

Step 2: Determine the Domain of \( \sin^{-1}(x) \)

The function \( \sin^{-1}(x) \) is defined for \( -1 \leq x \leq 1 \). This means that \( \sin(\sin^{-1}(x)) = x \) is valid only within this domain.

Step 3: Express the Function within the Domain

Since \( \sin(\sin^{-1}(x)) = x \) within the domain of \( \sin^{-1}(x) \), the function simplifies to \( y = x \) for \( -1 \leq x \leq 1 \).

Step 4: Sketch the Graph

The graph of \( y = x \) is a straight line passing through the origin, with a slope of 1. However, since this equality holds only for \( -1 \leq x \leq 1 \), the graph will be a segment of the line \( y = x \) from \( (-1, -1) \) to \( (1, 1) \).

Step 5: Finalize the Graph

Draw the line segment as described above. Ensure the ends of the graph at \( x = -1 \) and \( x = 1 \) are circles or dots to highlight that these points are included in the domain.

Key Concepts

Inverse Functions

Inverse functions can be a bit tricky to grasp at first, but they are essentially functions that reverse or "undo" the action of another function. In our exercise, we have the inverse sine function, denoted as \( \sin^{-1}(x) \), also commonly known as arcsin. The purpose of this function is to find an angle, say \( \theta \), whose sine value is \( x \). This means that when you input \( x \) into \( \sin^{-1}(x) \), the result is an angle \( \theta \) such that \( \sin(\theta) = x \). In the original equation \( y = \sin(\sin^{-1}(x)) \), the operation of sine and inverse sine cancel each other out. This simplifies the equation to \( y = x \) for all \( x \) that the inverse sine can take. This is due to the property of inverse functions where applying the function and its inverse "undoes" the operation, returning the original input if it’s within the allowed range. However, because inverse functions are related to angles in specific quadrants, \( \sin^{-1}(x) \) is explicitly defined only for \( -1 \leq x \leq 1 \). This ensures that the angle provided by \( \sin^{-1}(x) \) stays within the permissible domain for arcsin, which is typically \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) in terms of radians.

Graph Sketching

Graph sketching for the equation \( y = \sin(\sin^{-1}(x)) \) simplifies to visualizing the line \( y = x \). Essentially, this graph is a straight line passing diagonally through the origin and extending infinitely in both directions. For our specific function, however, the graph will not be complete as an infinite line; instead, it will be limited to a specific portion of the line. The range for the inverse sine function \( \sin^{-1}(x) \) restricts \( x \) between \(-1\) and \(1\), hence the complete function \( y = \sin(\sin^{-1}(x)) \) will only be valid between these values of \( x \). Within this interval:

• The function is linear.
• The slope of the line is \(1\).
• The y-intercept is \(0\).

This means that the sketch you will perform involves drawing a line segment rather than a full line. The segment starts at the point \((-1, -1)\) and ends at \((1, 1)\). When sketching, make sure to clearly mark these endpoints as the boundary beyond which the segment doesn't continue.

Domain and Range

The domain and range of a function define where the function is applicable and what values it can output, respectively. For \( y = \sin(\sin^{-1}(x)) \), it's essential to understand we're dealing with boundaries set by the inverse sine function. The domain of \( \sin^{-1}(x) \) is the closed interval \([-1, 1]\), indicating that only values \( x \) within this interval can be processed by \( \sin^{-1}(x) \). As a result, the domain of \( y = \sin(\sin^{-1}(x)) \) is also \([-1, 1]\).As for the range, since \( y = \sin(\sin^{-1}(x)) = x \), it's clear that the output \( y \) is also confined to the same interval \([-1, 1]\), as \( x \) itself is bounded within this domain. This concept ensures that:

• For every valid \( x \), there is a corresponding \( y \).
• The function produces a straight line within these boundary conditions.
• The range matches the domain exactly due to the linear nature of the expression \( y = x \).

Understanding domain and range helps in accurately sketching functions and knowing where they "live" on the graph.