Problem 48 - Precalculus with Calculus Previews

Question

Sketch the graph of the given function. $$ y=\cos (\arcsin x) $$

Step-by-Step Solution

Answer

The graph is the upper half of a unit circle from \((-1, 0)\) to \((1, 0)\).

Step 1: Interpret the Function

The function given is \( y = \cos(\arcsin x) \). Here, the \( \arcsin x \) represents the inverse sine function, which returns an angle whose sine is \( x \). The domain of \( \arcsin x \) is \([-1, 1]\), which means \( x \) must be within this range.

Step 2: Understand the Range of \( \arcsin x \)

The range of \( \arcsin x \) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This means the output of \( \arcsin x \), which becomes the input for the cosine function, will be angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

Step 3: Substitute and Simplify Using Trigonometric Identities

Utilize the identity \( \cos(\arcsin(x)) = \sqrt{1-x^2} \). This is derived from the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), where \( \sin(\theta) = x \). This identity helps simplify the function to \( y = \sqrt{1-x^2} \).

Step 4: Identify the Graph Characteristics

Identify that \( y = \sqrt{1-x^2} \) is the top half of the unit circle centered at the origin. This is because it represents the positive square root, meaning \( y \) is non-negative and peaks at \( y=1 \) when \( x=0 \).

Step 5: Sketch the Graph

Graph the function \( y = \sqrt{1-x^2} \) for \( x \) in the interval \([-1, 1]\). This is a semicircle above the x-axis with radius 1. Plot points, such as \((-1, 0), (0, 1), \) and \((1, 0)\), and draw a smooth curve connecting them.

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in simplifying expressions and solving equations involving trigonometric functions. In the exercise, we encounter the identity \( \cos(\arcsin(x)) = \sqrt{1-x^2} \), derived from the well-known Pythagorean identity:

• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)

Here, \( \theta = \arcsin(x) \), meaning \( \sin(\theta) = x \). To find \( \cos(\theta) \), we rearrange the Pythagorean identity:

• \( \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - x^2 \)

Taking the positive square root gives us \( \cos(\theta) = \sqrt{1 - x^2} \), because our domain for x is \([-1, 1]\) and we consider the principal value.
This is crucial because it simplifies our original function, making it easier to graph and understand.

Function Graphing

Graphing functions is a fundamental concept in visualizing mathematical relationships. For the function \( y = \sqrt{1-x^2} \), this involves recognizing that the function represents the upper half of a circle with a radius of 1. The graph is plotted by:

• Identifying critical points such as \((-1, 0), (0, 1), \text{ and } (1, 0)\)
• Understanding that \( y \) must be non-negative for real outputs, hence representing a semicircle above the x-axis

The entire graph visually emphasizes the unit circle's nature, showing its symmetry and constraint given by the domain [\([-1, 1]\). Also, note that smooth curves between the points ensure the circle's natural continuity is maintained.

Pythagorean Identities

Pythagorean identities are rooted in the geometry of right triangles and the unit circle, and they play a vital role in integrating sinusoids and their inverses. The identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) links the sine and cosine functions directly to the unit circle, reflecting the sum of squares of a triangle's sides being constant as 1. In this exercise:

• The identity helps in converting \( \cos(\arcsin(x)) \) to \( \sqrt{1-x^2} \)
• Highlights the connection between inverse trigonometric functions and circular behavior

This linkage is incredibly useful for simplifying complex expressions and aids in understanding the geometric interpretation of trigonometric equations.
When you see \( \sqrt{1-x^2} \), it instantly recalls the form of a semicircle, solving both trigonometric and geometric queries efficiently.