Problem 41 - Precalculus: Functions and Graphs

Question

Sketch the graph of the equation. $$y=\sin (\arccos x)$$

Step-by-Step Solution

Answer

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

Step 1: Understand the Function

The equation given is \( y = \sin( \arccos x) \). We identify that \( \arccos x \) is the angle whose cosine is \( x \), and we need to find its sine to solve for \( y \).

Step 2: Consider the Domain of \( \arccos x \)

The function \( \arccos x \) has a domain of \(-1 \leq x \leq 1\) because it is the range of possible values for the cosine function.

Step 3: Relate \( \sin \theta \) to \( \cos \theta \)

For an angle \( \theta \), if \( x = \cos \theta \), then \( \sin \theta = \sqrt{1 - x^2} \) by the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \).

Step 4: Transform the Equation for \( y \)

Substitute the expression from Step 3 into the function: \( y = \sqrt{1 - x^2} \). This simplifies our function to graph.

Step 5: Graph the Function \( y = \sqrt{1 - x^2} \)

This function represents a semicircle with radius 1 centered at the origin on the x-axis, but only the upper half where \( y \geq 0 \) and \( -1 \leq x \leq 1 \).

Step 6: Sketch the Graph

Draw the upper semicircle, starting from \((x, y) = (-1, 0)\), peaking at \((0, 1)\), and ending at \((1, 0)\), ensuring the curve is smooth and symmetric.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that allow us to find the angle when we are given the value of a trigonometric function. In our equation, we see the function \( \arccos x \), which reads as 'the angle whose cosine is \( x \)'. This is especially helpful when translating a range of values into angles.

• \( \arccos x \) has a range of \([0, \pi]\) or \([0^{\circ}, 180^{\circ}]\).
• This range covers angles starting from the positive x-axis all the way counterclockwise to the negative x-axis.

Inverse trigonometric functions often have stricter domain restrictions. For \( \arccos x \), the domain is \([-1, 1]\), capturing all possible cosine values. Recognizing these constraints helps in graphing functions like \( y = \sin(\arccos x) \). Here, our job was to find the 'sine' of \( \arccos x \)'s output angle, relying on established trigonometric relationships.

Graphing Techniques

Graphing trigonometric functions requires understanding both the component functions and their combinations. For \( y = \sin(\arccos x) \), we use the known shapes of these functions to create our graph.

• \( \sin \) function traditionally creates oscillatory waves, but in this context, it helps to create part of a circle.
• The range of \( \sin \) output for the upper arcs, which is \([0, 1]\), because the related angle \( \arccos x \) only outputs angles from the first quadrant.

With these basic graphing techniques, drawing the semicircle is simplified to ensuring endpoints are accurate: it runs from \((-1, 0)\) to \((1, 0)\), crossing through the highest point at \((0,1)\). This helps students visualize the transformation from the inverse equation into a graph.

Pythagorean Identity

The Pythagorean identity is a central concept in trigonometry that combines sine and cosine in a single equation: \( \sin^2 \theta + \cos^2 \theta = 1 \). It allows us to express one trigonometric function in terms of another.

• This is crucial for handling the expression \( \sin(\arccos x) \).
• We use: when \( \cos \theta = x \), then \( \sin \theta = \sqrt{1 - x^2} \).

Deploying this identity transforms the complex inverse trig equation into something recognizable—a perfect setting for trigonometric problem-solving. This particular identity is very powerful as it bridges the gap between different trigonometric functions, turning complex operations into straightforward calculations.

Function Transformation

Function transformation involves changing a basic function's appearance on a graph via shifts, stretches, or reflections. The move from \( y = \sin(\arccos x) \) to \( y = \sqrt{1 - x^2} \) is a transformation achieved through a trigonometric identity.

• This action redefines the problem from a potentially messy trig combination into a familiar geometric shape—part of a circle.
• By plotting \( y = \sqrt{1 - x^2} \), we apply this transformation, allowing us to graph half of a circle as seen from \(-1 \leq x \leq 1\).

Transformation is not only in algebraic expressions but is visual, shifting, rotating, or reshaping the way we interpret the equation's output. Seeing function transformation in action deepens the understanding of abstract math concepts, crucial for mastering any trigonometry segment.