Problem 72 - Precalculus : Building Concepts and Connections

Question

A sound wave has the form \(y=2 \cos \left(3 x-\frac{\pi}{4}\right)\) for \(x\) in the interval \(\left[\frac{\pi}{12}, \frac{5 \pi}{12}\right] .\) Express \(x\) as a function of \(y\) and state the domain of your function.

Step-by-Step Solution

Answer

The function of x in terms of y is \(x = \frac{\arccos \left(\frac{y}{2}\right) + \frac{\pi}{4}}{3}\), and the domain of this function is [-2, 2].

Step 1: Identify the Standard Form of a Cosine Function

First, understand that \(y=2 \cos \left(3 x-\frac{\pi}{4}\right)\) is a cosine function which is in the general form \(y = a \cos(bx - c)\) where a is the amplitude, b is the frequency and c is the phase shift.

Step 2: Express x as a function of y

To express x as a function of y, start by isolating the cosine term: \[ \cos \left(3 x-\frac{\pi}{4}\right) = \frac{y}{2}.\] Then, use the arc cosine function to isolate the term inside the cosine function: \[ 3x - \frac{\pi}{4} = \arccos \left(\frac{y}{2} \right).\] And finally, solve for x, yielding: \[x = \frac{\arccos \left(\frac{y}{2}\right) + \frac{\pi}{4}}{3}.\] Note that the arc cosine function is equivalent to the inverse of the cosine function.

Step 3: Determine the Domain

The domain of the function will be determined by the range of the inverse cosine function. The range of \(\arccos \left(\frac{y}{2}\right)\) is \([0, \pi]\). Since we have \(\frac{y}{2}\), the domain of \(y\) will be \([-2,2]\).

Key Concepts

Cosine Function

The cosine function is one of the fundamental trigonometric functions, often denoted as \( \cos(x) \). In the context of the given exercise, we are working with a function of the form \( y = 2 \cos(3x - \frac{\pi}{4}) \). This is referred to as a transformation of the standard cosine function.

• The parameter \( a = 2 \) represents the amplitude, which indicates the height of the wave's peaks above the center line. In this case, the wave oscillates between 2 and -2.
• The parameter \( b = 3 \) indicates the frequency, dictating how many complete cycles the wave completes in a given interval. It affects the period of the wave by compressing or stretching it.
• The term \( -\frac{\pi}{4} \) is a phase shift, moving the wave left or right on the x-axis. Here, the wave is shifted to the right by \( \frac{\pi}{4} \).

Understanding these transformations helps us interpret the behavior of the cosine function and, ultimately, to express \( x \) as a function of \( y \).

Inverse Trigonometric Functions

Inverse trigonometric functions are used to determine the angle that corresponds to a given trigonometric ratio. When we express \( x \) as a function of \( y \), we use the inverse cosine function, denoted \( \arccos \), which helps in finding the measure of angles for a known cosine value.

• The inverse cosine function is represented as \( \arccos(y) \) and is defined for values of \( y \) between -1 and 1 inclusive.
• In this scenario, we isolate the cosine term to form \( \cos(3x - \frac{\pi}{4}) = \frac{y}{2} \) and utilize \( \arccos \) to handle the inverse calculation, yielding the equation \( 3x - \frac{\pi}{4} = \arccos\left(\frac{y}{2}\right) \).
• Thus, by rearranging the terms, we eventually solve for \( x \) to find \( x = \frac{\arccos\left(\frac{y}{2}\right) + \frac{\pi}{4}}{3} \), enabling us to express \( x \) as a function of \( y \).

This transformation from the cosine function to the inverse cosine function reflects the core utility of inverse trigonometric functions - solving for the arc or angle based on cosine values.

Domain of a Function

The domain of a function refers to the complete set of possible values of the independent variable that will yield valid outputs for the function. For the function \( x = \frac{\arccos\left(\frac{y}{2}\right) + \frac{\pi}{4}}{3} \), the domain is determined by the range of the inverse trigonometric function involved.

• The range of \( \arccos(z) \) is \([0, \pi]\). For \( \arccos\left(\frac{y}{2}\right) \), it implies \( \frac{y}{2} \) must also lie within the range \([-1, 1]\).
• Thus, solving \( -2 \leq y \leq 2 \) ensures that the values of \( y \) remain valid within the range of \( \arccos \).
• The domain of \( y \) is derived as \([-2, 2] \), as it's this set of values that ensures the function remains real and well-defined.

Understanding the domain is crucial as it defines where the inverse function is operable and produces meaningful values, which ensures the function represents the scenario accurately.