Problem 37 - Algebra and Trigonometry

Question

\(29-42\) . Find the amplitude, period, and phase shift of the function, and graph one complete period. $$ y=\frac{1}{2}-\frac{1}{2} \cos \left(2 x-\frac{\pi}{3}\right) $$

Step-by-Step Solution

Answer

Amplitude: \(\frac{1}{2}\), Period: \(\pi\), Phase shift: \(\frac{\pi}{6}\).

Step 1: Identify the Function Parameters

The general form of a cosine function is \( y = a \cos(bx - c) + d \). For the function \( y=\frac{1}{2}-\frac{1}{2} \cos \left(2 x-\frac{\pi}{3}\right) \), compare coefficients to find \(a = -\frac{1}{2}\), \(b = 2\), \(c = \frac{\pi}{3}\), and \(d = \frac{1}{2}\).

Step 2: Find the Amplitude

The amplitude of a cosine function is the absolute value of \(a\). Thus, for this function, the amplitude is \( \left| -\frac{1}{2} \right| = \frac{1}{2} \).

Step 3: Determine the Period

The period of a cosine function is calculated as \( \frac{2\pi}{b} \). Here, \( b = 2 \), so the period is \( \frac{2\pi}{2} = \pi \).

Step 4: Calculate the Phase Shift

The phase shift is found using the formula \( \frac{c}{b} \). Given \( c = \frac{\pi}{3} \) and \( b = 2 \), the phase shift is \( \frac{\frac{\pi}{3}}{2} = \frac{\pi}{6} \). This shift is to the right because the sign in front of \(c\) is negative.

Step 5: Graph One Complete Period

Start at the phase shift \( x = \frac{\pi}{6} \). Because the period is \( \pi \), the complete cycle ends at \( x = \frac{\pi}{6} + \pi = \frac{7\pi}{6} \). The graph moves from a minimum at \( y = 0 \) to a maximum at \( y = 1 \), and the midline is at \( y = \frac{1}{2} \).

Key Concepts

Amplitude

In trigonometric functions, the amplitude represents how far the wave swings from its midpoint or center line. Think of it as the wave's height. In a cosine function such as the one given in this problem, the amplitude is determined by multiplying the peak of the wave by a vertical scaling factor.
For the function given, which is in the form of \[y = d - a \cos(bx - c),\] we extract the amplitude by taking the absolute value of the coefficient of the cosine function, \(a\).

• The function is \(y = \frac{1}{2} - \frac{1}{2} \cos(2x - \frac{\pi}{3})\).
• Here, \(a = -\frac{1}{2}\).
• The amplitude is thus \( \left| -\frac{1}{2} \right| = \frac{1}{2} \).

The amplitude tells us that the graph's peaks and valleys will reach \(\frac{1}{2}\) unit from the midline, which is at \(y = \frac{1}{2}\). This means it will oscillate between a maximum of 1 and a minimum of 0, considering the vertical shift.

Period

The period of a trigonometric function tells us how often the cycles or waves repeat. This is an essential characteristic as it depicts the function's frequency. For cosine functions, the period indicates the length of one complete wave cycle.
The formula to find the period \(P\) for functions of the form \[y = a \cos(bx - c) + d\] is given by: \[P = \frac{2\pi}{b}.\]
In the exercise function:

• \(b = 2\).
• The period is calculated as \(\frac{2\pi}{2} = \pi\).

This tells us that the cosine wave repeats every \(\pi\) units along the x-axis. Understanding the period helps us predict the behavior of the wave and effectively graph one complete cycle.

Phase Shift

Phase shift indicates how much a graph of a sine or cosine function is horizontally shifted from its usual position. In this problem, the phase shift determines where the start of a cycle occurs compared to the standard position.
For a cosine function of the form \[y = a \cos(bx - c) + d,\]phase shift \( \phi \) can be determined by using the formula:\[\phi = \frac{c}{b}.\]
In our problem, we have:

• \(c = \frac{\pi}{3}\).
• \(b = 2\).
• The phase shift is \(\frac{\frac{\pi}{3}}{2} = \frac{\pi}{6}\).

This tells us the cosine function graph is shifted \(\frac{\pi}{6}\) units to the right. Understanding the phase shift is crucial when aligning the wave with the correct starting point. This lets us accurately depict the graph over one complete period, ensuring that our depiction aligns with the characteristics prescribed by the function's parameters.