Problem 33 - Algebra and Trigonometry

Question

Find the amplitude, period, and phase shift of the function, and graph one complete period. $$ y=5 \cos \left(3 x-\frac{\pi}{4}\right) $$

Step-by-Step Solution

Answer

Amplitude: 5, Period: \(\frac{2\pi}{3}\), Phase Shift: \(\frac{\pi}{12}\) to the right.

Step 1: Identifying the Amplitude

The amplitude of a cosine function is given by the coefficient in front of the cosine term. Here, the function is \( y = 5 \cos(3x - \frac{\pi}{4}) \). Thus, the amplitude is \( 5 \).

Step 2: Finding the Period

The period of a cosine function \( y = a \cos(bx - c) \) is calculated using the formula \( \frac{2\pi}{b} \). In this case, \( b = 3 \), so the period is \( \frac{2\pi}{3} \).

Step 3: Calculating the Phase Shift

The phase shift is determined by \( \frac{c}{b} \) for the function \( y = a \cos(bx - c) \). Here, \( c = \frac{\pi}{4} \) and \( b = 3 \), so the phase shift is \( \frac{\pi}{4} \div 3 = \frac{\pi}{12} \). The shift is to the right since it is \( \frac{\pi}{4} \) to the right of the \( y \)-axis.

Step 4: Graphing One Complete Period

To graph one complete period of \( y = 5 \cos(3x - \frac{\pi}{4}) \), start at the phase shift \( x = \frac{\pi}{12} \), then mark the endpoint at one full period later: \( x = \frac{\pi}{12} + \frac{2\pi}{3} = \frac{9\pi}{12} + \frac{8\pi}{12} = \frac{17\pi}{12} \). Divide this interval into four equal parts, and plot the key points at these intervals showing the peak, zero crossing, trough, and another zero crossing of the cosine wave.

Key Concepts

Amplitude of a Function

The amplitude of a trigonometric function like a sine or cosine wave describes how tall or high the wave is, compared to the baseline. In simpler terms, it stretches or compresses the wave vertically. For the cosine function of the form \( y = a \cos(bx - c) \), the amplitude is determined by the absolute value of \( a \).
The coefficient \( a \) in front of the cosine function multiplies each value of the cosine by this number, effectively controlling how far the wave reaches from its central position. In our given function \( y = 5 \cos(3x - \frac{\pi}{4}) \), \( a \) is 5.

• Amplitude = \( |5| = 5 \)

This tells us that the maximum height of the wave is 5 units above and below the horizontal axis (centerline). Visualizing this, if you were to draw a line right through the middle of the wave, it would rise and dip 5 units from there.

Period of a Function

The period of a trigonometric function is the length of one complete cycle of the wave before it starts repeating itself. This cycle tells us how often the wave repeats over the x-axis. For the function \( y = a \cos(bx - c) \), the period \( T \) can be found using the formula \( T = \frac{2\pi}{b} \).
In our example function \( y = 5 \cos(3x - \frac{\pi}{4}) \), \( b = 3 \).
Therefore, the period is:

• Period = \( \frac{2\pi}{3} \)

This means that every \( \frac{2\pi}{3} \) units along the x-axis, the cosine wave completes a full cycle and starts over. This is shorter than the standard cosine function period of \( 2\pi \), because the wave is compressed horizontally by a factor of 3.

Phase Shift

Phase shift refers to a horizontal movement of a wave along the x-axis. In other words, it’s the measure of how much the entire graph of the function is shifted left or right from the usual position. It plays a crucial role in determining where each cycle of the wave begins.
For the function formula \( y = a \cos(bx - c) \), the phase shift is calculated as \( \frac{c}{b} \).
In the function \( y = 5 \cos(3x - \frac{\pi}{4}) \), \( c \) is \( \frac{\pi}{4} \) and \( b \) is 3.

• Phase Shift = \( \frac{\pi}{4} \div 3 = \frac{\pi}{12} \)

The positive value indicates that the wave shifts \( \frac{\pi}{12} \) units to the right. Understanding this helps to place the starting point of the cosine curve on the x-axis, aligning it according to the shift.