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Determine the amplitude and phase shift for each function, and sketch at least one cycle of the graph. Label five points as done in the examples. $$y=\frac{1}{2} \cos x$$

Short Answer

Expert verified
Amplitude is \frac{1}{2}, period is 2Ï€, no phase shift. Five points are (0, \frac{1}{2}), (\frac{Ï€}{2}, 0), (Ï€, - \frac{1}{2}), (\ \frac{3Ï€}{2}, 0), and (2Ï€,\frac{1}{2}). Sketch one cycle through these points.

Step by step solution

01

- Identify the amplitude

The general form of a cosine function is given by \[ y = A \cos(Bx - C) + D \]Here, the amplitude is the absolute value of the coefficient of the cosine function. In this case, the function is \[ y = \frac{1}{2} \cos(x) \]So the amplitude \(A\) is \[ \begin{align*} A &= \left| \frac{1}{2} \right| \&= \frac{1}{2} \end{align*} \]
02

- Determine the period

The period of a cosine function is given by \[ \frac{2\text{Ï€}}{|B|} \]In this function, \( B = 1 \).So the period \( T \) is \[ T = \frac{2\Ï€}{1} \ = 2\Ï€ \]
03

- Identify the phase shift

For the phase shift, we use the form \[ y = A \cos(Bx - C) + D \]The phase shift is given by \[ \frac{C}{B} \]In this case, there is no \(C\) term, so the phase shift is \[ \frac{0}{1} = 0 \]
04

- Sketch the graph

To sketch at least one cycle, start at the origin (0, 0) with the given amplitude of \frac{1}{2}Here are five key points:1. (0, \frac{1}{2}) - at x = 02. \(\ \left(\frac{\π}{2}, 0\right)\) - at \ x = \ \frac{\π}{2}\ 3. \(- \left( \ \π, - \frac{1}{2} \ )\) - Here x = \ \ π\4. \(\ \left\(\ \frac{\3π}{2}, 0\)\) - Here x =\ \frac{ 3π}{2}\5. \(\ \left(\2π, \frac{1}{2}\)\ - Back at the start of next cycle, x=2πPlot these points, drawing a smooth curve passing through them showing one cycle of cosine function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude
The amplitude of a trigonometric function like the cosine function is a measure of its maximum height from the middle value, also known as the equilibrium position.

In the function provided: \( y = \frac{1}{2} \, \cos(x) \), the amplitude is the absolute value of the coefficient in front of the cosine term. Here, the coefficient is \( \frac{1}{2} \).

Therefore, the amplitude \( A \) is:
\( \begin{align*} A & = \left| \frac{1}{2} \right| \ & = \frac{1}{2} \end{align*} \).
This means the graph of \( y = \frac{1}{2} \, \cos(x) \) will peak at \( \frac{1}{2} \) and bottom out at \( -\frac{1}{2} \). Understanding the amplitude helps you visualize how far the graph deviates from its centerline.
Period
The period of a trigonometric function like the cosine function is the length of one complete cycle of the wave. It tells you how frequently the pattern repeats along the x-axis.

For the general cosine function, \( y = A \, \cos(Bx - C) + D \), the period \( T \) can be found using the formula:
\( T = \frac{2\pi}{|B|} \).
In the function provided: \( y = \frac{1}{2} \, \cos(x) \), \( B = 1 \). Therefore, the period is:
\( T = \frac{2\pi}{1} = 2\pi \).
This means the graph completes one full cycle every \( 2\pi \) units along the x-axis. By knowing the period, you can accurately sketch and predict the behavior of the function over any interval.
Phase Shift
The phase shift of a trigonometric function like the cosine function is a horizontal shift along the x-axis. It tells us how the graph is translated left or right from its usual position.

For the general cosine function, \( y = A \, \cos(Bx - C) + D \), the phase shift is calculated as \( \frac{C}{B} \).
In the function provided: \( y = \frac{1}{2} \, \cos(x) \), there is no \( C \) term, which means \( C = 0 \). So the phase shift is:
\( \frac{C}{B} = \frac{0}{1} = 0 \).
This indicates that there is no horizontal shift, and the graph of \( y = \frac{1}{2} \, \cos(x) \) will start at the usual position of the cosine function. Understanding the phase shift helps correctly position the graph on the x-axis.

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