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Ch. 2 - Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Not the one you use?Change textbook
All textbooksBlitzer 3rd EditionCh. 2 - Graphs of the Trigonometric Functions; Inverse Trigonometric FunctionsProblem 47
Chapter 2, Problem 47

In Exercises 43–52, determine the amplitude, period, and phase shift of each function. Then graph one period of the function. y = 1/2 cos (3x + π/2)

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Identify the general form of the cosine function: \(y = A \cos(Bx + C)\), where \(A\) is the amplitude, \(B\) affects the period, and \(C\) affects the phase shift.
Find the amplitude by taking the absolute value of \(A\). In this case, \(A = \frac{1}{2}\), so the amplitude is \(|\frac{1}{2}|\).
Calculate the period using the formula \(\text{Period} = \frac{2\pi}{|B|}\). Here, \(B = 3\), so substitute to find the period.
Determine the phase shift using the formula \(\text{Phase shift} = -\frac{C}{B}\). Given \(C = \frac{\pi}{2}\) and \(B = 3\), substitute these values to find the phase shift.
To graph one period of the function, start at the phase shift on the x-axis, then plot points over one full period length, using the amplitude to determine the maximum and minimum values of the cosine wave.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Amplitude of a Trigonometric Function

Amplitude is the maximum absolute value of a trigonometric function, representing the height from the midline to the peak. For functions like y = A cos(Bx + C), the amplitude is |A|. In this example, the amplitude is 1/2, indicating the graph oscillates between 1/2 and -1/2.
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Period of a Trigonometric Function

The period is the length of one complete cycle of the function along the x-axis. For y = cos(Bx + C), the period is calculated as 2π divided by |B|. Here, B = 3, so the period is 2π/3, meaning the function repeats every 2π/3 units.
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Phase Shift of a Trigonometric Function

Phase shift refers to the horizontal translation of the graph, determined by solving Bx + C = 0 for x. It is given by -C/B. In this function, the phase shift is -π/2 divided by 3, or -π/6, indicating the graph shifts π/6 units to the left.
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Related Practice
Textbook Question

In Exercises 29–51, find the exact value of each expression. Do not use a calculator. tan [cos⁻¹ (− 4/5)]

Textbook Question

In Exercises 43–52, determine the amplitude, period, and phase shift of each function. Then graph one period of the function. y = 3 cos(2x − π)

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Textbook Question

In Exercises 39–54, find the exact value of each expression, if possible. Do not use a calculator. tan⁻¹ [tan(− π/6)]

Textbook Question

In Exercises 45–52, graph two periods of each function. y = sec(2x + π/2) − 1

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Textbook Question

In Exercises 43–52, determine the amplitude, period, and phase shift of each function. Then graph one period of the function. y = −3 cos (2x − π/2)

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Textbook Question

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function.

y = 4 cos(2x − π)

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