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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 29
Chapter 2, Problem 29

In Exercises 17–30, determine the amplitude, period, and phase shift of each function. Then graph one period of the function. y = −2 sin(2πx + 4π)

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Identify the general form of the sine function: \(y = A \sin(Bx + C)\), where \(A\) is the amplitude, \(B\) affects the period, and \(C\) affects the phase shift.
Find the amplitude by taking the absolute value of the coefficient in front of the sine function: \(\text{Amplitude} = |A| = |-2|\).
Calculate the period using the formula \(\text{Period} = \frac{2\pi}{|B|}\). Here, \(B\) is the coefficient of \(x\) inside the sine function, which is \(2\pi\).
Determine the phase shift using the formula \(\text{Phase shift} = -\frac{C}{B}\), where \(C\) is the constant added inside the sine function. In this case, \(C = 4\pi\) and \(B = 2\pi\).
Use the amplitude, period, and phase shift to sketch one full cycle of the sine function, starting from the phase shift and covering one period along the x-axis.

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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 the function's output, representing the height from the midline to the peak. For y = a sin(bx + c), the amplitude is |a|. In this case, the amplitude is 2, indicating the wave oscillates 2 units above and below the midline.
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Period of a Sine Function

The period is the length of one complete cycle of the sine wave. It is calculated as (2π) / |b| for y = sin(bx + c). Here, b = 2π, so the period is (2π) / (2π) = 1, meaning the function repeats every 1 unit along the x-axis.
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Phase Shift of a Trigonometric Function

Phase shift is the horizontal translation of the graph, found by solving bx + c = 0 for x, giving x = -c/b. For y = −2 sin(2πx + 4π), the phase shift is -4π / (2π) = -2, indicating the graph shifts 2 units to the left.
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Related Practice
Textbook Question

In Exercises 21–28, an object moves in simple harmonic motion described by the given equation, where t is measured in seconds and d in inches. In each exercise, find the following: a. the maximum displacement b. the frequency c. the time required for one cycle. d = −4 sin 3π/2 t

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Textbook Question

Graph two periods of the given cosecant or secant function.


y = 2 csc x

Textbook Question

In Exercises 29–44, graph two periods of the given cosecant or secant function. y = 3 csc x

Textbook Question

Use each graph to obtain the graph of the corresponding reciprocal function, cosecant or secant. Give the equation of the function for the graph that you obtain.


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Textbook Question

In Exercises 27–38, use a calculator to find the value of each expression rounded to two decimal places. sin⁻¹ (-0.32)

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Textbook Question

In Exercises 29–36, find the length x to the nearest whole unit.

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