Problem 54
Question
In Exercises \(49-60\), state the amplitude, period, and phase shift (including direction) of the given function. $$y=3 \sin \left[-\frac{\pi}{2}(x-1)\right]$$
Step-by-Step Solution
Verified Answer
Amplitude: 3, Period: 4, Phase shift: 1 unit to the right.
1Step 1: Identify the General Form
The general form of a sine function is \(y = a \, \sin(bx - c) + d\), where \(a\) is the amplitude, \(\frac{2\pi}{b}\) is the period, and \(\frac{c}{b}\) is the phase shift. The given function is \(y = 3 \sin\left[-\frac{\pi}{2}(x-1)\right]\).
2Step 2: Determine the Amplitude
The amplitude of a sine function is the absolute value of the coefficient of \(\sin\). Here, the coefficient \(a = 3\). Therefore, the amplitude is \(3\).
3Step 3: Find the Period
The period of a sine function is calculated as \(\frac{2\pi}{\left|b\right|}\), where \(b\) is the coefficient of \(x\) in the sine function. In this case, \(b = -\frac{\pi}{2}\). Thus, the period is calculated as: \[\frac{2\pi}{\left| -\frac{\pi}{2} \right|} = \frac{2\pi}{\frac{\pi}{2}} = 4.\] Hence, the period is \(4\).
4Step 4: Determine the Phase Shift and Direction
The phase shift is given by \(\frac{c}{b}\) from \(bx - c\). Here, \(-\frac{\pi}{2}(x-1)\) can be expanded to \(-\frac{\pi}{2}x + \frac{\pi}{2}\). Therefore, \(c = -\frac{\pi}{2}\cdot(-1) = \frac{\pi}{2}\). The phase shift is: \[\frac{c}{b} = \frac{\frac{\pi}{2}}{-\frac{\pi}{2}} = -1.\] This indicates a phase shift of 1 unit to the right.
Key Concepts
AmplitudePeriodPhase ShiftSine Function
Amplitude
Amplitude is an essential concept in understanding trigonometric functions, particularly the sine function. The amplitude of a sine function refers to its maximum displacement from its midline, or the central axis of the wave. In mathematical terms, amplitude is the absolute value of the coefficient in front of the sine function.
In the sine function provided:
In the sine function provided:
- Given Equation: \(y = 3 \sin\left[-\frac{\pi}{2}(x-1)\right]\)
- Amplitude (\(a\)): The coefficient of \(\sin\) is \(3\)
- Resulting Amplitude: \( |3| = 3\)
Period
The period of a sine function is a crucial property that defines the duration of one complete cycle of the wave. It is represented by how long the function takes to repeat itself. For a sine function in the form \(y = a \, \sin(bx - c) + d\), the period is calculated as \(\frac{2\pi}{|b|}\).
Let's explore the given function:
Let's explore the given function:
- Coefficient \(b\) of \(x\) is \(-\frac{\pi}{2}\)
- Period Calculation: \[\frac{2\pi}{\left| -\frac{\pi}{2} \right|} = \frac{2\pi}{\frac{\pi}{2}} = 4\]
- Conclusion: The period is 4 units.
Phase Shift
Phase shift in trigonometric functions refers to the horizontal displacement of the graph. It tells us how much the function is shifted left or right from the usual position. The phase shift is derived from \(\frac{c}{b}\) in the sine form \(bx - c\).
For our example:
For our example:
- Expression: \(-\frac{\pi}{2}(x-1)\) can be rewritten as \(-\frac{\pi}{2}x + \frac{\pi}{2}\)
- Values: Here, \(c = \frac{\pi}{2}\) and \(b = -\frac{\pi}{2}\)
- Phase Shift Calculation: \[\frac{\frac{\pi}{2}}{-\frac{\pi}{2}} = -1\]
- Outcome: The function shifts 1 unit to the right.
Sine Function
The sine function is one of the fundamental trigonometric functions, commonly expressed as \(y = \sin(x)\). It's known for its smooth, wave-like oscillation, making it pivotal in modeling periodic phenomena.
Key characteristics include:
Key characteristics include:
- Oscillation: The sine function produces a wave that repeatedly ascends and descends.
- Pattern: In its most basic form, it starts at the origin (0,0), peaks at (+1), descends to (-1), and completes a cycle at (2π).
- Amplitude \(a\): Adjusts the wave's height.
- Period \(\frac{2\pi}{b}\): Changes the cycle length.
- Phase Shift \(\frac{c}{b}\): Shifts the wave left or right.
- Vertical Shift \(d\): Moves the entire wave up or down.
Other exercises in this chapter
Problem 53
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Use the unit circle to find all of the exact values of \(\theta\) that make the equation true in the indicated interval. $$\tan \theta=\frac{\sqrt{3}}{3}, 0 \le
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