Chapter 5: Problem 8
Find the amplitude, the period, and the phase shift and sketch the graph of the equation. $$y=2 \sin \left(x-\frac{\pi}{2}\right)$$
Short Answer
Expert verified
Amplitude: 2, Period: \(2\pi\), Phase Shift: \(\frac{\pi}{2}\) to the right.
Step by step solution
01
Identify the Amplitude
The standard form of a sine function is \( y = a \, \sin(bx - c) + d \). In this equation, the coefficient \( a \) is the amplitude. For the given equation \( y = 2 \sin\left(x - \frac{\pi}{2}\right) \), the amplitude \( a \) is \( 2 \).
02
Determine the Period
The period of a sine function \( y = a \sin(bx - c) \) is given by \( \frac{2\pi}{b} \). Since the equation does not have a coefficient other than 1 next to \( x \) (meaning \( b = 1 \)), the period is \( \frac{2\pi}{1} = 2\pi \).
03
Find the Phase Shift
The phase shift of a sine function \( y = a \sin(bx - c) \) is calculated by \( \frac{c}{b} \). Here, \( c = \frac{\pi}{2} \) and \( b = 1 \), so the phase shift is \( \frac{\pi}{2} \). Since \( c \) is subtracted in the function, the shift is to the right by \( \frac{\pi}{2} \) units.
04
Sketch the Graph
To sketch the graph, start by plotting the basic sinusoidal shape. The sine wave usually starts at \( (0,0) \), peaks at \( \frac{\pi}{2} \), and returns to 0 at \( \pi \), but now it begins at \( x = \frac{\pi}{2} \) due to the phase shift. The amplitude stretches it vertically to a maximum of 2 and a minimum of -2. The period remains \( 2\pi \), so the next cycle will start at \( \frac{5\pi}{2} \).
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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 a sine function refers to its height from the centerline to its peak or trough. In the equation you provided, the term outside the sine function, here "2" in \( y = 2 \sin(x - \frac{\pi}{2}) \), determines the amplitude. Thus, the amplitude is 2. This means:
Amplitudes do not affect the period or phase shift, helping to focus purely on vertical stretching or compressing alone.
- The wave reaches a maximum height of 2 units above its midline.
- It also dips 2 units below the midline.
Amplitudes do not affect the period or phase shift, helping to focus purely on vertical stretching or compressing alone.
Period
The period of a trigonometric function describes how long it takes for the function to complete one full cycle of its pattern before repeating. For the standard sine function \( y = \sin(x) \), this period is \( 2\pi \).
In equations of the form \( y = a \sin(bx - c) \), the coefficient \( b \) affects the period, calculated by the formula \( \frac{2\pi}{b} \). In this case, our equation has \( b = 1 \), so the period is exactly \( 2\pi \), meaning:
In equations of the form \( y = a \sin(bx - c) \), the coefficient \( b \) affects the period, calculated by the formula \( \frac{2\pi}{b} \). In this case, our equation has \( b = 1 \), so the period is exactly \( 2\pi \), meaning:
- The wave completes one full repetition every \( 2\pi \) units along the x-axis.
- If \( b \) were greater than 1, the period would decrease, compacting the wave horizontally.
- If \( b \) were less than 1, the wave would spread out, increasing the period.
Phase shift
The phase shift refers to the horizontal shift of the graph of a trigonometric function. It essentially tells us where the wave starts along the x-axis. In the equation \( y = 2 \sin(x - \frac{\pi}{2}) \), the phase shift is calculated from the formula \( \frac{c}{b} \), where \( c \) is subtracted inside the function. For this equation:
- \( c = \frac{\pi}{2} \)
- \( b = 1 \)
- The entire sine wave is moved \( \frac{\pi}{2} \) units to the right.
- This causes the typical starting point of the sine wave from \((0, 0)\) to begin instead at \(x = \frac{\pi}{2}\).
- Such shifts can also be translated to the left if the value of \( c \) were subtractive.