Chapter 13: Problem 39
How many cycles does each sine function have in the interval from 0 to 2\(\pi ?\) Find the amplitude and period of each function. $$ y=-5 \sin 2 \pi \theta $$
Short Answer
Expert verified
The function \(y=-5 \sin 2\pi\theta\) has an amplitude of 5 and a period of \(2\pi\) (this implies it completes 1 cycle in the interval from 0 to \(2\pi\))
Step by step solution
01
Identify the Amplitude
From the equation \(y=-5 \sin 2\pi\theta\), we see the coefficient before the sine function is -5. The amplitude of a sine function is the absolute value of this coefficient. Therefore the amplitude is |-5| = 5.
02
Identify the Frequency
In \(y=-5 \sin 2\pi\theta\), the coefficient of \(\theta\) inside the sine function is \(2\pi\). Therefore, the frequency of the sine function is simply that coefficient divided by \(2\pi\) so that is \(B / 2\pi = 2\pi / 2\pi = 1\).
03
Calculate the number of cycles
Since the function completes 1 cycle in the interval from 0 to \(2\pi\), and we are looking for the numbers of cycles in the interval from 0 to \(2\pi\), this sine function completes 1 cycle in that interval.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Amplitude
Amplitude refers to the height of the wave in a trigonometric function, in this case, a sine function. It tells us how far the wave peaks or troughs are from the central axis (midline). Consider the function provided:
- In the equation \(y = -5 \sin 2\pi\theta\), the number in front of the sine is \(-5\).
- The amplitude is the absolute value of this coefficient, which is \(\left|-5\right| = 5\).
Period
The period of a trigonometric function indicates how long it takes for the function to complete one full cycle, or repetition, of the wave. For the basic sine function \(y = \sin \theta\), the period is \(2\pi\). This is because it takes \(2\pi\) radians to go through all values once, from 0, up and down and back to 0.
- In our exercise's function \(y = -5 \sin 2\pi\theta\), the period can be calculated as \(\frac{2\pi}{B}\), where \(B = 2\pi\).
- Substituting that value, we get: \(\frac{2\pi}{2\pi} = 1\).
Frequency
The frequency tells us how many cycles of the sine wave occur in a specific interval, usually measured over \(2\pi\) radians. Unlike the period, which tells us how long one cycle takes, frequency indicates how often these cycles occur.
- In the equation \(y = -5 \sin 2\pi\theta\), the frequency is the coefficient in front of \(\theta\), which is \(2\pi\).
- To find the frequency as it is typically defined over a \(2\pi\) interval, we divide this coefficient by \(2\pi\), which gives us \(\frac{2\pi}{2\pi} = 1\).
Sine Function
The sine function, represented as \(y = \sin \theta\), is fundamental in trigonometry and describes a smooth, periodic oscillation. It is important in modeling various cyclical phenomena, such as sound waves, tides, and light waves. The sine wave begins at zero, reaches a peak, descends back through zero to a trough, and returns back to zero. This full sequence is called a cycle.
- It is characterized by a consistent wave pattern that repeats every \(2\pi\) radians.
- The negative sign in the given function \(y = -5 \sin 2\pi\theta\) flips the wave vertically.
Cycle
A cycle of a sine wave represents one full wave repetition. In terms of the sine function, this means starting from zero, going to a maximum, then descending back through zero to a minimum, and returning to zero.
- In the exercise's function \(y = -5 \sin 2\pi\theta\), a full cycle encompasses the interval from 0 to 2\(\pi\), which is one complete round of the wave.
- Normally, a sine wave completes one cycle in an interval of \(2\pi\) radians.
- However, due to the frequency adjustment \(2\pi\) in the provided function, each unit on the \(\theta\)-axis corresponds to a full cycle.