19.2.3: Amplitude and Period
- Page ID
- 70765
- Understand amplitude and period.
- Graph the sine function with changes in amplitude and period.
- Graph the cosine function with changes in amplitude and period.
- Match a sine or cosine function to its graph and vice versa.
Introduction
You know how to graph the functions \(y = \sin x\) and \(y = \cos x\). Now you’ll learn how to graph a whole “family” of sine and cosine functions. These functions have the form \(y = a\sin bx\) or \(y = a\cos bx\), where \(a\) and \(b\) are constants.
Periodic Functions
We used the variable \(\theta\) previously to show an angle in standard position, and we also referred to the sine and cosine functions as \(y = \sin \theta\) and \(y = \cos \theta\). Often the sine and cosine functions are used in applications that have nothing to do with triangles or angles, and the letter \(x\) is used instead of \(\theta\) for the input (as well as to label the horizontal axis). So from this point forward, we’ll refer to these same functions as \(y = \sin x\) and \(y = \cos x\). This change does not affect the graphs; they remain the same.
You know that the graphs of the sine and cosine functions have a pattern of hills and valleys that repeat. The length of this repeating pattern is \(2\pi\). That is, the graph of \(y = \sin x\) (or \(y = \cos x\)) on the interval [0,\(2\pi\)] looks like the graph on the interval [\(2\pi, 4\pi\)] or [\(4\pi, 6\pi\)] or [\(-2\pi, 0\)]. This pattern continues in both directions forever.
The graph below shows four repetitions of a pattern of length \(2\pi\). Each one contains exactly one complete copy of the “hill and valley” pattern.
If a function has a repeating pattern like sine or cosine, it is called a periodic function. The period is the length of the smallest interval that contains exactly one copy of the repeating pattern. So the period of \(y = \sin x\) or \(y = \cos x\) is \(2\pi\). Any part of the graph that shows this pattern over one period is called a cycle. For example, the graph of \(y = \cos x\) on the interval [0,\(2\pi\)] is one cycle.
You know from graphing quadratic functions of the form \(y = ax^2\) that as you changed the value of \(a\), you changed the “width” of the graph. Now we’ll look at functions of the form \(y = \sin bx\) and see how changes to \(b\) will affect the graph. For example, is \(y = \sin 2x\) periodic, and if so, what is the period? Here is a table with some inputs and outputs for this function.
for this function.
| \(x\) (in radians) | \(2x\) (in radians) | \(\sin 2x\) |
| 0 | 0 | 0 |
| \(\frac{\pi}{6}\) | \(\frac{\pi}{3}\) | \(\frac{\sqrt{3}}{2}\) |
| \(\frac{\pi}{4}\) | \(\frac{\pi}{2}\) | 1 |
| \(\frac{\pi}{3}\) | \(\frac{2\pi}{3}\) | \(\frac{\sqrt{3}}{2}\) |
| \(\frac{\pi}{2}\) | \(\pi\) | 0 |
| \(\frac{2\pi}{3}\) | \(\frac{4\pi}{3}\) | \(-\frac{\sqrt{3}}{2}\) |
| \(\frac{3\pi}{4}\) | \(\frac{3\pi}{2}\) | -1 |
| \(\frac{5\pi}{6}\) | \(\frac{5\pi}{3}\) | \(-\frac{\sqrt{3}}{2}\) |
| \(\pi\) | \(2\pi\) | 0 |
As the values of \(x\) go from 0 to \(\pi\), the values of \(2x\) go from 0 to \(2\pi\). We can see from the graph that the function \(y = \sin 2x\) is a periodic function, and goes through one full cycle on the interval [0, \(\pi\)], so its period is \(\pi\). If you substituted values of \(x\) from \(\pi\) to \(2\pi\), the values of \(2x\) would go from\(2\pi\) to \(4\pi\), and \(\sin 2x\) would go through another complete cycle of the sine function.
Notice that \(y = \sin 2x\) has two cycles on the interval [0, \(2\pi\)] which is the interval \(y = \sin x\) needs to complete one full cycle.
What is the smallest positive value for \(x\) where \(y = \sin 2x\) is at its minimum?
- \(-\frac{\pi}{4}\)
- \(\frac{\pi}{4}\)
- \(\frac{3\pi}{4}\)
- \(\frac{\pi}{2}\)
- Answer
-
- \(-\frac{\pi}{4}\). Incorrect. The function does attain its minimum value at this point, but \(-\frac{\pi}{4}\) is not a positive value. The correct answer is \(\frac{3\pi}{4}\).
- \(\frac{\pi}{4}\). Incorrect. Perhaps you confused minimum and maximum. The correct answer is \(\frac{3\pi}{4}\).
- \(\frac{3\pi}{4}\)Correct. The minimum value for the sine function is -1. Look at the graph of \(y = \sin 2x\). It attains this minimum at the bottom of every valley. The bottom of the first valley where \(x\) is positive is at \(x = \frac{3\pi}{4}\).
- \(\frac{\pi}{2}\). Incorrect. You may have thought of 0 as the minimum value, but the sine function takes on negative values. The correct answer is \(\frac{3\pi}{4}\).
What is the period of the function \(y = \sin 3x\)? Here is a table with some inputs and outputs for this function.
| \(x\) (in radians) | \(3x\) (in radians) | \(\sin 3x\) |
| 0 | 0 | 0 |
| \(\frac{\pi}{9}\) | \(\frac{\pi}{3}\) | \(\frac{\sqrt{3}}{2}\) |
| \(\frac{\pi}{6}\) | \(\frac{\pi}{2}\) | 1 |
| \(\frac{2\pi}{9}\) | \(\frac{2\pi}{3}\) | \(\frac{\sqrt{3}}{2}\) |
| \(\frac{\pi}{3}\) | \(\pi\) | 0 |
| \(\frac{4\pi}{9}\) | \(\frac{4\pi}{3}\) | \(-\frac{\sqrt{3}}{2}\) |
| \(\frac{\pi}{2}\) | \(\frac{3\pi}{2}\) | -1 |
| \(\frac{5\pi}{9}\) | \(\frac{5\pi}{3}\) | \(-\frac{\sqrt{3}}{2}\) |
| \(\frac{2\pi}{3}\) | \(2\pi\) | 0 |
As the values of \(x\) go from 0 to \(\frac{2\pi}{3}\), the values of \(3x\) go from 0 to \(2\pi\). We can see from the graph that \(y = \sin 3x\) goes through one full cycle on the interval [0, \(\frac{2\pi}{3}\)], so its period is \(\frac{2\pi}{3}\).
Notice that \(y = \sin 3x\) has three cycles on the interval [0, \(2\pi\)], which is the interval \(y = \sin x\) needs to complete one full cycle.
What is the period of the function \(y=\sin \left(\frac{1}{2}\right) x\)? Here is a table with some inputs and outputs for this function.
| \(x\) (in radians) | \(\frac{1}{2}x\) (in radians) | \(\sin \left(\frac{1}{2}\right)\) |
| 0 | 0 | 0 |
| \(\frac{2\pi}{3}\) | \(\frac{\pi}{3}\) | \(\frac{\sqrt{3}}{2}\) |
| \(\pi\) | \(\frac{\pi}{2}\) | 1 |
| \(\frac{4\pi}{3}\) | \(\frac{2\pi}{3}\) | \(\frac{\sqrt{3}}{2}\) |
| \(2\pi\) | \(\pi\) | 0 |
| \(\frac{8\pi}{3}\) | \(\frac{4\pi}{3}\) | \(-\frac{\sqrt{3}}{2}\) |
| \(3\pi\) | \(\frac{3\pi}{2}\) | -1 |
| \(\frac{10\pi}{3}\) | \(\frac{5\pi}{3}\) | \(-\frac{\sqrt{3}}{2}\) |
| \(4\pi\) | \(2\pi\) | 0 |
As the values of \(x\) go from 0 to \(4\pi\), the values of \(\frac{1}{2}x\) go from 0 to \(2\pi\).
We can see from the graph that \(y=\sin \left(\frac{1}{2} x\right)\) goes through one full cycle on the interval \([0,4 \pi]\), so its period is \(4\pi\).
Notice that \(y=\sin \left(\frac{1}{2} x\right)\) has half of one full cycle on the interval \(\left[0,2^{\pi}\right]\), which is the interval \(y=\sin x\)
Let’s put these results into a table. For the first three functions, we have rewritten their periods with the numerator \(2\pi\) so that the pattern becomes clear. Can you see a relationship between the function and the denominator in the periods? needs to complete one full cycle.
| Function | Period |
|---|---|
| \(y=\sin \left(\frac{1}{2} x\right)\) | \(4 \pi=\frac{2 \pi}{\frac{1}{2}}\) |
| \(y=\sin 1 x\) | \(2 \pi=\frac{2 \pi}{1}\) |
| \(y=\sin 2 x\) | \(\pi=\frac{2 \pi}{2}\) |
| \(y=\sin (3 x)\) | \(\frac{2 \pi}{3}=\frac{2 \pi}{3}\) |
In each case, the period could be found by dividing \(2\pi\) by the coefficient of \(x\). In general, the period of \(y=\sin b x\) is \(\frac{2 \pi}{|b|}\), and the period of \(y=\cos b x\) is \(\frac{2 \pi}{|b|}\). Since the period is the length of an interval, it must always be a positive number. Since it is possible for \(b\) to be a negative number, we must use \(|b|\) in the formula to be sure the period, \(\frac{2 \pi}{|b|}\), is always a positive number.
The period of \(y=\sin b x\) or \(y=\cos b x\) is \(\frac{2 \pi}{|b|}\).
You can think of the different values of \(b\) as having an “accordion” (or a spring) effect on the graphs of sine and cosine. A large value of \(b\) squeezes them in and a small value of \(b\) stretches them out.
There is another way to describe this effect. In the interval [0, \(2\pi\)], \(y=\sin x\) goes through one cycle while \(y=\sin 2 x\) goes through two cycles. If you go back and check all of the examples above, you will see that \(y=\sin b x\) has \(|b|\) cycles in the interval \([0,2 \pi]\). Likewise, \(y=\cos b x\) has \(|b|\) cycles in the interval [0, \(2\pi\)].
Problem: What are the periods of \(y=\sin 4 x\) and \(y=\cos \left(-\frac{1}{2} x\right)\)?
Answer
For \(y=\sin 4 x\), \(b = 4\). Substitute this value into the formula.
\(\text { period }=\frac{2 \pi}{|b|}=\frac{2 \pi}{4}=\frac{\pi}{2}\)
For \(y=\cos \left(-\frac{1}{2} x\right)\), \(b = -\frac{1}{2}\).
\(\begin{array}{c}
\text { period }=\frac{2 \pi}{|b|}=\frac{2 \pi}{\left|-\frac{1}{2}\right|}=\frac{2 \pi}{\frac{1}{2}} \\
=\frac{2 \pi}{1} \cdot \frac{2}{1}=4 \pi
\end{array}\)
The period of \(y=\sin 4 x\) is \(\frac{\pi}{2}\), and the period of \(y=\cos \left(-\frac{1}{2} x\right)\) is \(4\pi\).
Amplitude
As you have seen, the graphs of all of these sine and cosine functions alternate between hills and valleys. The height of one hill (which equals the depth of one valley) is called the amplitude. You can see that for all the graphs we have looked at so far, the amplitude equals 1.
The formal way to say this for any periodic function is:
\(\text { amplitude }=\frac{\operatorname{maximum}-\operatorname{minimum}}{2}\)
You know that the maximum value of \(y=\sin x\) or \(y=\cos x\) is 1 and the minimum value of either is -1. So if you applied the above definition, you would get:
\(\text { amplitude }=\frac{1-(-1)}{2}=\frac{1+1}{2}=\frac{2}{2}=1\)
This result agrees with what was observed from the graph.
You have seen that changing the value of \(b\) in \(y=\sin b x\) or \(y=\cos b x\) either stretches or squeezes the graph like an accordion or a spring, but it does not change the maximum or minimum values. For all of these functions, the maximum is 1 and the minimum is -1. So if you applied the definition of amplitude, you would be doing the exact same calculation as we just did above. The amplitude of any of these functions is 1.
Let’s look at a different kind of change to a function by graphing the function \(y=2 \sin x\). Here’s a table with some values of this function.
| \(x\) (in radians) | \(\sin x\) | \(2 \sin x\) |
| 0 | 0 | 0 |
| \(\frac{\pi}{6}\) | \(\frac{1}{2}\) | 1 |
| \(\frac{\pi}{4}\) | \(\frac{\sqrt{2}}{2}\) | \(\sqrt{2}\) |
| \(\frac{\pi}{3}\) | \(\frac{\sqrt{3}}{2}\) | \(\sqrt{3}\) |
| \(\frac{\pi}{2}\) | 1 | 2 |
| \(\frac{2\pi}{3}\) | \(\frac{\sqrt{3}}{2}\) | \(\sqrt{3}\) |
| \(\frac{3\pi}{4}\) | \(\frac{\sqrt{2}}{2}\) | \(\sqrt{2}\) |
| \(\frac{5\pi}{6}\) | \(\frac{1}{2}\) | 1 |
| \(\pi\) | 0 | 0 |
We’ll take the first and third columns to make part of the graph and then extend that pattern to the left and to the right.
Now you can use this graph in the following example.
Problem: What is the amplitude of \(y = 2\sin x\)?
Answer
You can find the maximum and minimum values of the function from the graph. For example, at \(x = \frac{\pi}{2}\) the value is 2, and at \(x = -\frac{\pi}{2}\) the value is -2.
\(\begin{array}{c}
\text { maximum }=2 \\
\text { minimum }=-2
\end{array}\)
Use the definition of amplitude.
\(\text { amplitude }=\frac{2-(-2)}{2}=\frac{2+2}{2}=\frac{4}{2}=2\)
Notice that the height of each hill is 2, and the depth of each valley is 2. This is equal to the amplitude, as we mentioned at the start.
Notice also that the amplitude is equal to the coefficient of the function: \(y = 2\sin x\)
This is not a coincidence.
The amplitude is 2.
Let’s compare the graph of this function to the graph of the sine function.
The effect of multiplying \(\sin x\) by 2 is to stretch the graph vertically by a factor of 2. Because it has been stretched vertically by this factor, the amplitude is twice as much, or 2. If we had looked at \(y = 3\sin x\), the graph would have been stretched vertically by a factor of 3, and the amplitude of this function is 3. Again, this is equal to the coefficient of the function. In general, we have the following rule.
The amplitude of \(y = a\sin bx\) or \(y = a\cos bx\) is |\(a\)|.
As the last example, \(y = 2\sin x\), shows, multiplying by a constant on the outside affects the amplitude. If you multiply by a constant on the outside and on the inside, as in \(y = -3\sin 2x\), you will affect both the amplitude and the period. Here is the graph of \(y = -3\sin 2x\):
Problem: Determine the amplitude and period of \(y = -3\cos 2x\).
Answer
Use the formula for amplitude, with \(a = -3\).
\(\text { amplitude }=|a|=|-3|=3\)
Use the formula for period, with \(b = 2\).
\(\text { period }=\frac{2 \pi}{|b|}=\frac{2 \pi}{2}=\pi\)
The amplitude is 3 and the period is \(\pi\).
In this example, you could have found the period by looking at the graph above. One complete cycle is shown, for example, on the interval [0, \(\pi\)], so the period is \(\pi\).
In the functions \(y = a\sin bx\) and \(y = a\cos bx\), multiplying by the constant \(a\) only affects the amplitude, not the period. As we said earlier, changing the value of \(b\) only affects the period, not the amplitude. The general result is as follows.
The amplitude of \(y = a\sin bx\) or \(y = a\cos bx\) is |\(a\)|.
The period of \(y = a\sin bx\) or \(y = a\cos bx\) is \(\frac{2 \pi}{|b|}\).
Supplemental Interactive Activity
To help you understand changes in amplitude and period for both the sine function and cosine function, try the following interactive exercise.
*Insert Interactive Activity
What are the amplitude and period of \(y = \frac{1}{2}\sin 4x\)?
- The amplitude is \(\frac{1}{2}\), and the period is \(\frac{\pi}{2}\).
- The amplitude is \(\frac{1}{2}\), and the period is \(8\pi\).
- The amplitude is 1, and the period is \(\frac{\pi}{2}\).
- The amplitude is 1, and the period is \(8\pi\).
- Answer
-
- Correct. In this function, \(a = \frac{1}{2}\), so this is the amplitude. The period equals \(\frac{2 \pi}{|b|}=\frac{2 \pi}{4}=\frac{\pi}{2}\).
- Incorrect. The amplitude is correct, but the period is not. You probably multiplied \(2\pi\) by 4 instead of dividing. The correct answer is A.
- Incorrect. The period is correct, but the amplitude is not. You may have thought the amplitude is the maximum minus the minimum, but it is half of this. The correct answer is A.
- Incorrect. You may have thought the amplitude is the maximum minus the minimum, but it is half of this. You probably multiplied \(2\pi\) by 4, instead of dividing, to find the period. The correct answer is A.
Graphs of Sine Functions
You know the function \(y=a \sin b x\) has amplitude \(|a|\) and period \(\frac{2 \pi}{|b|}\). You can use these facts to draw the graph of any function in the form \(y=a \sin b x\) by starting with the graph of \(y = \sin x\) and modifying it.
For example, suppose you wanted the graph of \(y=4 \sin x\). Since \(b = 1\), this function has the same period as \(y = \sin x\). Since \(a = 4\), the amplitude is 4. Therefore, you would take the graph of \(y = \sin x\) and simply stretch it vertically by a factor of 4. Here is one cycle for these two functions.
Note that the points that were on the \(x\)-axis “remain” on the \(x\)-axis. At these points (where \(x=0, \pi, 2 \pi\)), the value of \(\sin x\) is 0. If you multiply 0 by 4 (or anything else), you will still have a value of 0. So the points will still be on the \(x\)-axis. On the other hand, the highest and lowest points have moved away from the \(x\)-axis. They had \(y\)-values of 1 and -1 for \(\sin x\), and they have \(y\)-values of 4 and -4 for \(4\sin x\).
Problem: Graph \(y = -\sin x\) on the interval [0, \(2\pi\)].
Answer
The value of \(b\) is 1, so the graph has a period of \(2\pi\), as does \(y = \sin x\).
The value of \(a\) is -1, so the graph has an amplitude of 1, as does \(y = \sin x\).
Though the amplitude and the period are the same as the function \(y = \sin x\), the graph is not exactly the same. The effect of multiplying by -1 is to replace \(y\)-values by their opposites. So the graph of \(y = \sin x\) gets reflected over the \(x\)-axis.
If you want to check these graphs with a graphing calculator, make sure that the graphing window has the correct settings. For this last example, you would use \(0 \leq x \leq 2 \pi\) and \(-1 \leq y \leq 1\). In general, you would probably want to adjust the \(x\)-values to show one full cycle and the \(y\)-values using the amplitude.
If the values of \(a\) and \(b\) are both different from 1, then you need to combine the effects of the two changes.
Problem: Graph \(y=3 \sin \left(\frac{1}{2} x\right)\) on the interval [0, \(4\pi\)].
Answer
The value of \(a\) is 3, so the graph has an amplitude of 3. This has the effect of taking the graph of \(y = \sin x\) and stretching it vertically by a factor of 3.
The value of \(b\) is \(\frac{1}{2}\), so the graph has a period of \(\frac{2 \pi}{\frac{1}{2}}=\frac{2 \pi}{1} \cdot \frac{2}{1}=4 \pi\). This is twice the period of \(y = \sin x\). This has the effect of taking the graph of \(y = \sin x\) and stretching it horizontally by a factor of 2. (The alternative way to say this is that \(y=3 \sin \left(\frac{1}{2} x\right)\) has \(\frac{1}{2}\) of a cycle on the interval [0, \(2\pi\)].)
To make the graph of \(y=3 \sin \left(\frac{1}{2} x\right)\), you must combine the two effects described above.
Sometimes you need to stretch the graph of the sine function, and sometimes you need to shrink it.
Problem: Graph two cycles of a sine function whose amplitude is \(\frac{1}{2}\) and whose period is \(\frac{2 \pi}{3}\).
Answer
There are different functions of the form \(y=a \sin b x\) that fit this description because \(a\) and \(b\) could be positive or negative. We will draw the graph assuming these are positive.
Because the amplitude is \(\frac{1}{2}\), this has the effect of taking the graph of \(y = \sin x\) and shrinking it vertically by a factor of 2.
The period is \(\frac{2 \pi}{3}=\frac{1}{3} \cdot 2 \pi\), which is \(\frac{1}{3}\) the period of \(y = \sin x\). This has the effect of taking the graph of \(y = \sin x\) and shrinking it horizontally by a factor of 3.
To make the graph, you must combine the two effects described above.
Even without knowing the specific value of a constant, you can sometimes still narrow down the possibilities for the shape of a graph.
The graph of a function \(y = a \sin x\), where \(a\) is a constant, is drawn on the interval \(0 \leq x \leq 2 \pi\). Which of the following options could be this graph?
- Answer
-
- Incorrect. This has the correct shape and period, but it is in the wrong position. Regardless of the value of \(a\), the graph must pass through the \(x\)-axis at \(x=0, \pi, 2 \pi\), which it does not. The correct answer is D.
- Incorrect. This is the graph of a cosine function. Regardless of the value of \(a\), the graph must pass through the \(x\)-axis at \(x=0, \pi, 2 \pi\). The correct answer is D.
- Incorrect. Because the coefficient of \(x\) is 1, the graph should have a period of \(2\pi\), but this graph has a period of \(\pi\). The correct answer is D.
- Correct. Because the coefficient of \(x\) is 1, the graph has a period of \(2\pi\), which this option has. The factor \(a\) could stretch or shrink the graph, but it must still pass through the \(x\)-axis at the points \(x=0, \pi, 2 \pi\), which it does.
Graphs of Cosine Functions
You know the function \(y=a \cos b x\) has amplitude \(|a|\) and period \(\frac{2 \pi}{|b|}\). Just as you did with sine functions, you can use these facts to draw the graph of any function in the form \(y=a \cos b x\) by starting with the graph of \(y = \cos x\) and modifying it.
For example, suppose you wanted the graph of \(y=\cos \left(\frac{1}{2} x\right)\). Since \(a = 1\), it has the same amplitude as \(y = \cos x\). And, because \(b=\frac{1}{2}\), the period is given by:
\(\frac{2 \pi}{|b|}=\frac{2 \pi}{\frac{1}{2}}=\frac{2 \pi}{1} \cdot \frac{2}{1}=4 \pi\)
Since this is twice the period of \(y = \cos x\), you would take the graph of \(y = \cos x\) and stretch it horizontally by a factor of 2. Here is a side-by-side comparison of these two graphs.
Note that in the interval [0, \(2\pi\)], the graph of \(y = \cos x\) has one full cycle. Since \(b=\frac{1}{2}\), the graph of \(y=\cos \left(\frac{1}{2} x\right)\) has \(\frac{1}{2}\) of a cycle in that interval.
If you are using a graphing calculator, you need to adjust the settings for each graph to get a graphing window that shows all the features of the graph. For the last example, you would use \(0 \leq x \leq 4 \pi\) and \(-1 \leq y \leq 1\). In general, you would probably want to adjust the \(x\)-values to show one full cycle and the \(y\)-values to show the highest and lowest points. In the next example, you would use \(-\pi \leq x \leq \pi\) and \(-2 \leq y \leq 2\) for the graphing window because you are specifically asked to graph it over the domain \([-\pi, \pi]\) and the graph will have an amplitude of 2, going as low as -2 and as high as +2.
Problem: Graph \(y = -2 \cos x\) on the interval \([-\pi, \pi]\).
Answer
The amplitude equals \(|a|=|-2|=2\). This has the effect of taking the graph of \(y = \cos x\) and stretching it vertically by a factor of 2. The negative sign on the “outside” has an additional effect: the \(y\)-values are replaced by their opposites, so the graph is also flipped over the \(x\)-axis.
The value of \(b\) is 1, so the graph has a period of \(2\pi\), as does \(y = \cos x\).
When the only change is a vertical stretch, compression, or flip, the \(x\)-intercepts remain the same. So the graph will pass through the \(x\)-axis at \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
Again, if the values of \(a\) and \(b\) are both different from 1, you need to combine the effects of the two changes. In the next example, you will see a variation that you have not seen before. Up to this point, all of the values of \(b\) have been rational numbers, but here we are using the irrational number \(\pi\). This situation does not really change the procedure, but you will see that it changes the scale on the \(x\)-axis in a new way.
Problem: Graph \(y = 4 \cos \pi x\) on the interval [0, 4].
Answer
The value of \(a\) is 4, so the graph has an amplitude of 4. This has the effect of taking the graph of \(y = \cos x\) and stretching it vertically by a factor of 4.
The value of \(b\) is \(\pi\), so the graph has a period of \(\frac{2 \pi}{\pi}=2\). This has the effect of shrinking the graph of \(y = \cos x\) horizontally by a factor of \(\pi\), causing it to complete one complete cycle on the interval [0, 2].
Note the effect on the \(x\)-values of the intercepts, the high points, and the low points. Because the period is 2, the first cycle of the graph will have high points at \(x = 0\) and 2. The low point will still be midway between these, so it is at \(x = 1\). The \(x\)-intercepts are still midway between the high and the low points, so they will be at \(x=\frac{1}{2}\) and \(\frac{3}{2}\). The second cycle of the graph has all of these points shifted to the right 2 units.
To make the graph of \(y = 4 \cos \pi x\), you must combine the effects described above.
In two previous examples (\(y = -\sin x\) and \(y = -\cos x\)), you saw that a negative sign on the outside (a negative value of \(a\)) has the effect of flipping the graph around the \(x\)-axis. In the next example, you will see the effect of a negative sign on the “inside” (a negative value of \(b\)).
One last hint: besides trying to figure out the overall effect of the value of \(a\) or \(b\) on the graph, you might want to check specific points. For example, just substitute \(x = 0\) into the function and see where that point will end up.
Which of the following options is the graph of \(y=\frac{3}{2} \cos (-x)\) on the interval \(-\pi \leq x \leq \pi\)?
- Answer
-
- Incorrect. This is the graph of \(y=\frac{3}{2} \sin (-x)\). Remember to check specific points like \( = 0\). At that point, \(y=\frac{3}{2} \cos (0)=\frac{3}{2} \cdot 1=\frac{3}{2}\). So the point \(\left(0, \frac{3}{2}\right)\) should be on the graph. The correct answer is C.
- Incorrect. You confused the effects of \(a\) and \(b\). This is the graph of \(y=-\cos \left(\frac{3}{2} x\right)\). The correct answer is C.
- Correct. The value of \(a\) is \(\frac{3}{2}\), which will stretch the graph vertically by a factor of \(\frac{3}{2}\). The period of the graph is \(\frac{2 \pi}{|-1|}=2 \pi\), as is the period of \(y = \cos x\). The effect of the negative sign on the inside is to replace \(x\)-values by their opposites. This will flip the graph around the \(y\)-axis. However, because the graph of cosine is symmetric about the \(y\)-axis, this has no effect at all. So the only change to the graph of \(y = \cos x\) is the vertical stretch.
- Incorrect. This graph has the correct period and amplitude. However, you have confused the effect of a minus sign on the inside with a minus sign on the outside. The correct answer is C.
Matching Graphs and Functions
Given any function of the form \(y=a \sin b x\) or \(y=a \cos b x\), you know how to find the amplitude and period and how to use this information to graph the functions.
Given a graph of a sine or cosine function, you also can determine the amplitude and period of the function. From this information, you can find values of \(a\) and \(b\), and then a function that matches the graph.
Remember that along with finding the amplitude and period, it’s a good idea to look at what is happening at \(x = 0\). If \(a\) and \(b\) are any nonzero constants, the functions \(y=a \sin b x\) and \(y=a \cos b x\) will have the following values at \(x = 0\):
\(\begin{array}{c}
y=a \sin (b \cdot 0)=a \sin 0=a \cdot 0=0 \\
y=a \cos (b \cdot 0)=a \cos 0=a \cdot 1=a \neq 0
\end{array}\)
This tells you that the graph of \(y=a \sin b x\) passes through (0, 0) regardless of the values of \(a\) and \(b\), and the graph of \(y=a \cos b x\) never passes through (0, 0) regardless of the values of \(a\) and \(b\). So, for example, if you are given a graph passing through the origin and are asked to determine which function it represents, you know right away that it is not in the form \(y=a \cos b x\).
You need to be careful about the sign of \(a\). You might determine that a function has an amplitude of 4, for example. Though it is possible that \(a = 4\), it is also possible that \(a = -4\). Here is an example of each of these two possibilities.
You will need to compare the graph to that of \(y = \sin x\) or \(y = \cos x\) to see if, in addition to any stretching or shrinking, there has been a reflection over the \(x\)-axis. The graph above on the right can be thought of as the result of stretching and reflecting the graph of \(y = \sin x\) across the \(x\)-axis. If there has been a reflection, then the value of \(a\) will be negative. Once you have determined if \(a\) is positive or negative, you can always choose a positive value of \(b\).
Here are some examples of this process.
Problem: Determine a function of the form \(y=a \sin b x\) or \(y=a \cos b x\) whose graph is shown below.
Answer
First, observe that the graph passes through the origin, so you are looking for a function of the form \(y=a \sin b x\).
Next, observe that the maximum value of the function is 2 and the minimum is -2, so the amplitude is 2. The graph has the same “orientation” as \(y = \sin x\). (It starts with a hill to the right of the \(y\)-axis.) This implies that \(a\) is positive, and in particular, \(a = 2\).
Finally, observe that the graph shows two cycles and that one entire cycle is contained in the interval \(\left[0, \frac{2 \pi}{3}\right]\). Therefore, the period is \(\frac{2 \pi}{3}\). Since \(\frac{2 \pi}{3}=\frac{2 \pi}{|b|}\), \(b=\pm 3\). According to our process, once you have determined if \(a\) is positive or negative, you can always choose a positive value of \(b\). So \(b = 3\).
Combine these three pieces of information.
\(y=2 \sin 3 x\)
Problem: Determine a function of the form \(y=a \sin b x\) or \(y=a \cos b x\) whose graph is shown below.
Answer
First, observe that the graph does not pass through the origin, but rather crests, reaching a maximum when \(x = 0\), so you are looking for a function of the form \(y=a \cos b x\).
Next, observe that the maximum value of the function is \(\frac{1}{2}\) and the minimum is \(-\frac{1}{2}\), so the amplitude is \(\frac{1}{2}\). The graph has the same “orientation” as \(y = \cos x\). (It has a hill with the \(y\)-axis running through the middle.) This implies that \(a\) is positive, and in particular, \(a = \frac{1}{2}\).
Finally, observe that the graph shows two cycles and that one entire cycle is contained in the interval [0, \(\pi\)]. Therefore, the period is \(\pi\). Because \(\pi=\frac{2 \pi}{|b|}\), \(b=\pm 2\). According to our process, once you have determined if \(a\) is positive or negative, you can always choose a positive value of \(b\). So \(b = 2\).
Combine these three pieces of information.
\(y=\frac{1}{2} \cos 2 x\)
Make sure that you recognize where a cycle starts and ends. The period is the length of the interval over which one cycle runs.
Problem: Which of the following functions is represented by the graph below?
- \(y=-\sin x\)
- \(y=\sin \left(\frac{1}{2} x\right)\)
- \(y=-\sin 2 x\)
- \(y=-\sin \left(\frac{1}{2} x\right)\)
- Answer
-
- \(y=-\sin x\). Incorrect. You correctly recognized the graph as a reflected sine function, but the period is incorrect. Perhaps you saw the \(2\pi\) on the right and used that as the length of one cycle. However, the entire graph is one cycle, and the period equals \(2 \pi-(-2 \pi)=4 \pi\). The correct answer is D.
- \(y=\sin \left(\frac{1}{2} x\right)\). Incorrect. You correctly found the amplitude and period of this sine function. However, you also need to check the orientation of the graph. Notice that to the right of the \(y\)-axis you have a valley instead of a hill. The correct answer is D.
- \(y=-\sin 2 x\). Incorrect. You correctly found the amplitude and the orientation of this sine function. However, the period is incorrect. Perhaps you recognized that the period of the graph is twice the period of \(y = \sin x\), and thought that the value of \(b\) would be 2. However, to find the value of \(b\), you must set the period equal to \(\frac{2 \pi}{|b|}\). The correct answer is D.
- \(y=-\sin \left(\frac{1}{2} x\right)\). Correct. The graph passes through the origin, so the function could have the form \(y=a \sin b x\), but not \(y=a \cos b x\). The amplitude is 1. The graph has a valley on the right, which could be the result of a reflection of \(y = \sin x\) over the \(x\)-axis. Therefore, \(a = -1\). The graph shows one cycle, so the period is \(4\pi\). Since \(4 \pi=\frac{2 \pi}{|b|}\), the value of \(b\) could be \(\frac{1}{2}\). So this could be the graph of \(y=-\sin \left(\frac{1}{2} x\right)\).
Remember that when writing a function you can use the notation \(f(x)\) in place of the variable \(y\).
Which of the following graphs represents \(f(x)=3 \cos \left(\frac{2}{3} x\right)\)?
- Answer
-
- Incorrect. This is the graph of a function of the form \(f(x)=a \sin b x\). Remember to check the value of the function at \(x = 0\). Since \(f(0) = 3\), the function \(f(x)=3 \cos \left(\frac{2}{3} x\right)\) passes through (0, 3), not the origin as shown in this graph. The correct answer is B.
- Correct. First, this graph has the shape of a cosine function. Second, because \(a = 3\) in the equation, the amplitude is 3. Finally, because \(b=\frac{2}{3},\), the period of this function is \(\frac{2 \pi}{\frac{2}{3}}=\frac{2 \pi}{1} \cdot \frac{3}{2}=3 \pi\). The graph in this answer completes one full cycle between \(x=-\frac{3}{2} \pi\) and \(x=\frac{3}{2} \pi\) so its period is \(\frac{3}{2} \pi-\left(-\frac{3}{2} \pi\right)=3 \pi\) as needed.
- Incorrect. This graph does have the shape of a cosine function, and the amplitude is 3, which is correct. However, the period is incorrect. You probably multiplied \(2\pi\) by \(\frac{2}{3}\) instead of dividing. The correct answer is B.
- Incorrect. This graph does have the shape of a cosine function. However, in determining the graph, it appears that you switched the values of \(a\) and \(b\). The correct answer is B.
Summary
The functions \(y=a \sin b x\) and \(y=a \cos b x\) are periodic functions: their graphs have a repeating pattern of hills and valleys that continues in both directions forever. The height of the hill or the depth of the valley is called the amplitude, and is equal to \(|a|\). Any one full pattern in the graph is called a cycle, and the length of an interval over which a cycle occurs is called the period. The period is equal to the value \(\frac{2 \pi}{|b|}\).
You can use this information to graph any of these functions by starting with the basic graph of \(y = \sin x\) or \(y = \cos x\) and then doing a combination of stretching or shrinking the graph vertically based on the value of \(a\), stretching or shrinking the graph horizontally based on the value of \(b\), or reflecting it based on the signs of \(a\) and \(b\).
You can also start with a graph, determine the values of \(a\) and \(b\), and then determine a function that it represents.