White light, such as the light from the sun, is not actually white at all. Instead, it is a composition of all the colors of the rainbow in the form of waves. The individual colors can be seen only when white light passes through an optical prism that separates the waves according to their wavelengths to form a rainbow.
Light waves can be represented graphically by the sine function. In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and create graphs of sine and cosine functions.
Recall that the sine and cosine functions relate real number values to the x- and y-coordinates of a point on the unit circle. So what do they look like on a graph on a coordinate plane? Let’s start with the sine function . We can create a table of values and use them to sketch a graph. Table 1 lists some of the values for the sine function on a unit circle.
| x x | 0 0 | π 6 π 6 | π 4 π 4 | π 3 π 3 | π 2 π 2 | 2 π 3 2 π 3 | 3 π 4 3 π 4 | 5 π 6 5 π 6 | π π |
| sin ( x ) sin ( x ) | 0 0 | 1 2 1 2 | 2 2 2 2 | 3 2 3 2 | 1 1 | 3 2 3 2 | 2 2 2 2 | 1 2 1 2 | 0 0 |
Plotting the points from the table and continuing along the x-axis gives the shape of the sine function. See Figure 2.
Notice how the sine values are positive between 0 and π , π , which correspond to the values of the sine function in quadrants I and II on the unit circle, and the sine values are negative between π π and 2 π , 2 π , which correspond to the values of the sine function in quadrants III and IV on the unit circle. See Figure 3.
Now let’s take a similar look at the cosine function . Again, we can create a table of values and use them to sketch a graph. Table 2 lists some of the values for the cosine function on a unit circle.
| x x | 0 0 | π 6 π 6 | π 4 π 4 | π 3 π 3 | π 2 π 2 | 2 π 3 2 π 3 | 3 π 4 3 π 4 | 5 π 6 5 π 6 | π π |
| cos ( x ) cos ( x ) | 1 1 | 3 2 3 2 | 2 2 2 2 | 1 2 1 2 | 0 0 | − 1 2 − 1 2 | − 2 2 − 2 2 | − 3 2 − 3 2 | − 1 − 1 |
As with the sine function, we can plots points to create a graph of the cosine function as in Figure 4.
Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval [ − 1 , 1 ] . [ − 1 , 1 ] .
In both graphs, the shape of the graph repeats after 2 π , 2 π , which means the functions are periodic with a period of 2 π . 2 π . A periodic function is a function for which a specific horizontal shift , P, results in a function equal to the original function: f ( x + P ) = f ( x ) f ( x + P ) = f ( x ) for all values of x x in the domain of f . f . When this occurs, we call the smallest such horizontal shift with P > 0 P > 0 the period of the function. Figure 5 shows several periods of the sine and cosine functions.
Looking again at the sine and cosine functions on a domain centered at the y-axis helps reveal symmetries. As we can see in Figure 6, the sine function is symmetric about the origin. Recall from The Other Trigonometric Functions that we determined from the unit circle that the sine function is an odd function because sin ( − x ) = − sin x . sin ( − x ) = − sin x . Now we can clearly see this property from the graph.
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Figure 7 shows that the cosine function is symmetric about the y-axis. Again, we determined that the cosine function is an even function. Now we can see from the graph that cos ( − x ) = cos x . cos ( − x ) = cos x .
The sine and cosine functions have several distinct characteristics:
As we can see, sine and cosine functions have a regular period and range. If we watch ocean waves or ripples on a pond, we will see that they resemble the sine or cosine functions. However, they are not necessarily identical. Some are taller or longer than others. A function that has the same general shape as a sine or cosine function is known as a sinusoidal function . The general forms of sinusoidal functions are
y = A sin ( B x − C ) + D and y = A cos ( B x − C ) + D y = A sin ( B x − C ) + D and y = A cos ( B x − C ) + D
Looking at the forms of sinusoidal functions, we can see that they are transformations of the sine and cosine functions. We can use what we know about transformations to determine the period.
If we let C = 0 C = 0 and D = 0 D = 0 in the general form equations of the sine and cosine functions, we obtain the forms
y = A sin ( B x ) y = A sin ( B x ) y = A cos ( B x ) y = A cos ( B x )The period is 2 π | B | . 2 π | B | .
Determine the period of the function f ( x ) = sin ( π 6 x ) . f ( x ) = sin ( π 6 x ) .
Let’s begin by comparing the equation to the general form y = A sin ( B x ) . y = A sin ( B x ) .
In the given equation, B = π 6 , B = π 6 , so the period will be
P = 2 π | B | = 2 π π 6 = 2 π ⋅ 6 π = 12 P = 2 π | B | = 2 π π 6 = 2 π ⋅ 6 π = 12Determine the period of the function g ( x ) = cos ( x 3 ) . g ( x ) = cos ( x 3 ) .
Returning to the general formula for a sinusoidal function, we have analyzed how the variable B B relates to the period. Now let’s turn to the variable A A so we can analyze how it is related to the amplitude, or greatest distance from rest. A A represents the vertical stretch factor, and its absolute value | A | | A | is the amplitude. The local maxima will be a distance | A | | A | above the horizontal midline of the graph, which is the line y = D ; y = D ; because D = 0 D = 0 in this case, the midline is the x-axis. The local minima will be the same distance below the midline. If | A | > 1 , | A | > 1 , the function is stretched. For example, the amplitude of f ( x ) = 4 sin x f ( x ) = 4 sin x is twice the amplitude of f ( x ) = 2 sin x . f ( x ) = 2 sin x . If | A | < 1 , | A | < 1 , the function is compressed. Figure 9 compares several sine functions with different amplitudes.
If we let C = 0 C = 0 and D = 0 D = 0 in the general form equations of the sine and cosine functions, we obtain the forms
y = A sin ( B x ) and y = A cos ( B x ) y = A sin ( B x ) and y = A cos ( B x )The amplitude is | A | , | A | , which is the vertical height from the midline . . In addition, notice in the example that
| A | = amplitude = 1 2 | maximum − minimum | | A | = amplitude = 1 2 | maximum − minimum |What is the amplitude of the sinusoidal function f ( x ) = −4 sin ( x ) ? f ( x ) = −4 sin ( x ) ? Is the function stretched or compressed vertically?
Let’s begin by comparing the function to the simplified form y = A sin ( B x ) . y = A sin ( B x ) .
In the given function, A = −4 , A = −4 , so the amplitude is | A | = | −4 | = 4. | A | = | −4 | = 4. The function is stretched.
The negative value of A A results in a reflection across the x-axis of the sine function , as shown in Figure 10.
What is the amplitude of the sinusoidal function f ( x ) = 1 2 sin ( x ) ? f ( x ) = 1 2 sin ( x ) ? Is the function stretched or compressed vertically?
Now that we understand how A A and B B relate to the general form equation for the sine and cosine functions, we will explore the variables C C and D . D . Recall the general form:
y = A sin ( B x − C ) + D and y = A cos ( B x − C ) + D o r y = A sin ( B ( x − C B ) ) + D and y = A cos ( B ( x − C B ) ) + D y = A sin ( B x − C ) + D and y = A cos ( B x − C ) + D o r y = A sin ( B ( x − C B ) ) + D and y = A cos ( B ( x − C B ) ) + D
While C C relates to the horizontal shift, D D indicates the vertical shift from the midline in the general formula for a sinusoidal function. See Figure 12. The function y = cos ( x ) + D y = cos ( x ) + D has its midline at y = D . y = D .
Any value of D D other than zero shifts the graph up or down. Figure 13 compares f ( x ) = sin ( x ) f ( x ) = sin ( x ) with f ( x ) = sin ( x ) + 2 , f ( x ) = sin ( x ) + 2 , which is shifted 2 units up on a graph.
Given an equation in the form f ( x ) = A sin ( B x − C ) + D f ( x ) = A sin ( B x − C ) + D or f ( x ) = A cos ( B x − C ) + D , f ( x ) = A cos ( B x − C ) + D , C B C B is the phase shift and D D is the vertical shift .
Determine the direction and magnitude of the phase shift for f ( x ) = sin ( x + π 6 ) − 2. f ( x ) = sin ( x + π 6 ) − 2.
Let’s begin by comparing the equation to the general form y = A sin ( B x − C ) + D . y = A sin ( B x − C ) + D .
In the given equation, notice that B = 1 B = 1 and C = − π 6 . C = − π 6 . So the phase shift is
C B = − π 6 1 = − π 6 C B = − π 6 1 = − π 6or π 6 π 6 units to the left.
We must pay attention to the sign in the equation for the general form of a sinusoidal function. The equation shows a minus sign before C . C . Therefore f ( x ) = sin ( x + π 6 ) − 2 f ( x ) = sin ( x + π 6 ) − 2 can be rewritten as f ( x ) = sin ( x − ( − π 6 ) ) − 2. f ( x ) = sin ( x − ( − π 6 ) ) − 2. If the value of C C is negative, the shift is to the left.
Determine the direction and magnitude of the phase shift for f ( x ) = 3 cos ( x − π 2 ) . f ( x ) = 3 cos ( x − π 2 ) .
Determine the direction and magnitude of the vertical shift for f ( x ) = cos ( x ) − 3. f ( x ) = cos ( x ) − 3.
Let’s begin by comparing the equation to the general form y = A cos ( B x − C ) + D . y = A cos ( B x − C ) + D .
In the given equation, D = −3 D = −3 so the shift is 3 units downward.
Determine the direction and magnitude of the vertical shift for f ( x ) = 3 sin ( x ) + 2. f ( x ) = 3 sin ( x ) + 2.
Given a sinusoidal function in the form f ( x ) = A sin ( B x − C ) + D , f ( x ) = A sin ( B x − C ) + D , identify the midline, amplitude, period, and phase shift.
Determine the midline, amplitude, period, and phase shift of the function y = 3 sin ( 2 x ) + 1. y = 3 sin ( 2 x ) + 1.
Let’s begin by comparing the equation to the general form y = A sin ( B x − C ) + D . y = A sin ( B x − C ) + D .
A = 3 , A = 3 , so the amplitude is | A | = 3. | A | = 3.
Next, B = 2 , B = 2 , so the period is P = 2 π | B | = 2 π 2 = π . P = 2 π | B | = 2 π 2 = π .
There is no added constant inside the parentheses, so C = 0 C = 0 and the phase shift is C B = 0 2 = 0. C B = 0 2 = 0.
Finally, D = 1 , D = 1 , so the midline is y = 1. y = 1.
Inspecting the graph, we can determine that the period is π , π , the midline is y = 1 , y = 1 , and the amplitude is 3. See Figure 14.
Determine the midline, amplitude, period, and phase shift of the function y = 1 2 cos ( x 3 − π 3 ) . y = 1 2 cos ( x 3 − π 3 ) .
Determine the formula for the cosine function in Figure 15.
To determine the equation, we need to identify each value in the general form of a sinusoidal function.
y = A sin ( B x − C ) + D y = A cos ( B x − C ) + D y = A sin ( B x − C ) + D y = A cos ( B x − C ) + D
The graph could represent either a sine or a cosine function that is shifted and/or reflected. When x = 0 , x = 0 , the graph has an extreme point, ( 0 , 0 ) . ( 0 , 0 ) . Since the cosine function has an extreme point for x = 0 , x = 0 , let us write our equation in terms of a cosine function.
Let’s start with the midline. We can see that the graph rises and falls an equal distance above and below y = 0.5. y = 0.5. This value, which is the midline, is D D in the equation, so D = 0.5. D = 0.5.
The greatest distance above and below the midline is the amplitude. The maxima are 0.5 units above the midline and the minima are 0.5 units below the midline. So | A | = 0.5. | A | = 0.5. Another way we could have determined the amplitude is by recognizing that the difference between the height of local maxima and minima is 1, so | A | = 1 2 = 0.5. | A | = 1 2 = 0.5. Also, the graph is reflected about the x-axis so that A = − 0.5. A = − 0.5.
The graph is not horizontally stretched or compressed, so B = 1; B = 1; and the graph is not shifted horizontally, so C = 0. C = 0.
Putting this all together,
