Table 1

Plotting the points from the table and continuing along the horizontal axis gives the shape of the sine function. See Figure 2.

Figure 2:  A graph of sin(t).

Notice how the sine values are positive between 0 and [latex]\pi ,[/latex] 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 [latex]\pi [/latex] and [latex]2\pi ,[/latex] which correspond to the values of the sine function in quadrants III and IV on the unit circle. See Figure 3.

Figure 3:  Plotting values of the sine function and relating them to the unit circle.

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.

Table 2

As with the sine function, we can plots points to create a graph of the cosine function as in Figure 4.

Figure 4: A graph of cos(t).

Domain and Range

We have created the graphs of the sine and cosine functions by following a point from the positive x axis one complete rotation counterclockwise around the unit circle.  This rotation generates  input values between 0 and [latex]2\pi. [/latex] We know that we can continue to travel counterclockwise around the unit circle over and over again, and generate larger and larger positive input values.  That means that we can continue to generate sine and cosine values for any positive input value.  Likewise, we can start at the positive x axis, and travel clockwise around the circle.  One revolution in the clockwise direction generates inputs starting at 0 and decreasing to [latex]-2\pi.[/latex]  Again, since we can continue to travel around the circle over and over again in the clockwise direction, we can generate sine and cosine values for any negative input value.  Because we can evaluate the sine and cosine of any real number, both of these functions have a domain of all real numbers.

What are the ranges of the sine and cosine functions?  By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that since the x and y coordinates of points on the circle must each be in the interval [latex]\left[-1,1\right][/latex], the range of both functions must be the interval [latex]\left[-1,1\right].[/latex]

Period

In both graphs, the shape of the graph repeats after [latex]2\pi ,[/latex] which means the functions are periodic with a period of [latex]2\pi .[/latex] A periodic function is a function for which a specific horizontal shift, P, results in a function equal to the original function: [latex]f\left(t+P\right)=f\left(t\right)[/latex] for all values of [latex]t[/latex] in the domain of [latex]f.[/latex] When this occurs, we call the smallest such horizontal shift with [latex]P>0[/latex] the period of the function. Figure 5a and 5b show several periods of the sine and cosine functions.

Figure 5a: Sine graph demonstrating a period.

Figure 5b: Cosine graph demonstrating a period.

Symmetries

Looking again at the sine and cosine functions on a domain centered at the vertical axis helps reveal symmetries. As we can see in Figure 6, the sine function is symmetric about the origin. Recall that in Section 3.3, we determined from the unit circle that the sine function is an odd function because [latex]\mathrm{sin}\left(-t\right)=-\mathrm{sin}\left(t\right).[/latex] Now we can clearly see this property from the graph.

Figure 6:  Odd symmetry of the sine function.

Figure 7 shows that the cosine function is symmetric about the y-axis. Again, we determined from the unit circle that the cosine function is an even function. Now we can see from the graph that [latex]\mathrm{cos}\left(-t\right)=\mathrm{cos}\left(t\right).[/latex]

Figure 7:  Even symmetry of the cosine function.

Recall when we first defined sine and cosine, we referenced an acute angle [latex]t[/latex] in standard position as the input and associated each function with either the [latex]x[/latex] or [latex]y[/latex] coordinate of a point on the unit circle.  In the material that follows, we switch back to typical function notation in the coordinate plane where [latex]y=f\left(x\right)[/latex] and [latex]x[/latex] will represent the angle and not the coordinate.

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:

Analyzing Shifts of Sinusoidal Functions

Now that we understand how [latex]A[/latex] and [latex]B[/latex] relate to the general form equation for the sine and cosine functions, we will explore the variables [latex]h[/latex] and [latex]k.[/latex] Recall the general form:

The value [latex]k[/latex] is the vertical shift of the function and for sinusoidal functions [latex]k[/latex] shifts the midline up or down [latex]k[/latex] units. The function [latex]y=A\mathrm{sin}\left(x\right)+k[/latex] has a midline of [latex]y=k[/latex]. See  Figure 11.

Figure 11: A graph of [latex]y=A\mathrm{sin}\left(x\right)+k.[/latex]

In the equation [latex]y=A\mathrm{sin}\left(B\left(x-h\right)\right)+k,[/latex] any value of [latex]k[/latex] other than zero shifts the graph of [latex]y=A\mathrm{sin}\left(B\left(x-h\right)\right)[/latex] up or down. Figure 12 compares [latex]f\left(x\right)=\mathrm{sin}\left(x\right)[/latex] with [latex]f\left(x\right)=\mathrm{sin}\left(x\right)+2,[/latex] which is shifted 2 units up on a graph.

Figure 12: Vertically shifted sine function.

The value [latex]h[/latex] for a sinusoidal function is called the the horizontal displacement of the basic sine or cosine function. If [latex]h>0,[/latex] the graph shifts to the right. If [latex]h<0,[/latex] the graph shifts to the left. The greater the value of [latex]|h|,[/latex] the more the graph is shifted. Figure 13 shows that the graph of [latex]g\left(x\right)=\mathrm{sin}\left(x-\pi \right)[/latex] shifts [latex]f\left(x\right)=\mathrm{sin}\left(x\right)[/latex] to the right by [latex]\pi [/latex] units, which is more than we see in the graph of [latex]p\left(x\right)=\mathrm{sin}\left(x-\frac{\pi }{4}\right),[/latex] which shifts [latex]f\left(x\right)=\mathrm{sin}\left(x\right)[/latex] to the right by [latex]\frac{\pi }{4}[/latex] units.

Figure 13: Horizontally shifted sine functions.

Throughout this section, we have learned about types of variations of sine and cosine functions and used that information to write equations from graphs. Now we can use the same information to create graphs from equations.

Instead of focusing on the general form equations

we will let [latex]h=0[/latex] and [latex]k=0[/latex] and work with a simplified form of the equations in the following examples.

Example 8:  Graphing a Function and Identifying the Amplitude and Period

Sketch a graph of [latex]f\left(x\right)=-2\mathrm{sin}\left(\frac{\pi x}{2}\right).[/latex]

Show Solution

Let’s begin by comparing the equation to the form [latex]y=A\mathrm{sin}\left(Bx\right).[/latex]

• • Step 1. We can see from the equation that [latex]A=-2,[/latex] so the amplitude is 2.
    [latex]|A|=2.[/latex]
  • Step 2. The equation shows that [latex]B=\frac{\pi }{2},[/latex] so the period is
    [latex]\begin{align*}P&=\frac{2\pi }{\frac{\pi }{2}} \\ &=2\pi\cdot \frac{2}{\pi } \\ &=4. \end{align*}[/latex]
  • Step 3. Because [latex]A[/latex] is negative, the graph descends as we move to the right of the origin.
  • Step 4–7. The x-intercepts are at the beginning of one period, [latex]x=0,[/latex] the horizontal midpoints are at [latex]x=2[/latex] and at the end of one period at [latex]x=4.[/latex]

The quarter points include the minimum at [latex]x=1[/latex] and the maximum at [latex]x=3.[/latex]  A local minimum will occur 2 units below the midline, at [latex]x=1,[/latex] and a local maximum will occur at 2 units above the midline, at [latex]x=3.[/latex]  Figure 20 shows the graph of the function.

Figure 20: A graph of [latex]y=2\mathrm{sin}\left(\frac{\pi x}{2}\right).[/latex]

The graph has range of [-2,2], period of 4, and amplitude of 2.

Using Five Key Points to Graph Sinusoidal Functions

One method of graphing sinusoidal functions is to find five key points. These points will correspond to intervals of equal length representing [latex]\frac{1}{4}[/latex] of the period. The key points will indicate the location of maximum and minimum values. If there is no vertical shift, they will also indicate x-intercepts. For example, suppose we want to graph the function [latex]y=\mathrm{cos}\text{ }\theta .[/latex] We know that the period is [latex]2\pi ,[/latex] so we find the interval between key points as follows.

[latex]\frac{2\pi }{4}=\frac{\pi }{2}[/latex][latex]\\[/latex]

Starting with [latex]\theta =0,[/latex] we calculate the first y-value, add the length of the interval [latex]\frac{\pi }{2}[/latex] to 0, and calculate the second y-value. We then add [latex]\frac{\pi }{2}[/latex] repeatedly until the five key points are determined. The last value should equal the first value, as the calculations cover one full period. Making a table similar to Table 3, we can see these key points clearly on the graph shown in  Figure 25.

Table 3

[latex]\theta [/latex]	[latex]0[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\pi [/latex]	[latex]\frac{3\pi }{2}[/latex]	[latex]2\pi [/latex]
[latex]y=\mathrm{cos}\text{ }\theta [/latex]	[latex]1[/latex]	[latex]0[/latex]	[latex]-1[/latex]	[latex]0[/latex]	[latex]1[/latex]

Figure 25

Example 11:  Graphing Sinusoidal Functions Using Key Points

Graph the function [latex]y=-4\mathrm{cos}\left(\pi x\right)[/latex] using amplitude, period, and key points.

Show Solution

The amplitude is[latex] |-4|=4. [/latex] The period is [latex]\frac{2\pi }{B}=\frac{2\pi }{\pi }=2.[/latex] One cycle of the graph can be drawn over the interval [latex]\left[0,2\right].[/latex] To find the key points, we divide the period by 4 which means we will increment [latex]x[/latex] by [latex]\frac{2}{4}=\frac{1}{2}[/latex]. Make a table similar to Table 4, starting with [latex]x=0[/latex] and then adding [latex]\frac{1}{2}[/latex] successively to [latex]x[/latex] and calculate [latex]y.[/latex] See the graph in Figure 26.

Table 4

[latex]x[/latex]	[latex]0[/latex]	[latex]\frac{1}{2}[/latex]	[latex]1[/latex]	[latex]\frac{3}{2}[/latex]	[latex]2[/latex]
[latex]y=-4\text{ }\mathrm{cos}\left(\pi x\right)[/latex]	[latex]-4[/latex]	[latex]0[/latex]	[latex]4[/latex]	[latex]0[/latex]	[latex]-4[/latex]

Figure 26

TRY IT #10

Graph the function [latex]y=3\mathrm{sin}\left(3x\right)[/latex] using the amplitude, period, and five key points.

Show Solution

Table 5

x	[latex]y=3\mathrm{sin}\left(3x\right)[/latex]
0	0
[latex]\frac{\pi }{6}[/latex]	3
[latex]\frac{\pi }{3}[/latex]	0
[latex]\frac{\pi }{2}[/latex]	[latex]-3[/latex]
[latex]\frac{2\pi }{3}[/latex]	0

Using Transformations of Sine and Cosine Functions

We can use the transformations of sine and cosine functions in numerous applications. As mentioned at the beginning of the chapter, circular motion can be modeled using either the sine or cosine function.

Example 12:  Finding the Vertical Component of Circular Motion

A point rotates around a circle of radius 3 centered at the origin. Sketch a graph of the y-coordinate of the point as a function of the angle of rotation.

Show Solution

Recall that we discussed in Section 3.3 that a point on a circle of radius r will have a y coordinate  of  [latex]y=r\mathrm{sin}\left(\theta\right).[/latex] In this case, we get the equation [latex]y\left(\theta\right)=3\mathrm{sin}\left(\theta\right).[/latex]

Figure 27: A graph of 3sin([latex]\theta[/latex]).

Notice that the period of the function is still [latex]2\pi ;[/latex] as we travel around the circle, we return to the point [latex]\left(3,0\right)[/latex] for [latex]x=2\pi ,4\pi ,6\pi ,….[/latex]

From our study of transformations of trigonometric functions, we also know that the constant 3 causes a vertical stretch of the sine function by a factor of 3, which we can see in the graph in Figure 27.

Because the outputs of the graph will now oscillate between [latex]-3[/latex] and [latex]3,[/latex] the amplitude of the sine wave is [latex]3.[/latex]  This means that the radius of the circle centered at the origin will correspond to the amplitude of the sine function.