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Graphs of the Sine and Cosine Functions

Graph variations of

Graphing 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. lists some of the values for the sine function on a unit circle.

\(x\)	\(0\)	\(\frac{\pi }{6}\)	\(\frac{\pi }{4}\)	\(\frac{\pi }{3}\)	\(\frac{\pi }{2}\)	\(\frac{2\pi }{3}\)	\(\frac{3\pi }{4}\)	\(\frac{5\pi }{6}\)	\(\pi\)
\(\sin (x)\)	\(0\)	\(\frac{1}{2}\)	\[\frac{\sqrt{2}}{2}\]	\[\frac{\sqrt{3}}{2}\]	\(1\)	\[\frac{\sqrt{3}}{2}\]	\[\frac{\sqrt{2}}{2}\]	\(\frac{1}{2}\)	\(0\)

Plotting the points from the table and continuing along the x-axis gives the shape of the sine function. See .

Notice how the sine values are positive between 0 and \(\pi ,\) 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 \(\pi\) and \(2\pi ,\) which correspond to the values of the sine function in quadrants III and IV on the unit circle. See .

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. lists some of the values for the cosine function on a unit circle.

\(x\)	\(0\)	\(\frac{\pi }{6}\)	\(\frac{\pi }{4}\)	\(\frac{\pi }{3}\)	\(\frac{\pi }{2}\)	\(\frac{2\pi }{3}\)	\(\frac{3\pi }{4}\)	\(\frac{5\pi }{6}\)	\(\pi\)
\(\cos (x)\)	\(1\)	\[\frac{\sqrt{3}}{2}\]	\[\frac{\sqrt{2}}{2}\]	\(\frac{1}{2}\)	\[0\]	\[-\frac{1}{2}\]	\[-\frac{\sqrt{2}}{2}\]	\[-\frac{\sqrt{3}}{2}\]	\(-1\)

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

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 \([-1,1].\)

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

Condensed — the full section is in OpenStax Precalculus 2e.

Investigating Sinusoidal Functions

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

\[\begin{array}{l}y=A\sin (Bx-C)+D \\ \text{and} \\ y=A\cos (Bx-C)+D\end{array}\]

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.

In the general formula, \(B\) is related to the period by \(P=\frac{2\pi }{|B|}.\) If \(|B|>1,\) then the period is less than \(2\pi\) and the function undergoes a horizontal compression, whereas if \(|B|<1,\) then the period is greater than \(2\pi\) and the function undergoes a horizontal stretch. For example, \(f(x)=\sin (x),\) \(B=1,\) so the period is \(2\pi ,\\) which we knew. If \(f(x)=\sin (2x),\) then \(B=2,\) so the period is \(\pi\) and the graph is compressed. If \(f(x)=\sin (\frac{x}{2}),\) then \(B=\frac{1}{2},\) so the period is \(4\pi\) and the graph is stretched. Notice in how the period is indirectly related to \(|B|.\)

Example

Try it.

Determine the period of the function \(f(x)=\sin (\frac{\pi }{6}x).\)

Solution

Let’s begin by comparing the equation to the general form \(y=A\sin (Bx).\)

In the given equation, \(B=\frac{\pi }{6},\) so the period will be

\[\begin{array}{l}\begin{array}{l} \\ P=\frac{2\pi }{|B|}\end{array} \\ =\frac{2\pi }{\frac{\pi }{6}} \\ =2\pi ⋅\frac{6}{\pi } \\ =12\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Analyzing Graphs of Variations of

Now that we understand how \(A\) and \(B\) relate to the general form equation for the sine and cosine functions, we will explore the variables \(C\) and \(D.\) Recall the general form:

\[\begin{array}{l}y=A\sin (Bx-C)+D\text{ and }y=A\cos (Bx-C)+D \\ or \\ y=A\sin (B(x-\frac{C}{B}))+D\text{ and }y=A\cos (B(x-\frac{C}{B}))+D\end{array}\]

The value \(\frac{C}{B}\) for a sinusoidal function is called the phase shift, or the horizontal displacement of the basic sine or cosine function. If \(C>0,\) the graph shifts to the right. If \(C<0,\) the graph shifts to the left. The greater the value of \(|C|,\) the more the graph is shifted. shows that the graph of \(f(x)=\sin (x-\pi )\) shifts to the right by \(\pi\) units, which is more than we see in the graph of \(f(x)=\sin (x-\frac{\pi }{4}),\) which shifts to the right by \(\frac{\pi }{4}\) units.

While \(C\) relates to the horizontal shift, \(D\) indicates the vertical shift from the midline in the general formula for a sinusoidal function. See . The function \(y=\cos (x)+D\) has its midline at \(y=D.\)

Any value of \(D\) other than zero shifts the graph up or down. compares \(f(x)=\sin \ (x)\) with \(f(x)=\sin \ (x)+2,\) which is shifted 2 units up on a graph.

Example

Try it.

Determine the direction and magnitude of the phase shift for \(f(x)=\sin (x+\frac{\pi }{6})-2.\)

Solution

Let’s begin by comparing the equation to the general form \(y=A\sin (Bx-C)+D.\)

In the given equation, notice that \(B=1\) and \(C=-\frac{\pi }{6}.\) So the phase shift is

\[\begin{array}{l} \\ \frac{C}{B}=-\frac{\frac{\pi }{6}}{1} \\ =-\frac{\pi }{6}\end{array}\]

or \(\frac{\pi }{6}\) units to the left.

Example

Try it.

Determine the direction and magnitude of the vertical shift for \(f(x)=\cos (x)-3.\)

Solution

Let’s begin by comparing the equation to the general form \(y=A\cos (Bx-C)+D.\)

In the given equation, \(D=-3\) so the shift is 3 units downward.

Condensed — the full section is in OpenStax Precalculus 2e.

Graphing Variations of

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

\[y=A\sin (Bx-C)+D\text{ and }y=A\cos (Bx-C)+D,\]

we will let \(C=0\) and \(D=0\) and work with a simplified form of the equations in the following examples.

Example

Try it.

Sketch a graph of \(f(x)=-2\sin (\frac{\pi x}{2}).\)

Solution

Let’s begin by comparing the equation to the form \(y=A\sin (Bx).\)

Step 1. We can see from the equation that \(A=-2,\) so the amplitude is 2. \[|A|=2\]
Step 2. The equation shows that \(B=\frac{\pi }{2},\) so the period is \[\begin{array}{l}P=\frac{2\pi }{\frac{\pi }{2}} \\ =2\pi ⋅\frac{2}{\pi } \\ =4\end{array}\]
Step 3. Because \(A\) 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, \(x=0,\) the horizontal midpoints are at \(x=2\) and at the end of one period at \(x=4.\)

The quarter points include the minimum at \(x=1\) and the maximum at \(x=3.\) A local minimum will occur 2 units below the midline, at \(x=1,\) and a local maximum will occur at 2 units above the midline, at \(x=3.\) shows the graph of the function.

Condensed — the full section is in OpenStax Precalculus 2e.

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

Try it.

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.

Solution

Recall that, for a point on a circle of radius r, the y-coordinate of the point is \(y=r\ \sin (x),\) so in this case, we get the equation \(y(x)=3\ \sin (x).\) The constant 3 causes a vertical stretch of the y-values of the function by a factor of 3, which we can see in the graph in .

Example

Try it.

A circle with radius 3 ft is mounted with its center 4 ft off the ground. The point closest to the ground is labeled P, as shown in . Sketch a graph of the height above the ground of the point \(P\) as the circle is rotated; then find a function that gives the height in terms of the angle of rotation.

Solution

Sketching the height, we note that it will start 1 ft above the ground, then increase up to 7 ft above the ground, and continue to oscillate 3 ft above and below the center value of 4 ft, as shown in .

Although we could use a transformation of either the sine or cosine function, we start by looking for characteristics that would make one function easier to use than the other. Let’s use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection.

Second, we see that the graph oscillates 3 above and below the center, while a basic cosine has an amplitude of 1, so this graph has been vertically stretched by 3, as in the last example.

Finally, to move the center of the circle up to a height of 4, the graph has been vertically shifted up by 4. Putting these transformations together, we find that

\[y=-3\cos (x)+4\]

Condensed — the full section is in OpenStax Precalculus 2e.

Key Concepts

Periodic functions repeat after a given value. The smallest such value is the period. The basic sine and cosine functions have a period of \(2\pi .\)
The function \(\sin \ x\) is odd, so its graph is symmetric about the origin. The function \(\cos \ x\) is even, so its graph is symmetric about the y-axis.
The graph of a sinusoidal function has the same general shape as a sine or cosine function.
In the general formula for a sinusoidal function, the period is \(P=\frac{2\pi }{|B|}.\) See .
In the general formula for a sinusoidal function, \(|A|\) represents amplitude. If \(|A|>1,\) the function is stretched, whereas if \(|A|<1,\) the function is compressed. See .
The value \(\frac{C}{B}\) in the general formula for a sinusoidal function indicates the phase shift. See .
The value \(D\) in the general formula for a sinusoidal function indicates the vertical shift from the midline. See .
Combinations of variations of sinusoidal functions can be detected from an equation. See .
The equation for a sinusoidal function can be determined from a graph. See and .
A function can be graphed by identifying its amplitude and period. See and .
A function can also be graphed by identifying its amplitude, period, phase shift, and horizontal shift. See .
Sinusoidal functions can be used to solve real-world problems. See , , and .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Determine the period of the function \(f(x)=\sin (\frac{\pi }{6}x).\)

Atskleisti atsakymą

Let’s begin by comparing the equation to the general form \(y=A\sin (Bx).\)
In the given equation, \(B=\frac{\pi }{6},\) so the period will be
\[\begin{array}{l}\begin{array}{l} \\ P=\frac{2\pi }{|B|}\end{array} \\ =\frac{2\pi }{\frac{\pi }{6}} \\ =2\pi ⋅\frac{6}{\pi } \\ =12\end{array}\]
Determine the period of the function \(g(x)=\cos (\frac{x}{3}).\)

Atskleisti atsakymą

\(6\pi\)
What is the amplitude of the sinusoidal function \(f(x)=-4\sin (x)?\) Is the function stretched or compressed vertically?

Atskleisti atsakymą

Let’s begin by comparing the function to the simplified form \(y=A\sin (Bx).\)
In the given function, \(A=-4,\) so the amplitude is \(|A|=|-4|=4.\) The function is stretched.
What is the amplitude of the sinusoidal function \(f(x)=\frac{1}{2}\sin (x)?\) Is the function stretched or compressed vertically?

Atskleisti atsakymą

\(\frac{1}{2}\) compressed
Determine the direction and magnitude of the phase shift for \(f(x)=\sin (x+\frac{\pi }{6})-2.\)

Atskleisti atsakymą

Let’s begin by comparing the equation to the general form \(y=A\sin (Bx-C)+D.\)
In the given equation, notice that \(B=1\) and \(C=-\frac{\pi }{6}.\) So the phase shift is
\[\begin{array}{l} \\ \frac{C}{B}=-\frac{\frac{\pi }{6}}{1} \\ =-\frac{\pi }{6}\end{array}\]
or \(\frac{\pi }{6}\) units to the left.
Determine the direction and magnitude of the phase shift for \(f(x)=3\cos (x-\frac{\pi }{2}).\)

Atskleisti atsakymą

\(\frac{\pi }{2};\) right
Determine the direction and magnitude of the vertical shift for \(f(x)=\cos (x)-3.\)

Atskleisti atsakymą

Let’s begin by comparing the equation to the general form \(y=A\cos (Bx-C)+D.\)
In the given equation, \(D=-3\) so the shift is 3 units downward.
Determine the direction and magnitude of the vertical shift for \(f(x)=3\sin (x)+2.\)

Atskleisti atsakymą

2 units up
Determine the midline, amplitude, period, and phase shift of the function \(y=3\sin (2x)+1.\)

Atskleisti atsakymą

Let’s begin by comparing the equation to the general form \(y=A\sin (Bx-C)+D.\)
\(A=3,\) so the amplitude is \(|A|=3.\)
Next, \(B=2,\) so the period is \(P=\frac{2\pi }{|B|}=\frac{2\pi }{2}=\pi .\)
There is no added constant inside the parentheses, so \(C=0\) and the phase shift is \(\frac{C}{B}=\frac{0}{2}=0.\)
Finally, \(D=1,\) so the midline is \(y=1.\)
Determine the midline, amplitude, period, and phase shift of the function \(y=\frac{1}{2}\cos (\frac{x}{3}-\frac{\pi }{3}).\)

Atskleisti atsakymą

midline: \(y=0;\) amplitude: \(|A|=\frac{1}{2};\) period: \(P=\frac{2\pi }{|B|}=6\pi ;\) phase shift: \(\frac{C}{B}=\pi\)
Determine the formula for the cosine function in .

Atskleisti atsakymą

To determine the equation, we need to identify each value in the general form of a sinusoidal function.
\[\begin{array}{l}y=A\sin (Bx-C)+D \\ y=A\cos (Bx-C)+D\end{array}\]
The graph could represent either a sine or a cosine function that is shifted and/or reflected. When \(x=0,\) the graph has an extreme point, \((0,0).\) Since the cosine function has an extreme point for \(x=0,\) 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 \(y=0.5.\) This value, which is the midline, is \(D\) in the equation, so \(D=0.5.\)
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 \(|A|=0.5.\) 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 \(|A|=\frac{1}{2}=0.5.\) Also, the graph is reflected about the x-axis so that \(A=-0.5.\)
The graph is not horizontally stretched or compressed, so \(B=1;\) and the graph is not shifted horizontally, so \(C=0.\)
Putting this all together,
\[g(x)=-0.5\cos (x)+0.5\]
Determine the formula for the sine function in .

Atskleisti atsakymą

\(f(x)=\sin (x)+2\)
Determine the equation for the sinusoidal function in .
Atskleisti atsakymą
With the highest value at 1 and the lowest value at \(-5,\) the midline will be halfway between at \(-2.\) So \(D=-2.\)
The distance from the midline to the highest or lowest value gives an amplitude of \(|A|=3.\)
The period of the graph is 6, which can be measured from the peak at \(x=1\) to the next peak at \(x=7,\) or from the distance between the lowest points. Therefore, \(P=\frac{2\pi }{|B|}=6.\) Using the positive value for \(B,\) we find that
\[B=\frac{2\pi }{P}=\frac{2\pi }{6}=\frac{\pi }{3}\]
So far, our equation is either \(y=3\sin (\frac{\pi }{3}x-C)-2\) or \(y=3\cos (\frac{\pi }{3}x-C)-2.\) For the shape and shift, we have more than one option. We could write this as any one of the following:
- a cosine shifted to the right
- a negative cosine shifted to the left
- a sine shifted to the left
- a negative sine shifted to the right
Choosing to use the cosine function, we observe that the peak, which would normally be at \(x=0\), is at \(x=1\), and given the horizontal compression factor of \(\frac{\pi }{3}\), we get \(C=1\cdot \frac{\pi }{3}=\frac{\pi }{3}\).
While any of these would be correct, the cosine shifts are easier to work with than the sine shifts in this case because they involve integer values. So our function becomes
\[y=3\cos (\frac{\pi }{3}x-\frac{\pi }{3})-2\text{ or }y=-3\cos (\frac{\pi }{3}x+\frac{2\pi }{3})-2\]
Again, these functions are equivalent, so both yield the same graph.
Write a formula for the function graphed in .

Atskleisti atsakymą

two possibilities: \(y=4\sin (\frac{\pi }{5}x-\frac{\pi }{5})+4\) or \(y=-4\sin (\frac{\pi }{5}x+\frac{4\pi }{5})+4\)
Sketch a graph of \(f(x)=-2\sin (\frac{\pi x}{2}).\)
Atskleisti atsakymą
Let’s begin by comparing the equation to the form \(y=A\sin (Bx).\)
- Step 1. We can see from the equation that \(A=-2,\) so the amplitude is 2. \[|A|=2\]
- Step 2. The equation shows that \(B=\frac{\pi }{2},\) so the period is \[\begin{array}{l}P=\frac{2\pi }{\frac{\pi }{2}} \\ =2\pi ⋅\frac{2}{\pi } \\ =4\end{array}\]
- Step 3. Because \(A\) 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, \(x=0,\) the horizontal midpoints are at \(x=2\) and at the end of one period at \(x=4.\)
The quarter points include the minimum at \(x=1\) and the maximum at \(x=3.\) A local minimum will occur 2 units below the midline, at \(x=1,\) and a local maximum will occur at 2 units above the midline, at \(x=3.\) shows the graph of the function.
Sketch a graph of \(g(x)=-0.8\cos (2x).\) Determine the midline, amplitude, period, and phase shift.

Atskleisti atsakymą

midline: \(y=0;\) amplitude: \(|A|=0.8;\) period: \(P=\frac{2\pi }{|B|}=\pi ;\) phase shift: \(\frac{C}{B}=0\) or none
Sketch a graph of \(f(x)=3\sin (\frac{\pi }{4}x-\frac{\pi }{4}).\)
Atskleisti atsakymą
- Step 1. The function is already written in general form: \(f(x)=3\sin (\frac{\pi }{4}x-\frac{\pi }{4}).\) This graph will have the shape of a sine function, starting at the midline and increasing to the right.
- Step 2. \(|A|=|3|=3.\) The amplitude is 3.
- Step 3. Since \(|B|=|\frac{\pi }{4}|=\frac{\pi }{4},\) we determine the period as follows. \[P=\frac{2\pi }{|B|}=\frac{2\pi }{\frac{\pi }{4}}=2\pi ⋅\frac{4}{\pi }=8\]
  The period is 8.
- Step 4. Since \(C=\frac{\pi }{4},\) the phase shift is \[\frac{C}{B}=\frac{\frac{\pi }{4}}{\frac{\pi }{4}}=1.\]
  The phase shift is 1 unit.
- Step 5. shows the graph of the function.
Draw a graph of \(g(x)=-2\cos (\frac{\pi }{3}x+\frac{\pi }{6}).\) Determine the midline, amplitude, period, and phase shift.

Atskleisti atsakymą

midline: \(y=0;\) amplitude: \(|A|=2;\) period: \(P=\frac{2\pi }{|B|}=6;\) phase shift: \(\frac{C}{B}=-\frac{1}{2}\)
Given \(y=-2\cos (\frac{\pi }{2}x+\pi )+3,\) determine the amplitude, period, phase shift, and vertical shift. Then graph the function.
Atskleisti atsakymą
Begin by comparing the equation to the general form and use the steps outlined in .
\[y=A\cos (Bx-C)+D\]
- Step 1. The function is already written in general form.
- Step 2. Since \(A=-2,\) the amplitude is \(|A|=2.\)
- Step 3. \(|B|=\frac{\pi }{2},\) so the period is \(P=\frac{2\pi }{|B|}=\frac{2\pi }{\frac{\pi }{2}}=2\pi ⋅\frac{2}{\pi }=4.\) The period is 4.
- Step 4. \(C=-\pi ,\) so we calculate the phase shift as \(\frac{C}{B}=\frac{-\pi ,}{\frac{\pi }{2}}=-\pi ⋅\frac{2}{\pi }=-2.\) The phase shift is \(-2.\)
- Step 5. \(D=3,\) so the midline is \(y=3,\) and the vertical shift is up 3.
Since \(A\) is negative, the graph of the cosine function has been reflected about the x-axis.
shows one cycle of the graph of the function.
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.

Atskleisti atsakymą

Recall that, for a point on a circle of radius r, the y-coordinate of the point is \(y=r\ \sin (x),\) so in this case, we get the equation \(y(x)=3\ \sin (x).\) The constant 3 causes a vertical stretch of the y-values of the function by a factor of 3, which we can see in the graph in .
What is the amplitude of the function \(f(x)=7\cos (x)?\) Sketch a graph of this function.

Atskleisti atsakymą

7
A circle with radius 3 ft is mounted with its center 4 ft off the ground. The point closest to the ground is labeled P, as shown in . Sketch a graph of the height above the ground of the point \(P\) as the circle is rotated; then find a function that gives the height in terms of the angle of rotation.

Atskleisti atsakymą

Sketching the height, we note that it will start 1 ft above the ground, then increase up to 7 ft above the ground, and continue to oscillate 3 ft above and below the center value of 4 ft, as shown in .
Although we could use a transformation of either the sine or cosine function, we start by looking for characteristics that would make one function easier to use than the other. Let’s use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection.
Second, we see that the graph oscillates 3 above and below the center, while a basic cosine has an amplitude of 1, so this graph has been vertically stretched by 3, as in the last example.
Finally, to move the center of the circle up to a height of 4, the graph has been vertically shifted up by 4. Putting these transformations together, we find that
\[y=-3\cos (x)+4\]
A weight is attached to a spring that is then hung from a board, as shown in . As the spring oscillates up and down, the position \(y\) of the weight relative to the board ranges from \(-1\) in. (at time \(x=0)\) to \(-7\) in. (at time \(x=\pi )\) below the board. Assume the position of \(y\) is given as a sinusoidal function of \(x.\) Sketch a graph of the function, and then find a cosine function that gives the position \(y\) in terms of \(x.\)

Atskleisti atsakymą

\(y=3\cos (x)-4\)
The London Eye is a huge Ferris wheel with a diameter of 135 meters (443 feet). It completes one rotation every 30 minutes. Riders board from a platform 2 meters above the ground. Express a rider’s height above ground as a function of time in minutes.
Atskleisti atsakymą
With a diameter of 135 m, the wheel has a radius of 67.5 m. The height will oscillate with amplitude 67.5 m above and below the center.
Passengers board 2 m above ground level, so the center of the wheel must be located \(67.5+2=69.5\) m above ground level. The midline of the oscillation will be at 69.5 m.
The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with a period of 30 minutes.
Lastly, because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve.
- Amplitude: \(67.5,\) so \(A=67.5\)
- Midline: \(69.5,\) so \(D=69.5\)
- Period: \(30,\) so \(B=\frac{2\pi }{30}=\frac{\pi }{15}\)
- Shape: \(-cos(t)\)
An equation for the rider’s height would be
\[y=-67.5\cos (\frac{\pi }{15}t)+69.5\]
where \(t\) is in minutes and \(y\) is measured in meters.
Why are the sine and cosine functions called periodic functions?

Atskleisti atsakymą

The sine and cosine functions have the property that \(f(x+P)=f(x)\) for a certain \(P.\) This means that the function values repeat for every \(P\) units on the x-axis.
How does the graph of \(y=\sin \ x\) compare with the graph of \(y=\cos \ x?\) Explain how you could horizontally translate the graph of \(y=\sin \ x\) to obtain \(y=\cos \ x.\)
For the equation \(A\ \cos (Bx+C)+D,\) what constants affect the range of the function and how do they affect the range?

Atskleisti atsakymą

The absolute value of the constant \(A\) (amplitude) increases the total range and the constant \(D\) (vertical shift) shifts the graph vertically.
How does the range of a translated sine function relate to the equation \(y=A\ \sin (Bx+C)+D?\)
How can the unit circle be used to construct the graph of \(f(t)=\sin \ t?\)

Atskleisti atsakymą

At the point where the terminal side of \(t\) intersects the unit circle, you can determine that the \(\sin \ t\) equals the y-coordinate of the point.
\(f(x)=2\sin \ x\)
\(f(x)=\frac{2}{3}\cos \ x\)

Atskleisti atsakymą

amplitude: \(\frac{2}{3};\) period: \(2\pi ;\) midline: \(y=0;\) maximum: \(y=\frac{2}{3}\) occurs at \(x=2\pi ;\) minimum: \(y=-\frac{2}{3}\) occurs at \(x=\pi ;\) for one period, the graph starts at 0 and ends at \(2\pi\)
\(f(x)=-3\sin \ x\)
\(f(x)=4\sin \ x\)

Atskleisti atsakymą

amplitude: 4; period: \(2\pi ;\) midline: \(y=0;\) maximum \(y=4\) occurs at \(x=\frac{\pi }{2};\) minimum: \(y=-4\) occurs at \(x=\frac{3\pi }{2};\) one full period occurs from \(x=0\) to \(x=2\pi\)
\(f(x)=2\cos \ x\)
\(f(x)=\cos (2x)\)

Atskleisti atsakymą

amplitude: 1; period: \(\pi ;\) midline: \(y=0;\) maximum: \(y=1\) occurs at \(x=\pi ;\) minimum: \(y=-1\) occurs at \(x=\frac{\pi }{2};\) one full period is graphed from \(x=0\) to \(x=\pi\)
\(f(x)=2\ \sin (\frac{1}{2}x)\)
\(f(x)=4\ \cos (\pi x)\)

Atskleisti atsakymą

amplitude: 4; period: 2; midline: \(y=0;\) maximum: \(y=4\) occurs at \(x=2;\) minimum: \(y=-4\) occurs at \(x=1\)
\(f(x)=3\ \cos (\frac{6}{5}x)\)
\(y=3\ \sin (8(x+4))+5\)

Atskleisti atsakymą

amplitude: 3; period: \(\frac{\pi }{4};\) midline: \(y=5;\) maximum: \(y=8\) occurs at \(x=0.12;\) minimum: \(y=2\) occurs at \(x=0.516;\) horizontal shift: \(-4;\) vertical translation 5; one period occurs from \(x=0\) to \(x=\frac{\pi }{4}\)
\(y=2\ \sin (3x-21)+4\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Graphs of the Sine and Cosine Functions

Graph variations of
Use phase shifts of sine and cosine curves.
They are periodic functions with a period of
The domain of each function is
The graph of
The graph of
Determine the amplitude as
Determine the period as

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Pabandyk savo pačių

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Daugiau informacijos Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates