<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 14.1: Graphing Trigonometric Functions

Difficulty Level: At Grade Created by: CK-12

Objective

To graph, translate, and reflect the sine, cosine, and tangent functions.

Review Queue

Find the exact values of the following expressions.

1. \begin{align*}\sin \frac{\pi}{4}\end{align*}

2. \begin{align*} \cos \frac{5 \pi}{6} \end{align*}

3. \begin{align*} \tan \frac{3 \pi}{4} \end{align*}

4. \begin{align*} \sin \frac{7 \pi}{3} \end{align*}

## Graphing Sine and Cosine

Objective

To graph and stretch the sine and cosine functions.

Guidance

In this concept, we will take the unit circle, introduced in the previous chapter, and graph it on the Cartesian plane.

To do this, we are going to “unravel” the unit circle. Recall that for the unit circle the coordinates are \begin{align*}(\cos \theta, \sin \theta)\end{align*} where \begin{align*}\theta\end{align*} is the central angle. To graph, \begin{align*}y=\sin x\end{align*} rewrite the coordinates as \begin{align*}(x, \sin x)\end{align*} where \begin{align*}x\end{align*} is the central angle, in radians. Below we expanded the sine coordinates for \begin{align*}\frac{3\pi}{4}\end{align*}.

Notice that the curve ranges from 1 to -1. The maximum value is 1, which is at \begin{align*}x=\frac{\pi}{2}\end{align*}. The minimum value is -1 at \begin{align*}x=\frac{3 \pi}{2}\end{align*}. This “height” of the sine function is called the amplitude. The amplitude is the absolute value of average between the highest and lowest points on the curve.

Now, look at the domain. It seems that, if we had continued the curve, it would repeat. This means that the sine curve is periodic. Look back at the unit circle, the sine value changes until it reaches \begin{align*}2 \pi\end{align*}. After \begin{align*}2 \pi\end{align*}, the sine values repeat. Therefore, the curve above will repeat every \begin{align*}2 \pi\end{align*} units, making the period \begin{align*}2 \pi\end{align*}. The domain is all real numbers.

Similarly, when we expand the cosine curve, \begin{align*}y=\cos x\end{align*}, from the unit circle, we have:

Notice that the range is also between 1 and -1 and the domain will be all real numbers. The cosine curve is also periodic, with a period of \begin{align*}2 \pi\end{align*}. If we draw the graph past \begin{align*}2\pi\end{align*}, it would look like:

Comparing \begin{align*}{\color{red}y=\sin x}\end{align*} and \begin{align*}{\color{blue}y=\cos x}\end{align*} (below), we see that the curves are almost identical, except that the sine curve starts at \begin{align*}y = 0\end{align*} and the cosine curve starts at \begin{align*}y = 1\end{align*}.

If we shift either curve \begin{align*}\frac{\pi}{2}\end{align*} units to the left or right, they will overlap. Any horizontal shift of a trigonometric function is called a phase shift. We will discuss phase shifts more in the upcoming concepts.

Example A

Identify the highlighted points on \begin{align*}y=\sin x\end{align*} and \begin{align*}y=\cos x\end{align*} below.

,

Solution: For each point, think about what the sine or cosine value is at those values. For point \begin{align*}A\end{align*}, \begin{align*}\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}\end{align*}, therefore the point is \begin{align*}\left(\frac{\pi}{4},\frac{\sqrt{2}}{2}\right)\end{align*}. For point \begin{align*}B\end{align*}, we have to work backwards because it is not exactly on a vertical line, but it is on a horizontal one. When is \begin{align*}\sin x=-\frac{1}{2}\end{align*}? When \begin{align*}x=\frac{7 \pi}{6}\end{align*} or \begin{align*}\frac{11 \pi}{6}\end{align*}. By looking at point \begin{align*}B\end{align*}’s location, we know it is the second option. Therefore, the point is \begin{align*}\left(\frac{11 \pi}{6},\frac{1}{2}\right)\end{align*}.

For the cosine curve, point \begin{align*}C\end{align*} is the same as point \begin{align*}A\end{align*} because the sine and cosine for \begin{align*}\frac{\pi}{4}\end{align*} is the same. As for point \begin{align*}D\end{align*}, we use the same logic as we did for point \begin{align*}B\end{align*}. When does \begin{align*}\cos x = -\frac{1}{2}\end{align*}? When \begin{align*}x=\frac{2 \pi}{3}\end{align*} or \begin{align*}\frac{4\pi}{3}\end{align*}. Again, looking at the location of point \begin{align*}D\end{align*}, we know it is the second option. The point is \begin{align*}\left(\frac{4\pi}{3},\frac{1}{2}\right)\end{align*}.

More Guidance

In addition to graphing \begin{align*}y=\sin x\end{align*} and \begin{align*}y=\cos x\end{align*}, we can stretch the graphs by placing a number in front of the sine or cosine, such as \begin{align*}y=a\sin x\end{align*} or \begin{align*}y=a\cos x\end{align*}. \begin{align*}|a|\end{align*} is the amplitude of the curve. In the next concept, we will shift the curves up, down, to the left and right.

Example B

Graph \begin{align*}y=3 \sin x\end{align*} over two periods.

Solution: Start with the basic sine curve. Recall that one period of the parent graph, \begin{align*}y=\sin x\end{align*}, is \begin{align*}2 \pi\end{align*}. Therefore, two periods will be \begin{align*}4 \pi \end{align*}. The 3 indicates that the range will now be from 3 to -3 and the curve will be stretched so that the maximum is 3 and the minimum is -3. The red curve is \begin{align*}y=3 \sin x\end{align*}.

Notice that the \begin{align*}x\end{align*}-intercepts are the same as the parent graph. Typically, when we graph a trigonometric function, we always show two full periods of the function to indicate that it does repeat.

Example C

Graph \begin{align*}y=\frac{1}{2}\cos x\end{align*} over two periods.

Solution: Now, the amplitude will be \begin{align*}\frac{1}{2}\end{align*} and the function will be “smooshed” rather than stretched.

Example D

Graph \begin{align*}y=-\sin x\end{align*} over two periods.

Solution: The last two examples dealt with changing \begin{align*}a\end{align*} and \begin{align*}a\end{align*} was positive. Now, \begin{align*}a\end{align*} is negative. Just like with other functions, when the leading coefficient is negative, the function is reflected over the \begin{align*}x\end{align*}-axis. \begin{align*}y=-\sin x\end{align*} is in red.

Guided Practice

1. Is the point \begin{align*}\left(\frac{5 \pi}{6},\frac{1}{2}\right)\end{align*} on \begin{align*}y=\sin x\end{align*}? How do you know?

Graph the following functions for two full periods.

2. \begin{align*}y=6 \cos x\end{align*}

3. \begin{align*}y=-3 \cos x\end{align*}

4. \begin{align*}y=\frac{3}{2} \sin x\end{align*}

1. Substitute in the point for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} and see if the equation holds true.

\begin{align*}\frac{1}{2}=\sin\left(\frac{5 \pi}{6}\right)\end{align*}

This is true, so \begin{align*}\left(\frac{5 \pi}{6},\frac{1}{2}\right)\end{align*} is on the graph.

2. Stretch the cosine curve so that the maximum is 6 and the minimum is -6.

3. The graph is reflected over the \begin{align*}x\end{align*}-axis and stretched so that the amplitude is 3.

4. The fraction is equivalent to 1.5, making 1.5 the amplitude.

Vocabulary

Trigonometric Function
When the sine, cosine, or tangent of an angle is plotted in the \begin{align*}x-y\end{align*} plane such that \begin{align*}(x,f(x))\end{align*}, where \begin{align*}x\end{align*} is the central angle from the unit circle and \begin{align*}f(x)\end{align*} is the sine, cosine, or tangent of that angle.
Amplitude
The height of a sine or cosine curve. In the equation, \begin{align*}y=a\sin x\end{align*} or \begin{align*}y = a\cos x\end{align*}, the amplitude is \begin{align*}|a|\end{align*}.
Periodic
When a function repeats its \begin{align*}y\end{align*}-values over a regular interval.
Period
The regular interval over which a periodic function repeats itself.
Phase Shift
The horizontal shift of a trigonometric function.

Problem Set

1. Determine the exact value of each point on \begin{align*}y=\sin x\end{align*} or \begin{align*}y=\cos x\end{align*}.
1. List all the points in the interval \begin{align*}[0,4 \pi]\end{align*} where \begin{align*}\sin x=\cos x\end{align*}. Use the graph from #1 to help you.
2. Draw from \begin{align*}y=\sin x\end{align*} from \begin{align*}[0, 2 \pi]\end{align*}. Find \begin{align*}f \left(\frac{\pi}{3}\right)\end{align*} and \begin{align*}f \left(\frac{5 \pi}{3}\right)\end{align*}. Plot these values on the curve.

For questions 4-12, graph the sine or cosine curve over two periods.

1. \begin{align*}y=2 \sin x\end{align*}
2. \begin{align*}y=-5 \cos x\end{align*}
3. \begin{align*}y=\frac{1}{4} \cos x\end{align*}
4. \begin{align*}y=- \frac{2}{3} \sin x\end{align*}
5. \begin{align*}y=4 \sin x\end{align*}
6. \begin{align*}y=-1.5 \cos x\end{align*}
7. \begin{align*}y=\frac{5}{3} \cos x\end{align*}
8. \begin{align*}y=10 \sin x\end{align*}
9. \begin{align*}y=-7.2 \sin x\end{align*}
10. Graph \begin{align*}y=\sin x\end{align*} and \begin{align*}y=\cos x\end{align*} on the same set of axes. How many units would you have to shift the sine curve (to the left or right) so that it perfectly overlaps the cosine curve?
11. Graph \begin{align*}y=\sin x\end{align*} and \begin{align*}y=-\cos x\end{align*} on the same set of axes. How many units would you have to shift the sine curve (to the left or right) so that it perfectly overlaps \begin{align*}y=- \cos x\end{align*}?

Write the equation for each sine or cosine curve below. \begin{align*}a>0\end{align*} for both questions.

## Translating Sine and Cosine Functions

Objective

To be able to graph a translated sine or cosine function.

Guidance

Just like other functions, sine and cosine curves can be moved to the left, right, up and down. The general equation for a sine and cosine curve is \begin{align*}y=a \sin (x-h)+k\end{align*} and \begin{align*}y=a \cos(x-h)+k\end{align*}, respectively. Also, just like in other functions, \begin{align*}h\end{align*} is the horizontal shift, also called a phase shift, and \begin{align*}k\end{align*} is the vertical shift. Notice, that because it is \begin{align*}x - h\end{align*} in the equation, \begin{align*}h\end{align*} will always shift in the opposite direction of what is in the equation.

Example A

Graph \begin{align*}y=\cos \left(x - \frac{\pi}{4}\right)\end{align*}.

Solution: This function will be shifted \begin{align*}\frac{\pi}{4}\end{align*} units to the right. The easiest way to sketch the curve, is to start with the parent graph and then move it to the right the correct number of units.

Example B

Graph \begin{align*}y=\sin (x+2)+3\end{align*}.

Solution: Because -2 is not written in terms of \begin{align*}\pi\end{align*} (like the \begin{align*}x\end{align*}-axis), we need to estimate where it would be on the axis. \begin{align*}- \frac{3 \pi}{4}=2.35 \ldots\end{align*} So, -2 will be shifted not quite to the \begin{align*}- \frac{3 \pi}{4}\end{align*} tic mark. Then, the entire function will be shifted up 3 units. The red graph is the final answer.

Example C

Find the equation of the sine curve below.

Solution: First, we know the amplitude is 1 because the average between 2 and 0 (the maximum and minimum) is 1. Next, we can find the vertical shift. Recall that the maximum is usually 1, in this equation it is 2. That means that the function is shifted up 1 unit \begin{align*}(2-1)\end{align*}. The horizontal shift is the hardest to find. Because sine curves are periodic, the horizontal shift can either be positive or negative.

Because \begin{align*}\pi\end{align*} is \begin{align*}3.14 \ldots\end{align*}, we can say that “moves back almost \begin{align*}\pi\end{align*} units” is -3 units. So, the equation is \begin{align*}y=\sin (x+3)+1\end{align*}. If we did the positive horizontal shift, we could say that the equation would be \begin{align*}y=\sin (x-3.28)+1\end{align*}.

To determine the value of the horizontal shift, you might have to estimate. For example, we estimated that the negative shift was -3 because the maximum value of the parent graph is at \begin{align*}x=\frac{\pi}{2}\end{align*} and the maximum to the left of it didn’t quite make it to \begin{align*}x=- \frac{\pi}{2}\end{align*} (the distance between \begin{align*}\frac{\pi}{2}\end{align*} and \begin{align*}-\frac{\pi}{2}\end{align*} is \begin{align*}\pi\end{align*}). Then, to determine the positive shift equation, recall that a period is \begin{align*}2 \pi\end{align*}, which is \begin{align*}6.28 \ldots\end{align*} So, the positive shift would be \begin{align*}2 \pi - 3\end{align*} or \begin{align*}6.28 - 3 = 3.28.\end{align*}.

Guided Practice

Graph the following functions from \begin{align*}[\pi, 3 \pi]\end{align*}.

1. \begin{align*}y=-1+\sin x\end{align*}

2. \begin{align*}y=\cos\left(x+\frac{\pi}{3}\right) - 2\end{align*}

3. Find the equation of the cosine curve below.

1. Shift the parent graph down one unit.

2. Shift the parent graph to the left \begin{align*}\frac{\pi}{3}\end{align*} units and down 2 units.

3. The parent graph is in green. It moves up 3 units and to the right \begin{align*}\frac{3 \pi}{4}\end{align*} units. Therefore, the equation is \begin{align*}y=\cos \left(x-\frac{3 \pi}{4}\right)+3\end{align*}.

If you moved the cosine curve backward, then the equation would be \begin{align*}y=\cos \left(x+\frac{5 \pi}{4}\right)+3\end{align*}.

Vocabulary

Phase Shift
The horizontal shift of a trigonometric function.

Problem Set

For questions 1-4, match the equation with its graph.

1. \begin{align*}y=\sin \left(x-\frac{\pi}{2}\right)\end{align*}
2. \begin{align*}y=\cos \left(x-\frac{\pi}{4}\right)+3\end{align*}
3. \begin{align*}y=\cos \left(x+\frac{\pi}{4}\right)-2\end{align*}
4. \begin{align*}y=\sin \left(x-\frac{\pi}{4}\right)+2\end{align*}
5. Which graph above represents:
1. \begin{align*}y=\cos(x-\pi)\end{align*}
2. \begin{align*}y=\sin \left(x+\frac{3 \pi}{4}\right)-2\end{align*}
6. Fill in the blanks below.
1. \begin{align*}\sin x=\cos(x-\underline{\;\;\;\;\;})\end{align*}
2. \begin{align*}\cos x=\sin(x-\underline{\;\;\;\;\;})\end{align*}

For questions 7-12, graph the following equations from \begin{align*}[-2\pi, 2\pi]\end{align*}.

1. \begin{align*}y=\sin \left(x+\frac{\pi}{4}\right)\end{align*}
2. \begin{align*}y=1+\cos x\end{align*}
3. \begin{align*}y=\cos(x+\pi)-2\end{align*}
4. \begin{align*}y=\sin(x+3)-4\end{align*}
5. \begin{align*}y=\sin \left(x-\frac{\pi}{6}\right)\end{align*}
6. \begin{align*}y=\cos(x-1)-3\end{align*}
7. Critical Thinking Is there a difference between \begin{align*}y=\sin x +1\end{align*} and \begin{align*}y=\sin(x+1)\end{align*}? Explain your answer.

## Putting it all Together

Objective

To graph sine and cosine functions where the amplitude is changed and horizontal and vertical shifts.

Guidance

In this concept, we will combine the previous two concepts and change the amplitude, the horizontal shifts, vertical shifts, and reflections.

Example A

Graph \begin{align*}y=4 \sin \left(x-\frac{\pi}{4}\right)\end{align*}. Find the domain and range.

Solution: First, stretch the curve so that the amplitude is 4, making the maximums and minimums 4 and -4. Then, shift the curve \begin{align*}\frac{\pi}{4}\end{align*} units to the right.

As for the domain, it is all real numbers because the sine curve is periodic and infinite. The range will be from the maximum to the minimum; \begin{align*}y\in [-4,4]\end{align*}.

Example B

Graph \begin{align*}y=-2\cos (x-1)+1\end{align*}. Find the domain and range.

Solution: The -2 indicates the cosine curve is flipped and stretched so that the amplitude is 2. Then, move the curve up one unit and to the right one unit.

The domain is all real numbers and the range is \begin{align*}y\in [-1,3]\end{align*}.

Example C

Find the equation of the sine curve to the right.

Solution: First, let’s find the amplitude. The range is from 1 to -5, which is a total distance of 6. Divided by 2, we find that the amplitude is 3. Halfway between 1 and -5 is \begin{align*}\frac{1+(-5)}{5}=-2\end{align*}, so that is our vertical shift. Lastly, we need to find the horizontal shift. The easiest way to do this is to superimpose the curve \begin{align*}y=3\sin (x)-2\end{align*} over this curve and determine the movement from one maximum to the closest maximum of this curve.

Subtracting \begin{align*}\frac{\pi}{2}\end{align*} and \begin{align*}\frac{\pi}{6}\end{align*}, we have:

\begin{align*}\frac{\pi}{2}-\frac{\pi}{6}=\frac{3 \pi}{6}-\frac{\pi}{6}=\frac{2 \pi}{6}=\frac{\pi}{3}\end{align*}

Making the equation \begin{align*}y=3 \sin \left(x+\frac{\pi}{3}\right)-2\end{align*}.

Guided Practice

Graph the following functions. State the domain and range. Show two full periods.

1. \begin{align*}y=-2\sin \left(x-\frac{\pi}{2}\right)\end{align*}

2. \begin{align*}y=\frac{1}{3}\cos (x+1)-2\end{align*}

3. Write one sine equation and one cosine equation for the curve below.

1. The domain is all real numbers and the range is \begin{align*}y\in [-2,2]\end{align*}.

2. The domain is all real numbers and the range is \begin{align*}y \in \left [-2\frac{1}{3},-1\frac{2}{3} \right ]\end{align*}.

3. The amplitude and vertical shift is the same, whether the equation is a sine or cosine curve. The vertical shift is -2 because that is the number that is halfway between the maximum and minimum. The difference between the maximum and minimum is 1, so the amplitude is half of that, or \begin{align*}\frac{1}{2}\end{align*}. As a sine curve, the function is \begin{align*}y=-2+\frac{1}{2}\sin x\end{align*}. As a cosine curve, there will be a shift of \begin{align*}\frac{\pi}{2}, y=\frac{1}{2} \cos \left(x-\frac{\pi}{2}\right)-2\end{align*}.

Problem Set

Determine if the following statements are true or false.

1. To change a cosine curve into a sine curve, shift the curve \begin{align*}\frac{\pi}{2}\end{align*} units.
2. For any given sine or cosine graph, there are infinitely many possible equations that can be written to represent the curve.
3. The amplitude is the same as the maximum value of the sine or cosine curve.
4. The horizontal shift is always in terms of \begin{align*}\pi\end{align*}.
5. The domain of any sine or cosine function is always all real numbers.

Graph the following sine or cosine functions such that \begin{align*}x \in [-2 \pi, 2 \pi]\end{align*}. State the domain and range.

1. \begin{align*}y=\sin \left(x+\frac{\pi}{4}\right)+1\end{align*}
2. \begin{align*}y=2-3\cos x\end{align*}
3. \begin{align*}y=\frac{3}{4} \sin \left(x-\frac{2 \pi}{3}\right)\end{align*}
4. \begin{align*}y=-5 \sin(x-3)-2\end{align*}
5. \begin{align*}y=2 \cos \left(x+\frac{5 \pi}{6}\right)-1.5\end{align*}
6. \begin{align*}y=-2.8 \cos(x-8)+4\end{align*}

Use the graph below to answer questions 12-15.

1. Write a sine equation for the function where the amplitude is positive.
2. Write a cosine equation for the function where the amplitude is positive.
3. How often does a sine or cosine curve repeat itself? How can you use this to help you write different equations for the same graph?
4. Write a second sine and cosine equation with different horizontal shifts.

Use the graph below to answer questions 16-20.

1. Write a sine equation for the function where the amplitude is positive.
2. Write a cosine equation for the function where the amplitude is positive.
3. Write a sine equation for the function where the amplitude is negative.
4. Write a cosine equation for the function where the amplitude is negative.
5. Describe the similarities and differences between the four equations from questions 16-19.

## Changes in the Period of a Sine and Cosine Function

Objective

Here you’ll learn how to change the period of a sine and cosine function.

Guidance

The last thing that we can manipulate on the sine and cosine curve is the period

The normal period of a sine or cosine curve is \begin{align*}2 \pi\end{align*}. To stretch out the curve, then the period would have to be longer than \begin{align*}2 \pi\end{align*}. Below we have sine curves with a period of \begin{align*}4 \pi\end{align*} and then the second has a period of \begin{align*}\pi\end{align*}.

To determine the period from an equation, we introduce \begin{align*}b\end{align*} into the general equation. So, the equations are \begin{align*}y=a\sin b(x-h)+k\end{align*} and \begin{align*}y=a\cos b(x-h)+k\end{align*}, where \begin{align*}a\end{align*} is the amplitude, \begin{align*}b\end{align*} is the frequency, \begin{align*}h\end{align*} is the phase shift, and \begin{align*}k\end{align*} is the vertical shift. The frequency is the number of times the sine or cosine curve repeats within \begin{align*}2 \pi\end{align*}. Therefore, the frequency and the period are indirectly related. For the first sine curve, there is half of a sine curve in \begin{align*}2 \pi\end{align*}. Therefore the equation would be \begin{align*}y=\sin \frac{1}{2}x\end{align*}. The second sine curve has two curves within \begin{align*}2 \pi\end{align*}, making the equation \begin{align*}y=\sin 2x\end{align*}. To find the period of any sine or cosine function, use \begin{align*}\frac{2 \pi}{|b|}\end{align*}, where \begin{align*}b\end{align*} is the frequency. Using the first graph above, this is a valid formula: \begin{align*}\frac{2 \pi}{\frac{1}{2}}=2 \pi \cdot 2=4 \pi\end{align*}.

Example A

Determine the period of the following sine and cosine functions.

a) \begin{align*}y=-3 \cos 6x\end{align*}

b) \begin{align*}y=2 \sin \frac{1}{4}x\end{align*}

c) \begin{align*}y=\sin \pi x -7\end{align*}

Solution: a) The 6 in the equation tells us that there are 6 repetitions within \begin{align*}2 \pi\end{align*}. So, the period is \begin{align*}\frac{2 \pi}{6}=\frac{\pi}{3}\end{align*}.

b) The \begin{align*}\frac{1}{4}\end{align*} in the equation tells us the frequency. The period is \begin{align*}\frac{2 \pi}{\frac{1}{4}}=2 \pi \cdot 4=8 \pi\end{align*}.

c) The \begin{align*}\pi\end{align*} is the frequency. The period is \begin{align*}\frac{2 \pi}{\pi}=2\end{align*}.

Example B

Graph part a) from the previous example from \begin{align*}[0, 2 \pi]\end{align*}. Determine where the maximum and minimum values occur. Then, state the domain and range.

Solution: The amplitude is -3, so it will be stretched and flipped. The period is \begin{align*}\frac{\pi}{3}\end{align*} (from above) and the curve should repeat itself 6 times from 0 to \begin{align*}2 \pi\end{align*}. The first maximum value is 3 and occurs at half the period, or \begin{align*}x=\frac{\pi}{6}\end{align*} and then repeats at \begin{align*}x=\frac{\pi}{2}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \ldots\end{align*} Writing this as a formula we start at \begin{align*}\frac{\pi}{6}\end{align*} and add \begin{align*}\frac{\pi}{3}\end{align*} to get the next maximum, so each point would be \begin{align*}\left(\frac{\pi}{6} \pm \frac{\pi}{3}n,3\right)\end{align*} where \begin{align*}n\end{align*} is any integer.

The minimums occur at -3 and the \begin{align*}x\end{align*}-values are multiples of \begin{align*}\frac{\pi}{3}\end{align*}. The points would be \begin{align*}\left(\pm \frac{\pi}{3}n, -3\right)\end{align*}, again \begin{align*}n\end{align*} is any integer. The domain is all real numbers and the range is \begin{align*}y \in [-3,3]\end{align*}.

Example C

Find all the solutions from the function in Example B from \begin{align*}[0, 2 \pi]\end{align*}.

Solution: Before this concept, the zeros didn’t change in the frequency because we hadn’t changed the period. Now that the period can be different, we can have a different number of zeros within \begin{align*}[0, 2\pi]\end{align*}. In this case, we will have 6 times the number of zeros that the parent function. To solve this function, set \begin{align*}y = 0\end{align*} and solve for \begin{align*}x\end{align*}.

\begin{align*}0 &=-3 \cos 6x \\ 0 &=\cos 6x\end{align*}

Now, use the inverse cosine function to determine when the cosine is zero. This occurs at the multiples of \begin{align*}\frac{\pi}{2}\end{align*}.

\begin{align*}6x=\cos^{-1}0=\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2},\frac{9\pi}{2}, \frac{11\pi}{2}, \frac{13\pi}{2}, \frac{15 \pi}{2}, \frac{17\pi}{2}, \frac{19\pi}{2}, \frac{21\pi}{2}, \frac{23\pi}{2}\end{align*}

We went much past \begin{align*}2 \pi\end{align*} because when we divide by 6, to get \begin{align*}x\end{align*} by itself, all of these answers are going to also be divided by 6 and smaller.

\begin{align*}x=\frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{5\pi}{4}, \frac{17 \pi}{12}, \frac{19\pi}{12}, \frac{21\pi}{2}, \frac{23\pi}{12}\end{align*}

\begin{align*}\frac{23 \pi}{12}<2\pi\end{align*} so we have found all the zeros in the range.

Guided Practice

1. Determine the period of the function \begin{align*}y=\frac{2}{3}\cos\frac{3}{4}x\end{align*}.

2. Find the zeros of the function from #1 from \begin{align*}[0, 2\pi]\end{align*}.

3. Determine the equation of the sine function with an amplitude of -3 and a period of \begin{align*}8\pi\end{align*}.

1. The period is \begin{align*}\frac{2 \pi}{\frac{3}{4}}=2 \pi \cdot \frac{4}{3}=\frac{8 \pi}{3}\end{align*}.

2. The zeros would be when \begin{align*}y\end{align*} is zero.

\begin{align*}0 &=\frac{2}{3} \cos \frac{3}{4}x \\ 0 &=\cos \frac{3}{4}x \\ \frac{3}{4}x &=\cos^{-1}0=\frac{\pi}{2},\frac{3 \pi}{2} \\ x &=\frac{4}{3}\left(\frac{\pi}{2},\frac{3 \pi}{2}\right) \\ x &=\frac{2\pi}{3},2\pi\end{align*}

3. The general equation of a sine curve is \begin{align*}y=a\sin bx\end{align*}. We know that \begin{align*}a = -3\end{align*} and that the period is \begin{align*}8 \pi\end{align*}. Let’s use this to find the frequency, or \begin{align*}b\end{align*}.

\begin{align*}\frac{2\pi}{b} &=8\pi \\ \frac{2\pi}{8\pi} &=b \\ \frac{1}{4} &=b\end{align*}

The equation of the curve is \begin{align*}y=-3\sin \frac{1}{4}x\end{align*}.

Vocabulary

Period
The length in which an entire sine or cosine curve is completed.
Frequency
The number of times a curve is repeated within \begin{align*}2\pi\end{align*}.

Problem Set

Find the period of the following sine and cosine functions.

1. \begin{align*}y=5\sin 3x\end{align*}
2. \begin{align*}y=-2\cos 4x\end{align*}
3. \begin{align*}y=-3\sin 2x\end{align*}
4. \begin{align*}y=\cos \frac{3}{4}x\end{align*}
5. \begin{align*}y=\frac{1}{2}\cos 2.5x\end{align*}
6. \begin{align*}y=4\sin 3x\end{align*}

Use the equation \begin{align*}y=5\sin 3x\end{align*} to answer the following questions.

1. Graph the function from \begin{align*}[0, 2\pi]\end{align*} and find the domain and range.
2. Determine the coordinates of the maximum and minimum values.
3. Find all the zeros from \begin{align*}[0, 2\pi]\end{align*}.

Use the equation \begin{align*}y=\cos \frac{3}{4}x\end{align*} to answer the following questions.

1. Graph the function from \begin{align*}[0, 4\pi]\end{align*} and find the domain and range.
2. Determine the coordinates of the maximum and minimum values.
3. Find all the zeros from \begin{align*}[0, 2\pi]\end{align*}.

Use the equation \begin{align*}y=-3\sin 2x\end{align*} to answer the following questions.

1. Graph the function from \begin{align*}[0, 2\pi]\end{align*} and find the domain and range.
2. Determine the coordinates of the maximum and minimum values.
3. Find all the zeros from \begin{align*}[0, 2\pi]\end{align*}.
4. What is the domain of every sine and cosine function? Can you make a general rule for the range? If so, state it.

Write the equation of the sine function, in the form \begin{align*}y=a\sin bx\end{align*}, with the given amplitude and period.

1. Amplitude: -2 Period: \begin{align*}\frac{3 \pi}{4}\end{align*}
2. Amplitude: \begin{align*}\frac{3}{5}\end{align*} Period: \begin{align*}5 \pi\end{align*}
3. Amplitude: 9 Period: 6
4. Challenge Find all the zeros from \begin{align*}[0, 2\pi]\end{align*} of \begin{align*}y=\frac{1}{2}\sin 3\left(x-\frac{\pi}{3}\right)\end{align*}.

## Graphing Tangent

Objective

Here you’ll learn how to graph a tangent function.

Guidance

The graph of the tangent function is very different from the sine and cosine functions. First, recall that the tangent ratio is \begin{align*}\tan \theta=\frac{\text{opposite}}{\text{hypotenuse}}\end{align*}. In radians, the coordinate for the tangent function would be \begin{align*}(\theta, \tan \theta)\end{align*}

\begin{align*}x && \theta && 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 \\ y && \tan \theta && 0 && \frac{\sqrt{3}}{3} && 1 && \sqrt{3} && \text{und.} && -\sqrt{3} && -1 && -\frac{\sqrt{3}}{3} && 0\end{align*}

After \begin{align*}\pi\end{align*}, the \begin{align*}y\end{align*}-values repeat, making the tangent function periodic with a period of \begin{align*}\pi\end{align*}.

The red portion of the graph represents the coordinates in the table above. Repeating this portion, we get the entire tangent graph. Notice that there are vertical asymptotes at \begin{align*}x=-\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}\end{align*} and \begin{align*}\frac{3\pi}{2}\end{align*}. If we were to extend the graph out in either direction, there would continue to be vertical asymptotes at the odd multiples of \begin{align*}\frac{\pi}{2}\end{align*}. Therefore, the domain is all real numbers, \begin{align*}x\ne n\pi \pm\frac{\pi}{2}\end{align*}, where \begin{align*}n\end{align*} is an integer. The range would be all real numbers. Just like with sine and cosine functions, you can change the amplitude, phase shift, and vertical shift.

The standard form of the equation is \begin{align*}y=a\tan b(x-h)+k\end{align*} where \begin{align*}a, b, h,\end{align*} and \begin{align*}k\end{align*} are the same as they are for the other trigonometric functions. For simplicity, we will not address phase shifts \begin{align*}(k)\end{align*} in this concept.

Example A

Graph \begin{align*}y=3 \tan x+1\end{align*} from \begin{align*}[-2\pi, 2\pi]\end{align*}. State the domain and range.

Solution: First, the amplitude is 3, which means each \begin{align*}y\end{align*}-value will be tripled. Then, we will shift the function up one unit.

Notice that the vertical asymptotes did not change. The period of this function is still \begin{align*}\pi\end{align*}. Therefore, if we were to change the period of a tangent function, we would use a different formula than what we used for sine and cosine. To change the period of a tangent function, use the formula \begin{align*}\frac{\pi}{|b|}\end{align*}.

The domain will be all real numbers, except where the asymptotes occur. Therefore, the domain of this function will be \begin{align*}x\in \mathbb{R}, x \notin n\pi \pm \frac{\pi}{2}\end{align*}. The range is all real numbers.

Example B

Graph \begin{align*}y=-\tan 2\pi\end{align*} from \begin{align*}[0, 2\pi]\end{align*} and state the domain and range. Find all zeros within this domain.

Solution: The period of this tangent function will be \begin{align*}\frac{\pi}{2}\end{align*} and the curves will be reflected over the \begin{align*}x\end{align*}-axis. The domain is all real numbers, \begin{align*}x \notin \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{\pi}{4}\pm \frac{\pi}{2}n\end{align*} where \begin{align*}n\end{align*} is any integer. The range is all real numbers. To find the zeros, set \begin{align*}y = 0\end{align*}.

\begin{align*}0 &=-\tan 2x \\ 0 &=\tan 2x \\ 2x &=\tan^{-1}0=0, \pi, 2\pi, 3\pi, 4\pi \\ x &=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\end{align*}

Example C

Graph \begin{align*}y=\frac{1}{4}\tan\frac{1}{4}x\end{align*} from \begin{align*}[0, 4\pi]\end{align*} and state the domain and range.

Solution: This function has a period of \begin{align*}\frac{\pi}{\frac{1}{4}}=4\pi\end{align*}. The domain is all real numbers, except \begin{align*}2\pi, 6\pi, 10\pi, 2\pi \pm 4\pi n\end{align*}, where \begin{align*}n\end{align*} is any integer. The range is all real numbers.

Guided Practice

1. Find the period of the function \begin{align*}y=-4\tan \frac{3}{2}x\end{align*}.

2. Find the zeros of the function from #1, from \begin{align*}[0, 2\pi]\end{align*}.

3. Find the equation of the tangent function with an amplitude of 8 and a period of \begin{align*}6\pi\end{align*}.

1. The period is \begin{align*}\frac{\pi}{\frac{3}{2}}=\pi \cdot \frac{2}{3}=\frac{2\pi}{3}\end{align*}.

2. The zeros are where \begin{align*}y\end{align*} is zero. \begin{align*}0 &=-4\tan \frac{3}{2}x \\ 0 &=\tan \frac{3}{2}x \\ \frac{3}{2}x &=\tan ^{-1}0=0, \pi, 2\pi, 3\pi \\ x &=\frac{2}{3}(0, \pi, 2\pi, 3\pi ) \\ x &=0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi\end{align*}

3. The general equation is \begin{align*}y=a\tan bx\end{align*}. We know that \begin{align*}a = 8\end{align*}. Let’s use the period to solve for the frequency, or \begin{align*}b\end{align*}.

\begin{align*}\frac{\pi}{b} &=6\pi \\ b &=\frac{\pi}{6\pi}=\frac{1}{6}\end{align*}

The equation is \begin{align*}y=8\tan \frac{1}{6}x\end{align*}.

Vocabulary

Tangent Function
Defined by the coordinates \begin{align*}(\theta, \tan \theta)\end{align*}, where \begin{align*}\theta\end{align*} is the central angle from the unit circle and tangent is the ratio of the sine and cosine functions.

Problem Set

Graph the following tangent functions over \begin{align*}[0, 4\pi]\end{align*}. Determine the period, domain, and range.

1. \begin{align*}y=2\tan x\end{align*}
2. \begin{align*}y=-\frac{1}{3}\tan x\end{align*}
3. \begin{align*}y=-\tan 3x\end{align*}
4. \begin{align*}y=4\tan 2x\end{align*}
5. \begin{align*}y=\frac{1}{2}\tan 4x\end{align*}
6. \begin{align*}y=-\tan \frac{1}{2}x\end{align*}
7. \begin{align*}y=4+\tan x\end{align*}
8. \begin{align*}y=-3+\tan 3x\end{align*}
9. \begin{align*}y=1+\frac{2}{3}\tan \frac{1}{2}x\end{align*}
10. Find the zeros of the function from #1.
11. Find the zeros of the function from #3.
12. Find the zeros of the function from #5.

Write the equation of the tangent function, in the form \begin{align*}y=a\tan bx\end{align*}, with the given amplitude and period.

1. Amplitude: 3 Period: \begin{align*}\frac{3\pi}{2}\end{align*}
2. Amplitude: \begin{align*}\frac{1}{4}\end{align*} Period: \begin{align*}2\pi\end{align*}
3. Amplitude: -2.5 Period: 8
4. Challenge Graph \begin{align*}y=2\tan \frac{1}{3}\left(x+\frac{\pi}{4}\right)-1\end{align*} over \begin{align*}[0, 6\pi]\end{align*}. Determine the domain and period.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: