# 2.6: General Sinusoidal Graphs

**At Grade**Created by: CK-12

## Learning Objectives

- Given any sinusoid in the form: \begin{align*}y=D \pm A \cos(B(x \pm C))\end{align*}
y=D±Acos(B(x±C)) or \begin{align*}y=D \pm A \sin(B(x \pm C))\end{align*}y=D±Asin(B(x±C)) identify the transformations performed by \begin{align*}A\end{align*}A , \begin{align*}B\end{align*}B , \begin{align*}C\end{align*}, and \begin{align*}D\end{align*}. - Graph any sinusoid given an equation in the form \begin{align*}y=D \pm A \cos(B(x \pm C))\end{align*} or \begin{align*}y=D \pm A \sin(B(x \pm C))\end{align*}.
- Identify the equation of any sinusoid given a graph and critical values.

## The Generalized Equations

In the previous two sections, you learned how to translate and dilate sine and cosine waves both horizontally and vertically. Combining all the information learned, the general equations are: \begin{align*}y=D \pm A \cos(B(x \pm C))\end{align*} or \begin{align*}y=D \pm A \sin(B(x \pm C))\end{align*}, where \begin{align*}A\end{align*} is the amplitude, \begin{align*}B\end{align*} is the frequency, \begin{align*}C\end{align*} is the horizontal translation, and \begin{align*}D\end{align*} is the vertical translation.

Recall the relationship between period, \begin{align*}p\end{align*}, and frequency, \begin{align*}B\end{align*}.

\begin{align*}p=\frac{2\pi}{B} \ \text{and} \ B=\frac{2\pi}{p}\end{align*}

With this knowledge, we should be able to sketch any sine or cosine function as well as write an equation given its graph.

## Drawing Sketches/Identifying Transformations from the Equation

**Example 1:** Given the function: \begin{align*}f(x) = 1 + 2 \sin(2(x + \pi))\end{align*}

a. Identify the period, amplitude, and frequency.

b. Explain any vertical or horizontal translations present in the equation.

c. Sketch the graph from \begin{align*}-2\pi\end{align*} to \begin{align*}2\pi\end{align*}.

**Solution:** a. From the equation, the amplitude is 2 and the frequency is also 2. To find the period we use:

\begin{align*}p=\frac{2\pi}{B} \rightarrow p=\frac{2\pi}{2}=\pi\end{align*}

So, there are two complete waves from \begin{align*}[0, 2\pi]\end{align*} and each individual wave requires \begin{align*}\pi\end{align*} radians to complete.

b. \begin{align*}D = 1\end{align*} and \begin{align*}C = -\pi\end{align*}, so this graph has been translated 1 unit up, and \begin{align*}\pi\end{align*} units to the left.

c. To sketch the graph, start with the graph of \begin{align*}y = \sin(x)\end{align*}

Translate the graph \begin{align*}\pi\end{align*} units to the left (the \begin{align*}C\end{align*} value).

Next, move the graph 1 unit up (\begin{align*}D\end{align*} value)

Now we can add the dilations. Remember that the “starting point” of the wave is \begin{align*}-\pi\end{align*} because of the horizontal translation. A normal sine wave takes \begin{align*}2\pi\end{align*} units to complete a cycle, but this wave completes one cycle in \begin{align*}\pi\end{align*} units. The first wave will complete at 0, then we will see a second wave from 0 to \begin{align*}\pi\end{align*} and a third from \begin{align*}\pi\end{align*} to \begin{align*}2\pi\end{align*}. Start by placing points at these values:

Using symmetry, each interval needs to cross the line \begin{align*}y = 1\end{align*} through the center of the wave.

One sine wave contains a “mountain” and a “valley”. The mountain “peak” and the valley low point must occur halfway between the points above.

Extend the curve through the domain.

Finally, extend the minimum and maximum points to match the amplitude of 2.

**Example 2:** Given the function: \begin{align*}f(x)=3+3 \cos \left(\frac{1}{2}(x-\frac{\pi}{2}\right))\end{align*}

a. Identify the period, amplitude, and frequency.

b. Explain any vertical or horizontal translations present in the equation.

c. Sketch the graph from \begin{align*}-2\pi\end{align*} to \begin{align*}2\pi\end{align*}.

**Solution:** a. From the equation, the amplitude is 3 and the frequency is \begin{align*}\frac{1}{2}\end{align*}. To find the period we use:

\begin{align*}period=\frac{2\pi}{\frac{1}{2}}=4\pi\end{align*}

So, there is only one half of a cosine curve from 0 to \begin{align*}2\pi\end{align*} and each individual wave requires \begin{align*}4\pi\end{align*} radians to complete.

b. \begin{align*}D = 3\end{align*} and \begin{align*}C=\frac{\pi}{2}\end{align*}, so this graph has been translated 3 units up, and \begin{align*}\frac{\pi}{2}\end{align*} units to the right.

c. To sketch the graph, start with the graph of \begin{align*}y = \cos(x)\end{align*}

Adjust the amplitude so the cosine wave reaches up to 3 and down to negative three. This affects the maximum points, but the points on the \begin{align*}x-\end{align*}axis remain the same. These points are sometimes called **nodes.**

According to the period, we should see one of these shapes every \begin{align*}4\pi\end{align*} units. Because the interval specified is \begin{align*}[-2\pi, 2\pi]\end{align*} and the cosine curve “starts” at the \begin{align*}y-\end{align*}axis, at (0, 3) and at \begin{align*}2\pi\end{align*} the value is -3. Conversely, at \begin{align*}-2\pi\end{align*}, the function is also -3.

Now, shift the graph \begin{align*}\frac{\pi}{2}\end{align*} units to the right.

Finally, we need to adjust for the vertical shift by moving it up 3 units.

## Writing the Equation from a Sketch

In order to write the equation from a graph, you need to be provided with enough information to find the four constants.

**Example 3:** Find the equation of the sinusoid graphed here.

**Solution:** First of all, remember that either sine or cosine could be used to model these graphs. However, it is usually easier to use cosine because the horizontal shift is easier to locate in most cases. Therefore, the model that we will be using is \begin{align*}y=D \pm A \cos(B(x \pm C))\end{align*} .

First, if we think of the graph as a cosine function, it has a horizontal translation of zero. The maximum point is also the \begin{align*}y-\end{align*}intercept of the graph, so there is no need to shift the graph horizontally and therefore, \begin{align*}C = 0\end{align*}. The amplitude is the height from the center of the wave. If you can’t find the center of the wave by sight, you can calculate it. The center should be halfway between the highest and the lowest points, which is really the **average** of the maximum and minimum. This value will actually be the vertical shift, or \begin{align*}D\end{align*} value.

\begin{align*}D=center=\frac{60+-20}{2}=\frac{40}{2}=20\end{align*}

The amplitude is the height from the center line, or vertical shift, to either the minimum or the maximum. So, \begin{align*}A=60-20=40\end{align*}.

The last value to find is the frequency. In order to do so, we must first find the period. The period is the distance required for one complete wave. To find this value, look at the horizontal distance between two consecutive maximum points.

On our graph, from maximum to maximum is 3.

Therefore, the period is 3, so the frequency is \begin{align*}B=\frac{2\pi}{3}\end{align*}.

We have now calculated each of the four parameters necessary to write the equation. Replacing them in the equation gives:

\begin{align*}y=20+40 \cos \frac{2\pi}{3}x\end{align*}

If we had chosen to model this curve with a sine function instead, the amplitude, period and frequency, as well as the vertical shift would all be the same. The only difference would be the horizontal shift. The sine wave starts in the middle of an upward sloped section of the curve as shown by the red circle.

This point intersects with the vertical translation line and is a third of the distance back to -3. So, in this case, the sine wave has been translated 1 unit *to the left*. The equation using a sine function instead would have been: \begin{align*}y=20+40 \sin \left(\frac{2\pi}{3}(x+1) \right)\end{align*}

## Points to Consider

- When using either sine or cosine to model a graph, why is only the phase shift different?
- How would you write \begin{align*}y=\sin x\end{align*} in the form \begin{align*}y=D \pm A \sin(B(x \pm C))\end{align*}? What are \begin{align*}A\end{align*}, \begin{align*}B\end{align*}, \begin{align*}C\end{align*}, and \begin{align*}D\end{align*}?
- Is it possible to solve \begin{align*}y=D \pm A \sin(B(x \pm C))\end{align*} for \begin{align*}x\end{align*}?

## Review Questions

For problems 1-5, identify the amplitude, period, frequency, maximum and minimum points, vertical shift, and horizontal shift.

- \begin{align*}y=2+3 \sin(2(x-1))\end{align*}
- \begin{align*}y=-1+ \sin \left(\pi (x+\frac{\pi}{3}\right))\end{align*}
- \begin{align*}y=\cos (40(x-120))+5\end{align*}
- \begin{align*}y=-\cos \left(\frac{1}{2}(x+\frac{5\pi}{4}\right))\end{align*}
- \begin{align*}y=2 \cos(-x)+3\end{align*}

For problems 6-10, write the equation of each graph. Recall that cosine might be an easier model, but you may write your answer in terms of cosine or sine.

## Review Answers

- This is a sine wave that has been translated 1 unit to the right and 2 units up. The amplitude is 3 and the frequency is 2. The period of the graph is \begin{align*}\pi\end{align*}. The function reaches a maximum point of 5 and a minimum of -1.
- This is a sine wave that has been translated 1 unit down and \begin{align*}\frac{\pi}{3}\end{align*} radians to the left. The amplitude is 1 and the period is 2. The frequency of the graph is \begin{align*}\pi\end{align*}. The function reaches a maximum point of 0 and a minimum of -2.
- This is a cosine wave that has been translated 5 units up and 120 radians to the right. The amplitude is 1 and the frequency is 40. The period of the graph is \begin{align*}\frac{\pi}{20}\end{align*}. The function reaches a maximum point of 6 and a minimum of 4.
- This is a cosine wave that has not been translated vertically. It has been translated \begin{align*}\frac{5\pi}{4}\end{align*} radians to the left. The amplitude is 1 and the frequency is \begin{align*}\frac{1}{2}\end{align*}. The period of the graph is \begin{align*}4\pi\end{align*}. The function reaches a maximum point of 1 and a minimum of -1. The negative in front of the cosine function does not change the amplitude, it simply reflects the graph across the \begin{align*}x-\end{align*}axis.
- This is a cosine wave that has been translate up 3 units and has an amplitude of 2. The frequency is 1 and the period is \begin{align*}2\pi\end{align*}. There is no horizontal translation. Putting a negative in front of the \begin{align*}x-\end{align*}value reflects the function across the \begin{align*}y-\end{align*}axis. A cosine wave that has not been translated horizontally is symmetric to the \begin{align*}y-\end{align*}axis so this reflection will have no visible effect on the graph. The function reaches a maximum of 5 and a minimum of 1. \begin{align*}^{***} \text{other answers are possible given different horizontal translations of sine/cosine}\end{align*}
- \begin{align*}y=3+2 \cos \left(4(x-\frac{\pi}{6}\right))\end{align*}
- \begin{align*}y=2+ \sin x \ \text{or} \ y=2+ \cos \left(x-\frac{\pi}{2}\right)\end{align*}
- \begin{align*}y=10+20 \cos(6(x-30))\end{align*}
- \begin{align*}y=3+\frac{3}{4} \cos \left(\frac{1}{2}(x+\pi \right))\end{align*}
- \begin{align*}y=3+7 \cos \left(\frac{1}{3}(x-\frac{\pi}{4}\right))\end{align*}

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