While working on an assignment about sound in your science class, your Instructor informs you that what you know as the "pitch" of a sound is, in fact, the frequency of the sound waves. He then plays a note on a musical instrument, and the pattern of the sound wave on a graph looks like this:

He then tells you to find the frequency of the sound wave from the graph? Can you do it?

### Period and Frequency

The **period** of a trigonometric function is the horizontal distance traversed before the \begin{align*}y-\end{align*}values begin to repeat. For both graphs, \begin{align*}y = \sin x\end{align*} and \begin{align*}y = \cos x\end{align*}, the period is \begin{align*}2\pi .\end{align*} As you may remember, after completing one rotation of the unit circle, these values are the same.

**Frequency** is a measurement that is closely related to period. In science, the frequency of a sound or light wave is the number of complete waves for a given time period (like seconds). In trigonometry, because all of these periodic functions are based on the unit circle, we usually measure frequency as the number of complete waves every \begin{align*}2\pi\end{align*} units. Because \begin{align*}y = \sin x\end{align*} and \begin{align*}y = \cos x\end{align*} cover exactly one complete wave over this interval, their frequency is 1.

Period and frequency are inversely related. That is, the higher the frequency (more waves over \begin{align*}2\pi\end{align*} units), the lower the period (shorter distance on the \begin{align*}x-\end{align*}axis for each complete cycle).

After observing the transformations that result from multiplying a number *in front of* the sinusoid, it seems natural to look at what happens if we multiply a constant *inside* the argument of the function, or in other words, by the \begin{align*}x\end{align*} value. In general, the equation would be \begin{align*}y=\sin Bx\end{align*} or \begin{align*}y=\cos Bx\end{align*}. For example, look at the graphs of \begin{align*}y = \cos 2x\end{align*} and \begin{align*}y = \cos x\end{align*}.

Notice that the number of waves for \begin{align*}y = \cos 2x\end{align*} has increased, in the same interval as \begin{align*}y = \cos x\end{align*}. There are now **2 waves** over the interval from 0 to \begin{align*}2\pi\end{align*}. Consider that you are doubling each of the \begin{align*}x\end{align*} values because the function is \begin{align*}2x\end{align*}. When \begin{align*}\pi\end{align*} is plugged in, for example, the function becomes \begin{align*}2\pi\end{align*}. So the portion of the graph that normally corresponds to \begin{align*}2\pi\end{align*} units on the \begin{align*}x-\end{align*}axis, now corresponds to *half* that distance—so the graph has been “scrunched” horizontally. The frequency of this graph is therefore 2, or the same as the constant we multiplied by in the argument. The period (the length for each complete wave) is \begin{align*}\pi\end{align*}.

#### Finding the Period and Frequency

1. What is the frequency and period of \begin{align*}y = \sin 3x\end{align*}?

If we follow the pattern from the previous example, multiplying the angle by 3 should result in the sine wave completing a cycle **three times** as often as \begin{align*}y = \sin x\end{align*}. So, there will be three complete waves if we graph it from 0 to \begin{align*}2\pi\end{align*}. The frequency is therefore 3. Similarly, if there are 3 complete waves in \begin{align*}2\pi\end{align*} units, one wave will be a third of that distance, or \begin{align*}\frac{2\pi}{3}\end{align*} radians. Here is the graph:

This number that is multiplied by \begin{align*}x\end{align*}, called \begin{align*}B\end{align*}, will create a horizontal dilation. The larger the value of \begin{align*}B\end{align*}, the more compressed the waves will be horizontally. To stretch out the graph horizontally, we would need to *decrease* the frequency, or multiply by a number that is less than 1. Remember that this dilation factor is *inversely* related to the period of the graph.

Adding, one last time to our equations from before, we now have: \begin{align*}y=D \pm A \sin (B(x \pm C))\end{align*} or \begin{align*}y=D \pm A \cos (B(x \pm C))\end{align*}, where \begin{align*}B\end{align*} is the frequency, the period is equal to \begin{align*}\frac{2\pi}{B}\end{align*}, and everything else is as defined before.

2. What is the frequency and period of \begin{align*}y=\cos \frac{1}{4}x\end{align*}?

Using the generalization above, the frequency must be \begin{align*}\frac{1}{4}\end{align*} and therefore the period is \begin{align*}\frac{\frac{2\pi}{1}}{\frac{1}{4}}\end{align*}, which simplifies to: \begin{align*}\frac{2\pi}{\frac{1}{4}}=\frac{\frac{2\pi}{1}}{\frac{1}{4}} \cdot \frac{\frac{4}{1}}{\frac{4}{1}} = \frac{8\pi}{1}=8\pi\end{align*}

Thinking of it as a transformation, the graph is stretched horizontally. We would only see \begin{align*}\frac{1}{4}\end{align*} of the curve if we graphed the function from 0 to \begin{align*}2\pi\end{align*}. To see a complete wave, therefore, we would have to go four times as far, or all the way from 0 to \begin{align*}8\pi\end{align*}.

3. What is the frequency and period of \begin{align*}y = \sin \frac{1}{2}x\end{align*}?

Like the previous two problems, we can see that the frequency is \begin{align*}\frac{1}{2}\end{align*}, and so the period is \begin{align*}\frac{\frac{2\pi}{1}}{\frac{1}{2}}\end{align*}, which becomes \begin{align*}2\pi \times \frac{2}{1} = 4\pi\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the frequency of the sound wave from the graph.

By inspecting the graph

You can see that the wave takes about 6.2 seconds to make one complete cycle. This means that the frequency of the wave is approximately 1 cycle per second (since \begin{align*}2 \pi\end{align*} is approximately 6.28). (You should note that in a real wave of sound, you would need to use the speed of the wave and so the calculation would be different. But if you read the graph the same way you read trigonometric functions to find the frequency, this is the result you would find.)

#### Example 2

Draw a sketch of \begin{align*}y = 3 \sin2x\end{align*} from 0 to \begin{align*}2\pi\end{align*}.

The "2" inside the sine function makes the function "squashed" by a factor of 2 in the horizontal direction.

#### Example 3

Draw a sketch of \begin{align*}y = 2.5 \cos \pi x\end{align*} from 0 to \begin{align*}2\pi\end{align*}.

The \begin{align*}\pi\end{align*} inside the sine function makes the function "squashed" by a factor of \begin{align*}\pi\end{align*} in the horizontal direction.

#### Example 4

Draw a sketch of \begin{align*}y=4 \sin \frac{1}{2} x\end{align*} from 0 to \begin{align*}2\pi\end{align*}.

The \begin{align*}\frac{1}{2}\end{align*} inside the sine function makes the function "stretched" by a factor of \begin{align*}\frac{1}{2}\end{align*} in the horizontal direction.

### Review

Find the period and frequency of each function below.

- \begin{align*}y=\sin(4x)\end{align*}
- \begin{align*}y=\cos(2x)\end{align*}
- \begin{align*}y=\cos(\frac{1}{2}x)\end{align*}
- \begin{align*}y=\sin(\frac{3}{4}x)\end{align*}
- \begin{align*}y=\sin(3x)\end{align*}

Draw a sketch of each function from 0 to \begin{align*}2\pi\end{align*}.

- \begin{align*}y=\sin(3x)\end{align*}
- \begin{align*}y=\cos(5x)\end{align*}
- \begin{align*}y=3\cos(\frac{2}{5}x)\end{align*}
- \begin{align*}y=\frac{1}{2}\sin(\frac{3}{4}x)\end{align*}
- \begin{align*}y=-\sin(2x)\end{align*}
- \begin{align*}y=\tan(3x)\end{align*}
- \begin{align*}y=\sec(2x)\end{align*}
- \begin{align*}y=\csc(4x)\end{align*}

Find the equation of each function.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.15.