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# Period

## Determine vertical asymptotes, period, domain, and range of trig function.

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Changes in the Period of a Sine and Cosine Function

What is the period of the cosine function \begin{align*}y=\cos [\pi (2x + 4)]\end{align*}?

### 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*}.

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*}.

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

Concept Problem Revisit First, we need to get the function in the form \begin{align*}y=a\cos b(x-h)+k\end{align*}. Therefore we need to factor out the 2.

The \begin{align*}2\pi\end{align*} is the frequency. The period is therefore \begin{align*}\frac{2 \pi}{2\pi}=1\end{align*}.

### 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.

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*}.

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*}.

### Practice

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*}.