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Tangent Graphs

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Changes in the Period of a Sine and Cosine Function

What is the period of the cosine function y=\cos [\pi (2x + 4)] ?

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 2 \pi . To stretch out the curve, then the period would have to be longer than 2 \pi . Below we have sine curves with a period of 4 \pi and then the second has a period of \pi .

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

Example A

Determine the period of the following sine and cosine functions.

a) y=-3 \cos 6x

b) y=2 \sin \frac{1}{4}x

c) y=\sin \pi x -7

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

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

c) The \pi is the frequency. The period is \frac{2 \pi}{\pi}=2 .

Example B

Graph part a) from the previous example from [0, 2 \pi] . 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 \frac{\pi}{3} (from above) and the curve should repeat itself 6 times from 0 to 2 \pi . The first maximum value is 3 and occurs at half the period, or x=\frac{\pi}{6} and then repeats at x=\frac{\pi}{2}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \ldots Writing this as a formula we start at \frac{\pi}{6} and add \frac{\pi}{3} to get the next maximum, so each point would be \left(\frac{\pi}{6} \pm \frac{\pi}{3}n,3\right) where n is any integer.

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

Example C

Find all the solutions from the function in Example B from [0, 2 \pi] .

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 [0, 2\pi] . In this case, we will have 6 times the number of zeros that the parent function. To solve this function, set y = 0 and solve for x .

0 &=-3 \cos 6x \\0 &=\cos 6x

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

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}

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

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}

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

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

y=\cos [\pi (2x + 4)]\\y = \cos [2\pi(x + 2)]

The 2\pi is the frequency. The period is therefore \frac{2 \pi}{2\pi}=1 .

Guided Practice

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

2. Find the zeros of the function from #1 from [0, 2\pi] .

3. Determine the equation of the sine function with an amplitude of -3 and a period of 8\pi .

Answers

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

2. The zeros would be when y is zero.

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

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

\frac{2\pi}{b} &=8\pi \\\frac{2\pi}{8\pi} &=b \\\frac{1}{4} &=b

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

Vocabulary

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

Practice

Find the period of the following sine and cosine functions.

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

Use the equation y=5\sin 3x to answer the following questions.

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

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

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

Use the equation y=-3\sin 2x to answer the following questions.

  1. Graph the function from [0, 2\pi] and find the domain and range.
  2. Determine the coordinates of the maximum and minimum values.
  3. Find all the zeros from [0, 2\pi] .
  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 y=a\sin bx , with the given amplitude and period.

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

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