Lesson Objectives
• Learn how to sketch the graph of the sine function
• Learn how to sketch the graph of the cosine function
• Learn how to sketch transformations of the graphs of sine and cosine

## How to Graph the Sine and Cosine Functions

In this lesson, we will learn how to sketch the graphs of the sine and cosine functions. The functions sine and cosine are known as periodic functions. Their values repeat in a regular pattern.

### Periodic Function

A periodic function is a function f such that: $$f(x)=f(x + np)$$ Where x is any real number in the domain of our function f, n is an integer, and p is some positive real number. The least possible positive value of p is known as the period of the function. If we revisit our unit circle, where our radius is 1, the circumference is: $$C=2πr=2π$$ This tells us that the lowest value of p, for our sine and cosine functions to repeat is 2$π$. From this, we can state that the sine and cosine functions are periodic functions with a period of 2$π$. $$\text{sin}\hspace{.15em}x=\text{sin}(x + n \cdot 2π)$$ $$\text{cos}\hspace{.15em}x=\text{cos}(x + n \cdot 2π)$$ Where n is any integer. Adding or subtracting multiples of 2$π$ to an angle measure produces a coterminal angle. Therefore, the values of sine and cosine will be unchanged. For example: $$\text{sin}\hspace{.15em}\frac{5π}{6}=\text{sin}\hspace{.15em}\left(\frac{5π}{6}+ 2π\right)$$ $$\text{sin}\hspace{.15em}\frac{5π}{6}=\text{sin}\hspace{.15em}\left(\frac{5π}{6}+ \frac{12π}{6}\right)$$ $$\text{sin}\hspace{.15em}\frac{5π}{6}=\text{sin}\hspace{.15em}\left(\frac{17π}{6}\right)$$ $$\frac{1}{2}=\frac{1}{2}$$ $$\text{cos}\hspace{.15em}\frac{2π}{3}=\text{cos}\hspace{.15em}\left(\frac{2π}{3}- 2π\right)$$ $$\text{cos}\hspace{.15em}\frac{2π}{3}=\text{cos}\hspace{.15em}\left(\frac{2π}{3}- \frac{6π}{3}\right)$$ $$\text{cos}\hspace{.15em}\frac{2π}{3}=\text{cos}\hspace{.15em}\left(-\frac{4π}{3}\right)$$ $$-\frac{1}{2}=-\frac{1}{2}$$

### Graph of the Sine Function

We previously introduced the concept of the circular functions. Using the unit circle, s, the arc length of the intercepted angle was equal to θ, the angle measure. We used this fact to replace θ with s in our definitions for the six trigonometric functions. On the unit circle, where the radius is 1: $$\text{sin}\hspace{.15em}s=\frac{y}{r}=y$$ $$\text{cos}\hspace{.15em}s=\frac{x}{r}=x$$ When graphing the sine and cosine function, we will use x instead of s. This will allow us to think about our graphs with the traditional xy coordinate plane. In other words, if we choose to graph sin x, we will plug in real number values for x (measured in radians) and obtain the corresponding values for y or sin x. From our unit circle, we can observe that the range of our sine function is from -1 to 1.
x Increases From: sin x:
0 to $\frac{π}{2}$Increases from 0 to 1
$\frac{π}{2}$ to $π$Decreases from 1 to 0
$π$ to $\frac{3π}{2}$Decreases from 0 to -1
$\frac{3π}{2}$ to $2π$Increases from -1 to 0
Let's create a table of quadrantal angles and angles in the first quadrant with their associated sine values.

### f(x) = sin x, 0 ≤ x ≤ 2$π$

x y
$0$$0 \frac{π}{6}$$\frac{1}{2}$
$\frac{π}{4}$$\frac{\sqrt{2}}{2} \frac{π}{3}$$\frac{\sqrt{3}}{2}$
$\frac{π}{2}$$1 π$$0$
$\frac{3π}{2}$$-1 2π$$0$
Let's first graph f(x) = sin x over 1 period.
We will place notches in increments of $\frac{π}{6}$ on our x-axis. Let's plot a few points and join them with a smooth curve. $$f(x)=\text{sin}\hspace{.1em}x, 0 ≤ x ≤ 2π$$ The graph will continue in both directions following the same pattern. We may hear this graph referred to as a sine wave or sinusoid. $$f(x)=\text{sin}\hspace{.1em}x$$
• $f(x)=\text{sin}\hspace{.1em}x$ is continuous over its entire domain: (-∞, ∞)
• x-intercepts are of the form n$π$, where n is an integer
• The period is 2$π$
• The range is: $[-1, 1]$
• The graph is symmetric with respect to the origin:
• $\text{sin}(-x)=-\text{sin}\hspace{.1em}x$
• $f(x)=\text{sin}\hspace{.1em}x$ is an odd function

### Graphing Variations of the Sine Graph: f(x) = sin x

Let's begin by thinking about examples where our sine graph will be vertically stretched or compressed (shrunk) when compared to the graphs of: $$f(x)=\text{sin}\hspace{.1em}x$$ In order to graph variations by hand, it is helpful to have the following:
• x-intercepts
• maximum points
• minimum points
When we look at one period or one cycle of our sine curve, we obtain three x-intercepts at: $$(0, 0)$$ $$(π, 0)$$ $$(2π, 0)$$ We also have 1 maximum point: $$\left(\frac{π}{2}, 1\right)$$ And 1 minimum point: $$\left(\frac{3π}{2}, -1\right)$$ Our x-intercepts occur at the start, middle, and end of our period, whereas, the maximum occurs at 1/4 of the period, and the minimum occurs at 3/4 of the period. From this, we can say that the key points for graphing a sine function are found by dividing the period by 4. This will give us 5 key x-coordinates to work with: $$x_{1}\hspace{.1em}» \hspace{.1em}\text{cycle begins}$$ $$x_{2}=x_{1}+ \frac{\text{period}}{4}$$ $$x_{3}=x_{2}+ \frac{\text{period}}{4}$$ $$x_{4}=x_{3}+ \frac{\text{period}}{4}$$ $$x_{5}=x_{4}+ \frac{\text{period}}{4}$$ Our y-coordinate for each of our five key points can be obtained by simply evaluating the function for the given values of x.
The graph of f(x) = sin x forms the basis for graphing functions of the form: $$f(x)=a \cdot \text{sin}\hspace{.1em}x$$

### Vertical Stretch and Compression of the Sine Graph

Let's consider the graph of: $$f(x)=2 \cdot \text{sin}\hspace{.1em}x$$ We can obtain our graph by multiplying each y-coordinate from our basic sine graph by 2. This graph will be vertically stretched by a factor of 2 when compared to our basic sine graph. Our range is now from -2 to 2. The period however is still unchanged.
Let's get our five key points and sketch the graph: $$(0, 0)$$ $$\left(\frac{π}{2}, 2\right)$$ $$(π, 0)$$ $$\left(\frac{3π}{2}, -2\right)$$ $$(2π, 0)$$ Let's now consider the graph of: $$f(x)=\frac{1}{2}\cdot \text{sin}\hspace{.1em}x$$ We can obtain our graph by multiplying each y-coordinate from our basic sine graph by 1/2. This graph will be vertically compressed by a factor of 1/2 when compared to our basic sine graph. Our range is now from -1/2 to 1/2. The period however is still unchanged.
Let's get our five key points and sketch the graph: $$(0, 0)$$ $$\left(\frac{π}{2}, \frac{1}{2}\right)$$ $$(π, 0)$$ $$\left(\frac{3π}{2}, -\frac{1}{2}\right)$$ $$(2π, 0)$$ Let's also consider the graph of: $$f(x)=-2 \cdot \text{sin}\hspace{.1em}x$$ We can obtain our graph by multiplying each y-coordinate from our basic sine graph by -2. Our range is now from -2 to 2. The period however is still unchanged. Notice that this graph can be obtained by sketching the graph of f(x) = 2 sin x and then reflecting across the x-axis. Recall this is the case when we have -f(x), this is the graph of f(x) reflected across the x-axis. In other words, when compared to our basic sine graph we have reflected across the x-axis and vertically stretched by a factor of 2.
Let's get our five key points and sketch the graph: $$(0, 0)$$ $$\left(\frac{π}{2}, -2\right)$$ $$(π, 0)$$ $$\left(\frac{3π}{2}, 2\right)$$ $$(2π, 0)$$ In general, the graph of f(x) = a sin x will have the following properties: $$\text{range}: \left[-|a|, |a|\right]$$ If |a| > 1, the basic sine curve is vertically stretched.
If |a| < 1, the basic sine curve is vertically compressed (shrunk).

### Amplitude

The amplitude of a periodic function is found by taking half of the difference between the maximum and minimum values. The amplitude will describe the height of the graph both above and below the horizontal line passing through the middle of the graph. Finding the amplitude for the basic sine and cosine functions:
Maximum value for sine and cosine is 1.
Minimum value for sine and cosine is -1.
$$\frac{1}{2}(1 - (-1))=\frac{1}{2}(2)=1$$ Therefore, the amplitude for the basic sine and cosine functions is 1. Let's revisit our basic sine graph: Let's now think about a different form of our graphs: $$f(x)=a \cdot \text{sin}\hspace{.1em}x, a ≠ 0$$ What's the amplitude? We just saw that the new range would be from -|a| to |a|. $$\frac{1}{2}(a - (-a))=\frac{1}{2}\cdot 2a=a$$ Since a could be negative, we will include absolute value bars in our final result: $$f(x)=a \cdot \text{sin}\hspace{.1em}x, a ≠ 0$$ Amplitude: |a|
If we return to our function: $$f(x) = 2 \cdot \text{sin}\hspace{.1em}x$$ The amplitude is just |2|, which is 2. If we think about this our range is from -2 to 2, so the difference between the two values is 4: $$2 - (-2)=4$$ Half of this amount would put us back to 2. $$4 \cdot \frac{1}{2}=2$$

### Finding the Period

Let's now consider the case where our period for sine is not 2$π$: $$f(x)=a \cdot \text{sin}\hspace{.1em}bx, b > 0$$ We know if b is 1, then our period is 2$π$.
This happens over 1 cycle as x is increased from 0 to 2$π$.
Now, let's consider when b is positive and not equal to 1:
Now, our function will complete one cycle as bx increases from 0 to 2$π$:
Let's set this up as an inequality: $$0 ≤ bx ≤ 2π$$ Divide all parts by b: $$0 ≤ x ≤ \frac{2π}{b}$$ This tells us that our function: $$f(x)=a \cdot \text{sin}\hspace{.1em}bx, b > 0$$ will complete one cycle from 0 to $\frac{2π}{b}$:
Our period will be $\frac{2π}{b}$.
Our graph will be horizontally compressed (shrunk) by a factor of 1/b when b > 1
Our graph will be horizontally stretched by a factor of 1/b when 0 < b < 1 When the b-value is negative, we can find the period with absolute value bars:
Our period will be $\frac{2π}{|b|}$
Additionally, since sine is an odd function, it can be rewritten. $$f(x)=\text{sin}(-2x)$$ $$f(x)=-\text{sin}(2x)$$

### Steps to Graph Variations of f(x) = sin x

1. Identify the amplitude and period
2. Find the values of x for the five key points (min, max, and x-intercepts):
• Add quarter periods or period/4 to each previous x-value
3. Find the values for y by evaluating the function for the values of x in the last step
4. Connect the five key points with a smooth curve to graph one cycle of the sine function
5. Optional: extend the graph left or right as much as needed
Let's look at an example.
Example #1: Sketch the graph. $$f(x)=3 \cdot \text{sin}\hspace{.1em}2x$$ Compare this to our form: $$f(x)=a \cdot \text{sin}\hspace{.1em}bx$$ Step 1) Identify the amplitude and period:
Here a is 3 and b is 2.
Amplitude: |3| = 3
Period: $\frac{2π}{2}=π$
Step 2) Find the x-values for the five key points:
Divide the period by 4: $$\frac{π}{4}$$ Start with zero and add quarter periods:
x-values: $$x_{1}=0$$ $$x_{2}=\frac{π}{4}$$ $$x_{3}=\frac{π}{2}$$ $$x_{4}=\frac{3π}{4}$$ $$x_{5}=π$$ Step 3) Find the values for y by evaluating the function for the values of x in the last step: $$f(x)=3 \cdot \text{sin}\hspace{.1em}2x$$ Plug in for x: $$f(0)=3 \cdot \text{sin}\hspace{.1em}2(0)=0$$ $$f\left(\frac{π}{4}\right)=3 \cdot \text{sin}\hspace{.1em}2\left(\frac{π}{4}\right)=3$$ $$f\left(\frac{π}{2}\right)=3 \cdot \text{sin}\hspace{.1em}2\left(\frac{π}{2}\right)=0$$ $$f\left(\frac{3π}{4}\right)=3 \cdot \text{sin}\hspace{.1em}2\left(\frac{3π}{4}\right)={-3}$$ $$f(π)=3 \cdot \text{sin}\hspace{.1em}2(π)=0$$ Our coordinates of key points: $$(0, 0)$$ $$\left(\frac{π}{4}, 3\right)$$ $$\left(\frac{π}{2}, 0\right)$$ $$\left(\frac{3π}{4}, -3\right)$$ $$(π, 0)$$ 4) Connect the five key points with a smooth curve to graph one cycle of the sine function: 5) Optional: extend the graph left or right as much as needed. Let's extend our graph one full period right: We can see that based on the basic sine function the graph is vertically stretched by a factor of 3 and horizontally compressed by a factor of 1/2.
Example #2: Sketch the graph. $$f(x)=3 \cdot \text{sin}\hspace{.1em}\frac{x}{3}$$ Compare this to our form: $$f(x)=a \cdot \text{sin}\hspace{.1em}bx$$ Step 1) Identify the amplitude and period:
Here a is 3 and b is 1/3.
Amplitude: |3| = 3
Period: $\frac{2π}{\frac{1}{3}}=6π$
Step 2) Find the x-values for the five key points:
Divide the period by 4: $$\frac{6π}{4}=\frac{3π}{2}$$ Start with zero and add quarter periods:
x-values: $$x_{1}=0$$ $$x_{2}=\frac{3π}{2}$$ $$x_{3}=3π$$ $$x_{4}=\frac{9π}{2}$$ $$x_{5}=6π$$ Step 3) Find the values for y by evaluating the function for the values of x in the last step: $$f(x)=3 \cdot \text{sin}\hspace{.1em}\frac{x}{3}$$ Plug in for x: $$f(0)=3 \cdot \text{sin}\hspace{.1em}\left(\frac{0}{3}\right)=0$$ $$f\left(\frac{3π}{2}\right)=3 \cdot \text{sin}\hspace{.1em}\left(\frac{\frac{3π}{2}}{3}\right)=3$$ $$f(3π)=3 \cdot \text{sin}\hspace{.1em}\left(\frac{3π}{3}\right)=0$$ $$f\left(\frac{9π}{2}\right)=3 \cdot \text{sin}\hspace{.1em}\left(\frac{\frac{9π}{2}}{3}\right)=-3$$ $$f(6π)=3 \cdot \text{sin}\hspace{.1em}\left(\frac{6π}{3}\right)=0$$ Our coordinates of key points: $$(0, 0)$$ $$\left(\frac{3π}{2}, 3\right)$$ $$(3π, 0)$$ $$\left(\frac{9π}{2}, -3\right)$$ $$(6π, 0)$$ 4 & 5) Connect the five key points with a smooth curve to graph the sine function: We can see that based on the basic sine function the graph is vertically and horizontally stretched by a factor of 3.

### Phase Shift

When we work with circular functions, a horizontal shift is known as a phase shift. Let's start out with our sine function of the form: $$f(x)=a \cdot \text{sin}\hspace{.1em}bx, b > 0$$ Now, let's consider the horizontal shift or phase shift of the following: $$f(x)=a \cdot \text{sin}\left(bx - c\right), b > 0$$ Our function above will be shifted right by c/b if c/b is positive.
Our function above will be shifted left by |c/b| if c/b is negative.
It may also be easier to follow if the b is factored out: $$f(x)=a \cdot \text{sin}\hspace{.1em}b\left(x - \frac{c}{b}\right), b > 0$$ If c/b is positive, we will shift right by c/b, if c/b is negative, (meaning we have plus a positive number), then we will shift left by |c/b|. Let's look at an example.
Example #3: Sketch the graph. $$f(x)=4 \cdot \text{sin}\left(2x - \frac{2π}{3}\right)$$ We will factor out a 2 (coefficient of x): $$f(x)=4 \cdot \text{sin}\hspace{.1em}2 \left(x - \frac{π}{3}\right)$$ Step 1) Identify the amplitude, period, and phase shift:
Here a is 4 and b is 2.
Amplitude: |4| = 4
Period: $\frac{2π}{2}=π$
Phase Shift: right $\frac{π}{3}$ units
Step 2) Find the x-values for the five key points:
Divide the period by 4: $$\frac{π}{4}$$ Start with zero, consider the phase shift, then add quarter periods:
x-values:
Our first x-value now occurs at $\frac{π}{3}$ since we had a shift right, $$x_{1}=\frac{π}{3}$$ $$x_{2}=\frac{7π}{12}$$ $$x_{3}=\frac{5π}{6}$$ $$x_{4}=\frac{13π}{12}$$ $$x_{5}=\frac{4π}{3}$$ Step 3) Find the values for y by evaluating the function for the values of x in the last step: $$f(x)=4 \cdot \text{sin}\hspace{.1em}2 \left(x - \frac{π}{3}\right)$$ Plug in for x: $$f\left(\frac{π}{3}\right)=4 \cdot \text{sin}\hspace{.1em}2 \left(\frac{π}{3}- \frac{π}{3}\right)=0$$ $$f\left(\frac{7π}{12}\right)=4 \cdot \text{sin}\hspace{.1em}2 \left(\frac{7π}{12}- \frac{π}{3}\right)=4$$ $$f\left(\frac{5π}{6}\right)=4 \cdot \text{sin}\hspace{.1em}2 \left(\frac{5π}{6}- \frac{π}{3}\right)=0$$ $$f\left(\frac{13π}{12}\right)=4 \cdot \text{sin}\hspace{.1em}2 \left(\frac{13π}{12}- \frac{π}{3}\right)=-4$$ $$f\left(\frac{4π}{3}\right)=4 \cdot \text{sin}\hspace{.1em}2 \left(\frac{4π}{3}- \frac{π}{3}\right)=0$$ Our coordinates of key points: $$\left(\frac{π}{3}, 0\right)$$ $$\left(\frac{7π}{12}, 4\right)$$ $$\left(\frac{5π}{6}, 0\right)$$ $$\left(\frac{13π}{12}, -4\right)$$ $$\left(\frac{4π}{3}, 0\right)$$ 4 & 5) Connect the five key points with a smooth curve to graph the sine function: We can see that based on the basic sine function the graph is vertically stretched by a factor of 4, horizontally compressed by a factor of 1/2 and shifted right by $\frac{π}{3}$ units.

### Vertical Shift

Additionally, we will encounter a sine function with a vertical shift. $$f(x)=a \cdot \text{sin}(bx - c) + d, b > 0$$ Amplitude: |a|
Period: $\frac{2π}{b}$
Phase Shift: $\frac{c}{b}$
Vertical Shift: d
When we think about the vertical shift, if d is positive, we will shift up by d units, when d is negative, we will shift down by the absolute value of d units. Let's look at an example.
Example #4: Sketch the graph. $$f(x)=4 \cdot \text{sin}\left(3x + \frac{3π}{4}\right) - 1$$ We will factor out a 3 (coefficient of x): $$f(x)=4 \cdot \text{sin}\hspace{.1em}3 \left(x + \frac{π}{4}\right) - 1$$ Step 1) Identify the amplitude, period, phase shift, and vertical shift:
Here a is 4 and b is 3.
Amplitude: |4| = 4
Period: $\frac{2π}{3}$
Phase Shift: left $\frac{π}{4}$ units
Vertical Shift: down 1 unit
Step 2) Find the x-values for the five key points:
Divide the period by 4: $$\frac{2π}{3}\cdot \frac{1}{4}=\frac{π}{6}$$ Start with zero, consider the phase shift, then add quarter periods:
x-values:
Our first x-value now occurs at $-\frac{π}{4}$ since we had a shift left, $$x_{1}=-\frac{π}{4}$$ $$x_{2}=-\frac{π}{12}$$ $$x_{3}=\frac{π}{12}$$ $$x_{4}=\frac{π}{4}$$ $$x_{5}=\frac{5π}{12}$$ Step 3) Find the values for y by evaluating the function for the values of x in the last step: $$f(x)=4 \cdot \text{sin}\hspace{.1em}3 \left(x + \frac{π}{4}\right) - 1$$ Plug in for x: $$f\left(-\frac{π}{4}\right)=4 \cdot \text{sin}\hspace{.1em}3 \left(-\frac{π}{4}+ \frac{π}{4}\right) - 1=-1$$ $$f\left(-\frac{π}{12}\right)=4 \cdot \text{sin}\hspace{.1em}3 \left(-\frac{π}{12}+ \frac{π}{4}\right) - 1=3$$ $$f\left(\frac{π}{12}\right)=4 \cdot \text{sin}\hspace{.1em}3 \left(\frac{π}{12}+ \frac{π}{4}\right) - 1=-1$$ $$f\left(\frac{π}{4}\right)=4 \cdot \text{sin}\hspace{.1em}3 \left(\frac{π}{4}+ \frac{π}{4}\right) - 1=-5$$ $$f\left(\frac{5π}{12}\right)=4 \cdot \text{sin}\hspace{.1em}3 \left(\frac{5π}{12}+ \frac{π}{4}\right) - 1=-1$$ Our coordinates of key points: $$\left(-\frac{π}{4}, -1\right)$$ $$\left(-\frac{π}{12}, 3\right)$$ $$\left(\frac{π}{12}, -1\right)$$ $$\left(\frac{π}{4}, -5\right)$$ $$\left(\frac{5π}{12}, -1\right)$$ 4 & 5) Connect the five key points with a smooth curve to graph the sine function: We can see that based on the basic sine function the graph is vertically stretched by a factor of 4, horizontally compressed by a factor of 1/3, shifted left by $\frac{π}{4}$ units and shifted down by 1 unit.

### Graph of the Cosine Function

The graph of the cosine function f(x) = cos x has the same general shape as the graph of f(x) = sin x.
The graph of the cosine function is the graph of the sine function with a phase shift of $\frac{π}{2}$ units to the left. $$\text{cos} \hspace{.1em} x = \text{sin}\left(x + \frac{π}{2}\right)$$Again, from our unit circle, we can observe that the range of our cosine function is from -1 to 1.
x Increases From: cos x:
0 to $\frac{π}{2}$Decreases from 1 to 0
$\frac{π}{2}$ to $π$Decreases from 0 to -1
$π$ to $\frac{3π}{2}$Increases from -1 to 0
$\frac{3π}{2}$ to $2π$Increases from 0 to 1
Let's create a table of quadrantal angles and angles in the first quadrant with their associated cosine values.

### f(x) = cos x, 0 ≤ x ≤ 2$π$

$f(x)=\text{cos}\hspace{.15em}x$
xy
$0$$1 \frac{π}{6}$$\frac{\sqrt{3}}{2}$
$\frac{π}{4}$$\frac{\sqrt{2}}{2} \frac{π}{3}$$\frac{1}{2}$
$\frac{π}{2}$$0 π$$-1$
$\frac{3π}{2}$$0 2π$$1$
Let's first graph f(x) = cos x over 1 period.
We will place notches in increments of $\frac{π}{2}$ on our x-axis. Let's plot a few points and join them with a smooth curve. $$f(x)=\text{cos}\hspace{.1em}x$$ Just like with our graph of f(x) = sin x, the graph of f(x) = cos x will continue in both directions following the same pattern.
• $f(x)=\text{cos}\hspace{.1em}x$ is continuous over its entire domain: (-∞, ∞)
• x-intercepts are of the form (2n + 1)$\frac{π}{2}$, where n is an integer
• The period is 2$π$
• The range is: $[-1, 1]$
• The graph is symmetric with respect to the y-axis:
• $\text{cos}(-x)=\text{cos}\hspace{.1em}x$
• $f(x)=\text{cos}\hspace{.1em}x$ is an even function
Let's look at an example of how to graph our cosine function with graphing transformations.
Example 5: Sketch the graph. $$f(x)=3 \cdot \text{cos}\left(3x + \frac{2π}{3}\right) + 2$$ We will factor out a 3 (coefficient of x): $$f(x)=3 \cdot \text{cos}\hspace{.1em}3 \left(x + \frac{2π}{9}\right) + 2$$ Step 1) Identify the amplitude, period, phase shift, and vertical shift:
Here a and b are both 3.
Amplitude: |3| = 3
Period: $\frac{2π}{3}$
Phase Shift: left $\frac{2π}{9}$ units
Vertical Shift: up 2 units
Step 2) Find the x-values for the five key points:
Divide the period by 4: $$\frac{2π}{3}\cdot \frac{1}{4}=\frac{π}{6}$$ Start with zero, consider the phase shift, then add quarter periods:
x-values:
Our first x-value now occurs at $-\frac{2π}{9}$ since we had a shift left, $$x_{1}=-\frac{2π}{9}$$ $$x_{2}=-\frac{π}{18}$$ $$x_{3}=\frac{π}{9}$$ $$x_{4}=\frac{5π}{18}$$ $$x_{5}=\frac{4π}{9}$$ Step 3) Find the values for y by evaluating the function for the values of x in the last step: $$f(x)=3 \cdot \text{cos}\hspace{.1em}3 \left(x + \frac{2π}{9}\right) + 2$$ Plug in for x: $$f\left(-\frac{2π}{9}\right)=3 \cdot \text{cos}\hspace{.1em}3 \left(-\frac{2π}{9}+ \frac{2π}{9}\right) + 2=5$$ $$f\left(-\frac{π}{18}\right)=3 \cdot \text{cos}\hspace{.1em}3 \left(-\frac{π}{18}+ \frac{2π}{9}\right) + 2=2$$ $$f\left(\frac{π}{9}\right)=3 \cdot \text{cos}\hspace{.1em}3 \left(\frac{π}{9}+ \frac{2π}{9}\right) + 2=-1$$ $$f\left(\frac{5π}{18}\right)=3 \cdot \text{cos}\hspace{.1em}3 \left(\frac{5π}{18}+ \frac{2π}{9}\right) + 2=2$$ $$f\left(\frac{4π}{9}\right)=3 \cdot \text{cos}\hspace{.1em}3 \left(\frac{4π}{9}+ \frac{2π}{9}\right) + 2=5$$ Our coordinates of key points: $$\left(-\frac{2π}{9}, 5\right)$$ $$\left(-\frac{π}{18}, 2\right)$$ $$\left(\frac{π}{9}, -1\right)$$ $$\left(\frac{5π}{18}, 2\right)$$ $$\left(\frac{4π}{9}, 5\right)$$ 4 & 5) Connect the five key points with a smooth curve to graph the cosine function: We can see that based on the basic cosine function the graph is vertically stretched by a factor of 3, horizontally compressed by a factor of 1/3, shifted left by $\frac{2π}{9}$ units and shifted up by 2 units.

#### Skills Check:

Example #1

Find the Period. $$f(x)=4 \text{sin}\left(4x + \frac{11π}{6}\right) - 1$$

A
$$\frac{π}{12}$$
B
$$\frac{π}{3}$$
C
$$\frac{π}{6}$$
D
$$\frac{π}{2}$$
E
$$\frac{11π}{24}$$

Example #2

Find the phase shift. $$f(x)=2 \text{cos}\left(3x - \frac{5π}{6}\right) + 2$$

A
$$\text{right}\hspace{.1em}\frac{5π}{18}$$
B
$$\text{left}\hspace{.1em}\frac{5π}{18}$$
C
$$\text{left}\hspace{.1em}\frac{12π}{13}$$
D
$$\text{left}\hspace{.1em}\frac{6π}{5}$$
E
$$\text{right}\hspace{.1em}4π$$

Example #3

Find the amplitude. $$f(x)=\frac{1}{2}\text{cos}\left(2x + \frac{π}{3}\right)$$

A
$$2$$
B
$$\frac{1}{2}$$
C
$$4π$$
D
$$\frac{π}{2}$$
E
$$\frac{π}{6}$$